Research Article Exact Evaluation of Infinite Series Using Double Laplace...
7
Research Article Exact Evaluation of Infinite Series Using Double Laplace Transform Technique Hassan Eltayeb, 1 Adem KJlJçman, 2 and Said Mesloub 1 1 Mathematics Department, College of Science, King Saud University, P.O. Box 2455, Riyadh 11451, Saudi Arabia 2 Department of Mathematics and Institute for Mathematical Research, University Putra Malaysia, 43400 Serdang, Selangor, Malaysia Correspondence should be addressed to Adem Kılıc ¸man; [email protected] Received 10 October 2013; Accepted 28 December 2013; Published 20 January 2014 Academic Editor: S. A. Mohiuddine Copyright © 2014 Hassan Eltayeb et al. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Double Laplace transform method was applied to evaluate the exact value of double infinite series. Further we generalize the current existing methods and provide some examples to illustrate and verify that the present method is a more general technique. Finding the exact value of infinite series is not an easy task. us, it is a common way to estimate by using the certain test or methods. It is also known that there are certain conditions to apply the estimation methods. For example, some methods are only applicable for the series having only positive terms. However, there is no general method to estimate values of all the type series. One of the most valuable approaches to summing certain infinite series is the use of Laplace transforms in conjunction with the geometric series. Eſthimiou presented a method [1] that uses the Laplace transform and allows one to find exact values for a large class of convergent series of rational terms. Lesko and Smith [2] revisited the method and demonstrated an extension of the original idea to additional infinite series. Sofo [3] used a forced differential difference equation and by the use of Laplace Transform eory generated nonhyperge- ometric type series. Dacunha, in [4], introduced a finite series representation of the matrix exponential using the Laplace transform for time scales. In [5], infinite series and complex numbers were applied to derive formulas. Abate and Whitt in their paper [6] applied infinite series and Laplace transform in study of probability density function by numerical inversion; see [7] for application to summing series arising from integrodifferential difference equations; also see Eltayeb in [8] who studied the relation between double differential transform and double Laplace transform by using power series. Our intention in this note is to illustrate the power of the technique in the case of double series. e infinite series, whose summand (, ) = which is given by ∑ ∞ =1 ∑ ∞ =1 , can be realized as double Laplace transform as =∬ ∞ 0 −− () () . (1) In such a case, what appeared to be a sum of numbers is now written as a sum of integrals. is may not seem like progress, but interchanging the order of summation and integration yields a sum that we can evaluate easily, namely, the way of a geometric series consider ∞ ∑ =1 ∞ ∑ =1 =∫ ∞ 0 () ∞ ∑ =1 ( − ) ∫ ∞ 0 () ∞ ∑ =1 ( − ) =∫ ∞ 0 () ( − 1− − ) ∫ ∞ 0 () ( − 1− − ) . (2) We first illustrate the method for a special case. We then de- scribe the general result pointing out further generalizations of the method, and we finally end with a brief discussion. Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2014, Article ID 327429, 6 pages http://dx.doi.org/10.1155/2014/327429
Transcript of Research Article Exact Evaluation of Infinite Series Using Double Laplace...
Research ArticleExact Evaluation of Infinite Series Using DoubleLaplace Transform Technique
Hassan Eltayeb1 Adem KJlJccedilman2 and Said Mesloub1
1 Mathematics Department College of Science King Saud University PO Box 2455 Riyadh 11451 Saudi Arabia2Department of Mathematics and Institute for Mathematical Research University Putra Malaysia43400 Serdang Selangor Malaysia
Correspondence should be addressed to Adem Kılıcman akilicupmedumy
Received 10 October 2013 Accepted 28 December 2013 Published 20 January 2014
Academic Editor S A Mohiuddine
Copyright copy 2014 Hassan Eltayeb et alThis is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
Double Laplace transformmethodwas applied to evaluate the exact value of double infinite series Further we generalize the currentexisting methods and provide some examples to illustrate and verify that the present method is a more general technique
Finding the exact value of infinite series is not an easytask Thus it is a common way to estimate by using thecertain test or methods It is also known that there are certainconditions to apply the estimation methods For examplesome methods are only applicable for the series having onlypositive terms However there is no general method toestimate values of all the type series One of themost valuableapproaches to summing certain infinite series is the use ofLaplace transforms in conjunction with the geometric series
Efthimiou presented a method [1] that uses the Laplacetransform and allows one to find exact values for a largeclass of convergent series of rational terms Lesko and Smith[2] revisited the method and demonstrated an extensionof the original idea to additional infinite series Sofo [3]used a forced differential difference equation and by theuse of Laplace Transform Theory generated nonhyperge-ometric type series Dacunha in [4] introduced a finiteseries representation of the matrix exponential using theLaplace transform for time scales In [5] infinite series andcomplex numbers were applied to derive formulas Abateand Whitt in their paper [6] applied infinite series andLaplace transform in study of probability density function bynumerical inversion see [7] for application to summing seriesarising from integrodifferential difference equations alsosee Eltayeb in [8] who studied the relation between doubledifferential transform and double Laplace transform by using
power series Our intention in this note is to illustrate thepower of the technique in the case of double series
The infinite series whose summand 119876(119899119898) = 119876119899119876119898
which is given bysuminfin119899=1
suminfin
119898=1119876119899119876119898 can be realized as double
Laplace transform as
119876119899119876119898= ∬
infin
0
119890minus119899119905minus119898119909
119891 (119909) 119891 (119905) 119889119905 119889119909 (1)
In such a case what appeared to be a sum of numbersis now written as a sum of integrals This may not seemlike progress but interchanging the order of summation andintegration yields a sum that we can evaluate easily namelythe way of a geometric series considerinfin
sum119899=1
infin
sum119898=1
119876119899119876119898
= int
infin
0
119891 (119909)
infin
sum119898=1
(119890minus119909
)119898
119889119909int
infin
0
119891 (119905)
infin
sum119899=1
(119890minus119905
)119899
119889119905
= int
infin
0
119891 (119909) (119890minus119909
1 minus 119890minus119909)119889119909int
infin
0
119891 (119905) (119890minus119905
1 minus 119890minus119905)119889119905
(2)We first illustrate the method for a special case We then de-scribe the general result pointing out further generalizationsof the method and we finally end with a brief discussion
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 327429 6 pageshttpdxdoiorg1011552014327429
2 Abstract and Applied Analysis
Consider the series
119876 (120572 120573 119888 119889) =
infin
sum119899=1
infin
sum119898=1
1
(119898 + 120572) (119898 + 120573)
1
(119899 + 119888) (119899 + 119889) (3)
where 120572 = 120573 and 119888 = 119889 and neither 120572 nor 120573 is a negative integerand also 119888 119889 Sums of this form arise often in problems deal-ing with Quantum Field Theory By using partial fractionswe have
119876 (120572 120573 119888 119889) =1
120572 minus 120573
infin
sum119898=1
(1
(119898 + 120573)minus
1
(119898 + 120572))
sdot1
119888 minus 119889
infin
sum119899=1
(1
(119899 + 119889)minus
1
(119899 + 119888))
(4)
By using double Laplace transform of function 119891(119905 119909) = 1
∬
infin
0
119890minus119860119905minus119861119909
119889119905 119889119909 =1
119860119861 119860 119861 gt 0 (5)
Therefore for 120572 120573 gt minus1 and 119888 119889 gt minus1 we have
119876 (120572 120573 119888 119889)
= (1
(120572 minus 120573)lim119872rarrinfin
119872
sum119898=1
int
infin
0
119890minus119898119905
(119890minus120573119905
minus 119890minus120572119905
) 119889119905)
times (1
(119888 minus 119889)lim119873rarrinfin
119873
sum119898=1
int
infin
0
119890minus119898119909
(119890minus119889119909
minus 119890minus119888119909
) 119889119909)
= ( lim119872rarrinfin
int
infin
0
(119890minus120573119905
minus 119890minus120572119905
120572 minus 120573) 119890minus119905(1 minus 119890
minus119872119905
)
1 minus 119890minus119905119889119905)
times ( lim119873rarrinfin
int
infin
0
(119890minus119889119909
minus 119890minus119888119909
119888 minus 119889) 119890minus119909
(1 minus 119890minus119873119909
)
1 minus 119890minus119909119889119909)
= (int
infin
0
(119890minus120573119905
minus 119890minus120572119905
120572 minus 120573)
119890minus119905
1 minus 119890minus119905119889119905)
times (int
infin
0
(119890minus119889119909
minus 119890minus119888119909
119888 minus 119889)
119890minus119909
1 minus 119890minus119909119889119909)
(6)
Using change of variables 119910 = 119890minus119905 and 119911 = 119890
minus119909 in the integralof (6) gives a more symmetric result
119876 (120572 120573 119888 119889) =1
(120572 minus 120573) (119888 minus 119889)∬
1
0
119910120573
minus 119910120572
1 minus 119910
119911119889
minus 119911119888
1 minus 119911119889119910 119889119911
(7)
The integral in (7) converges for 120572 = 120573 119888 = 119889 and 120572 120573 gt minus1
119888 119889 gt minus1 In case 120572 = 12 119888 = 12 and 120573 = 0 119889 = 0 (7)becomes
infin
sum119899=1
infin
sum119898=1
1
119898119899 (119898 + 12)
1
(119899 + 12)
= 4∬
1
0
1 minus radic119910
1 minus 119910
1 minus radic119911
1 minus 119911119889119910 119889119911
= 4∬
1
0
1
(1 + radic119910) (1 + radic119911)119889119910 119889119911
(8)
By using substitution method we haveinfin
sum119899=1
infin
sum119898=1
1
119898119899 (119898 + (12))
1
(119899 + (12))= (4 (1 minus ln 2))2 (9)
For another example take 120572 = 0 119888 = 0 and 120573 119889 as positiveintegers then (7) becomesinfin
sum119899=1
infin
sum119898=1
1
119898 (119898 + 120573)
1
119899 (119899 + 119889)
=1
120573119889∬
1
0
1 minus 119910120573
1 minus 119910
1 minus 119911119889
1 minus 119911119889119910 119889119911
=1
120573119889∬
1
0
(1 + 119910 + 1199102
+ sdot sdot sdot + 119910120573minus1
)
times (1 + 119911 + 1199112
+ sdot sdot sdot + 119911119889minus1
) 119889119910 119889119911
=1
120573119889(1 +
1
2+1
3+ sdot sdot sdot +
1
120573)(1 +
1
2+1
3+ sdot sdot sdot +
1
119889)
(10)Example 1 Find closed-form expressions for series of theform
infin
sum119898=2
1
(1198982 minus 1)
infin
sum119899=2
1
(1198992 minus 1) (11)
We observe from the definition of double Laplace transformthat
1
(1199012 minus 1) (1199042 minus 1)= ∬
infin
0
119890minus119904119905minus119901119909 sinh (119905) sinh (119909) 119889119905 119889119909 (12)
The hyperbolic sine of 119905 and 119909 can be expressed exponentiallyas follows
sinh 119905 = 119890119905
2minus119890minus119905
2 sinh119909 =
119890119909
2minus119890minus119909
2 (13)
Then (12) can be written as1
(1199012 minus 1) (1199042 minus 1)
= (1
2int
infin
0
119890119905
119890minus119904119905
119889119905 minus1
2int
infin
0
119890minus119905
119890minus119904119905
119889119905)
times (1
2int
infin
0
119890119909
119890minus119901119909
119889119909 minus1
2int
infin
0
119890minus119909
119890minus119901119909
119889119909)
(14)
By using the general summation for (14) this becomesinfin
sum119904=2
1
(1199042 minus 1)
infin
sum119901=2
1
(1199012 minus 1)
= (1
2int
infin
0
119890119905
infin
sum119904=2
(119890minus119905
)119904
119889119905 minus1
2int
infin
0
119890minus119905
infin
sum119904=2
(119890minus119905
)119904
119889119905)
times (1
2int
infin
0
119890119909
infin
sum119904=2
(119890minus119909
)119901
119889119909 minus1
2int
infin
0
119890minus119909
infin
sum119904=2
(119890minus119909
)119901
119889119909)
(15)
Abstract and Applied Analysis 3
Let us make substitution of variables 119910 = 119890minus119909 and 119911 = 119890
minus119905 wehaveinfin
sum119904=2
1
(1199042 minus 1)
infin
sum119901=2
1
(1199012 minus 1)
= (1
2int
0
1
1
119911
minus1199112
119889119911
(1 minus 119911) 119911minus1
2int
0
1
minus1199112
119911 119889119911
(1 minus 119911) 119911)
times (1
2int
0
1
1
119910
minus1199102
119889119910
(1 minus 119910) 119910minus1
2int
0
1
minus1199102
119910119889119910
(1 minus 119910) 119910)
(16)
Simplification of the above equation yieldsinfin
sum119904=2
1
(1199042 minus 1)
infin
sum119901=2
1
(1199012 minus 1)
= (1
2int
1
0
(1 + 119911) 119889119911)(1
2int
1
0
(1 + 119910) 119889119910) =9
16
(17)
In the next example we use the same method
Example 2 As a variation let us choose a general term withanalog function of 119904 and 119901 A typical function of this typemight be
119865 (119898 119899) =16119898119899
(1198982 minus 1)2
(1198992 minus 1)2 (18)
It is desired to evaluate this series from 119898 = 2 119899 = 2 to 119898 =
infin 119899 = infin by using table of Laplace transform we have4119901
(1199012 minus 1)2
4119904
(1199042 minus 1)2
= 4∬
infin
0
119890minus119904119905minus119901119909
119909119905 sinh (119909) sinh (119905) 119889119905 119889119909
(19)
When the two numbers 120572 120573 and 119888 119889 differ by an integer119896 119903 that is 120572 = 120573 + 119896 and 119888 = 119889 + 119903 respectively then thesum of (3) can be easily calculated from (4)
119876 (120572 120572 minus 119896 119888 119888 minus 119903) =1
119896119903
minus119896
sum
119895=1
minus119903
sum
119894=1
(1
119895 + 120572)(
1
119894 + 119888) (20)
The previous equation can be checked from (7) as follows
119876 (120572 120572 minus 119896 119888 119888 minus 119903) =1
119896119903∬
1
0
119910119896
minus 1
119910 minus 1119910120572119911119903
minus 1
119911 minus 1119911119888
119889119910119889119911
=1
119896119903∬
1
0
minus119896minus1
sum
ℎ=1
minus119903minus1
sum
119897=1
119910ℎ+120572
119911119897+119888
119889119910119889119911
=1
119896
119896minus1
sum
ℎ=1
(119910ℎ+120572+1
ℎ + 120572 + 1
100381610038161003816100381610038161003816100381610038161003816
1
0
1
119903
119903minus1
sum
119897=1
(119911119897+119888+1
119897 + 119888 + 1
100381610038161003816100381610038161003816100381610038161003816
1
0
(21)
Then
119876 (120572 120572 minus 119896 119888 119888 minus 119903) =1
119896119903
minus119896
sum
119895=1
minus119903
sum
119894=1
(1
119895 + 120572)(
1
119894 + 119888) (22)
We now generalize Efthirnioursquos technique to the series ofthe form
infin
sum119899=1
infin
sum119898=1
119876119899119875119899119906119898V119898 (23)
where it is convenient to write that only
119876119899= int
infin
0
119890minus119899119905
119891 (119905) 119889119905 119906119898= int
infin
0
119890minus119898119909
119891 (119909) 119889119909 (24)
are Laplace transform as follows
infin
sum119899=1
119876119899119875119899
infin
sum119898=1
119906119898V119898
= int
infin
0
119891 (119905) (
infin
sum119899=1
119875119899119890minus119899119905
)119889119905int
infin
0
119891 (119909)(
infin
sum119898=1
V119898119890minus119898119909
)119889119909
(25)
For example consider
infin
sum119899=1
infin
sum119898=1
119903119899
120572119899 + 120573
119896119898
119886119898 + 119887 (26)
where 119903 isin [minus1 1) 119896 isin [minus1 1) 120572 119886 gt 0 120573 119887 ge 0By using Laplace transform
∬
infin
0
119890minus119899119905minus119898119909
(119890minus(120573120572)119905
119890minus(119887119886)119909
)119889119909119889119905
120572119886 (27)
where 119903 = minus 1 119896 = minus 1 the partial sum of
infin
sum119899=1
infin
sum119898=1
119903119899
119896119898
119890minus119899119905minus119898119909
(119890minus(120573120572)119905
119890minus(119887119886)119909
)1
120572119886(28)
is dominated above by
infin
sum119899=1
1003816100381610038161003816100381610038161003816119903119899
119890minus119899119905
(1
120572119890minus(120573120572)119905
)1003816100381610038161003816100381610038161003816
infin
sum119898=1
1003816100381610038161003816100381610038161003816119896119898
119890minus119898119909
(1
119886119890minus(119887119886)119909
)1003816100381610038161003816100381610038161003816
=1
120572119890minus(120573120572)119905
(|119903| 119890minus119905
1 minus |119903| 119890minus119905)1
119886119890minus(119887119886)119909
times (|119896| 119890minus119909
1 minus |119896| 119890minus119909)
(29)
where
int
infin
0
1
120572119890minus(120573120572)119905
(|119903| 119890minus119905
1 minus |119903| 119890minus119905)119889119905int
infin
0
1
119886119890minus(119887119886)119909
(|119896| 119890minus119909
1 minus |119896| 119890minus119909)119889119909
le1
120572119886int
infin
0
(|119903| 119890minus119905
1 minus |119903| 119890minus119905)119889119905
times int
infin
0
(|119896| 119890minus119909
1 minus |119896| 119890minus119909)119889119909 lt infin
(30)
4 Abstract and Applied Analysis
We apply the Lebesgue dominated convergence theoremto have
infin
sum119899=1
infin
sum119898=1
119903119899
120572119899 + 120573
119896119898
119886119898 + 119887
=
infin
sum119899=1
int
infin
0
119903119899
119890minus119899119905
(1
120572119890minus(120573120572)119905
)119889119905
infin
sum119898=1
int
infin
0
119896119898
119890minus119898119909
times (1
119886119890minus(119887119886)119909
)119889119909
= int
infin
0
(1
120572119890minus(120573120572)119905
)
infin
sum119899=1
119903119899
119890minus119899119905
119889119905int
infin
0
(1
119886119890minus(119887119886)119909
)
times
infin
sum119898=1
119896119898
119890minus119898119909
119889119909
=1
120572119886int
infin
0
119890minus(120573120572)119905
(119903119890minus119905
1 minus 119903119890minus119905)119889119905int
infin
0
119890minus(119887119886)119909
times (119896119890minus119909
1 minus 119896119890minus119909)119889119909
=1
120572119886∬
1
0
119903119906120573120572
1 minus 119903119906
119896V119887119886
1 minus 119896V119889119906 119889V
(31)
Now let 119886 = 1 119887 = 0 and 120572 = 1 120573 = 0 yield
infin
sum119899=1
infin
sum119898=1
119903119899
119899
119896119898
119898= ∬
1
0
119903
1 minus 119903119906
119896
1 minus 119896V119889119906 119889V
= ln( 1
1 minus 119903) ln( 1
1 minus 119896)
(32)
and when 119903 = minus1 119896 = minus1 119886 = 1 120572 = 1 and 119887 = 12 120573 = 12we have
infin
sum119899=1
infin
sum119898=1
(minus1)119899
119899 + (12)
(minus1)119898
119898 + (12)= ∬
1
0
11990612
1 + 119906
V12
1 + V119889119906 119889V
= (120587
2minus 2)2
(33)
Consider one more example of this technique
119876 (120572 120573 119886 119887) =
infin
sum119899=1
infin
sum119898=1
119896119898
(119898 + 119886) (119898 + 119887)
119903119899
(119899 + 120572) (119899 + 120573)
= (
infin
sum119899=1
119903119899
int
infin
0
119890minus119899119905
(119890minus120572119905
minus 119890minus120573119905
120573 minus 120572)119889119905)
times (
infin
sum119898=1
119896119898
int
infin
0
119890minus119898119909
(119890minus119886119909
minus 119890minus119887119909
119887 minus 119886)119889119909)
= (int
infin
0
(119890minus120572119905
minus 119890minus120573119905
120573 minus 120572)
infin
sum119899=1
119903119899
119890minus119899119905
119889119905)
times (int
infin
0
(119890minus119886119909
minus 119890minus119887119909
119887 minus 119886)
infin
sum119898=1
119896119898
119890minus119898119909
119889119909)
= (1
120573 minus 120572int
infin
0
(119890minus120572119905
minus 119890minus120573119905
) (119903119890minus119905
1 minus 119903119890minus119905)119889119905)
times (1
119887 minus 119886int
infin
0
(119890minus119886119909
minus 119890minus119887119909
) (119896119890minus119909
1 minus 119896119890minus119909)119889119909)
=119903119896
(120573 minus 120572) (119887 minus 119886)∬
1
0
119906120572
minus 119906120573
1 minus 119903119906
V119886 minus V119887
1 minus 119896V119889119906 119889V
(34)
Now let 119886 = 1 119887 = 0 and 120572 = 1 120573 = 0 and 119903 119896 isin (minus1 1)then (34) becomesinfin
sum119899=1
infin
sum119898=1
119896119898
119898(119898 + 1)
119903119899
119899 (119899 + 1)
= 119903119896∬
1
0
1 minus 119906
1 minus 119903119906
1 minus V1 minus 119896V
119889119906 119889V
= (1 + (1 minus 119903
119903) ln (1 minus 119903)) (1 + (
1 minus 119896
119896) ln (1 minus 119896))
(35)
To apply the method to trigonometric series we need tobe able to handle series of the form
119878 =
infin
sum119899=1
infin
sum119898=1
sin (119898119910) sin (119899119911) 119890minus119898119909minus119899119905
119862 =
infin
sum119899=1
infin
sum119898=1
cos (119898119910) cos (119899119911) 119890minus119898119909minus119899119905
(36)
In particular we assume 119910 and 119911 are real number and 119909 gt
0 and 119905 gt 0 whereinfin
sum119899=1
infin
sum119898=1
sin (119898119910) sin (119899119911) 119890minus119898119909minus119899119905
=119890minus119909 sin119910
1 minus 2 cos119910119890minus119909 + 119890minus2119909
119890minus119905 sin 119911
1 minus 2 cos 119911119890minus119905 + 119890minus2119905
infin
sum119899=1
infin
sum119898=1
cos (119898119910) cos (119899119911) 119890minus119898119909minus119899119905
=119890minus119909
(cos119910 minus 119890minus119909
)
1 minus 2 cos119910119890minus119909 + 119890minus2119909
119890minus119905
(cos 119911 minus 119890minus119905
)
1 minus 2 cos 119911119890minus119905 + 119890minus2119905
(37)
We start by the seriesinfin
sum119899=1
infin
sum119898=1
cos (119899119911) cos (119898119910)119899]119898120583
(38)
where ] 120583 isin N and 0 lt 119910 119911 lt 2120587 if 120583 ] = 1 or 0 le 119910 119911 le 2120587
if 120583 ] gt 1 Using
1
119899]119898120583=
1
(] minus 1) (120583 minus 1)∬
infin
0
119890minus119898119909minus119899119905
119905]minus1
119909120583minus1
119889119905 119889119909 (39)
Abstract and Applied Analysis 5
we can write (38) asinfin
sum119899=1
infin
sum119898=1
cos (119899119911) cos (119898119910)119899]119898120583
= (1
(] minus 1)
infin
sum119899=1
cos (119899119911) intinfin
0
119890minus119899119905
119905]minus1
119889119905)
times (1
(120583 minus 1)
infin
sum119898=1
cos (119898119910)intinfin
0
119890minus119898119909
119909120583minus1
119889119909)
= (1
(] minus 1)int
infin
0
(
infin
sum119899=1
cos (119899119911) 119890minus119899119905) 119905]minus1
119889119905)
times (1
(120583 minus 1)int
infin
0
(
infin
sum119898=1
cos (119898119910) 119890minus119898119909)119909120583minus1
119889119909)
= (1
(] minus 1)int
infin
0
119890minus119905
(cos 119911 minus 119890minus119905
)
1 minus 2 cos 119911119890minus119905 + 119890minus2119905119905]minus1
119889119905)
times1
(120583 minus 1)int
infin
0
119890minus119909
(cos119910 minus 119890minus119909
)
1 minus 2 cos119910119890minus119909 + 119890minus2119909119909120583minus1
119889119909
(40)
With the change of variables 119906 = 119890minus119905 and V = 119890
minus119909 (40)becomes
infin
sum119899=1
infin
sum119898=1
cos (119899119911) cos (119898119910)119899]119898120583
= ((minus1)
]minus1
(] minus 1)int
1
0
(cos 119911 minus 119906)
1 minus 2 cos 119911119906 + 1199062(ln 119906)]minus1119889119906)
times(minus1)120583minus1
(120583 minus 1)int
1
0
(cos119910 minus V)1 minus 2 cos119910V + V2
(ln V)120583minus1119889V
(41)
In the special case let 120583 = 1 ] = 1 we haveinfin
sum119899=1
infin
sum119898=1
cos (119899119911) cos (119898119910)119899119898
= (minus1
2int
1
0
119889 (1 minus 2 cos 119911119906 + 1199062
)
1 minus 2 cos 119911119906 + 1199062)
times (minus1
2int
1
0
119889 (1 minus 2 cos119910V + V2)
1 minus 2 cos119910V + V2)
= (minus1
2[ln (1 minus 2 cos 119911119906 + 119906
2
)]1
0
)
times (minus1
2[ln (1 minus 2 cos119910V + V2)]
1
0
)
= ln(2 sin 119911
2) ln(2 sin
119910
2)
(42)
We use the same steps for the seriesinfin
sum119899=1
infin
sum119898=1
sin (119899119911) sin (119898119910)119899]119898120583
(43)
we arrive at the integral representation
infin
sum119899=1
infin
sum119898=1
sin (119899119911) sin (119898119910)119899]119898120583
= ((minus1)
]minus1
(] minus 1)sin 119911int
1
0
(ln 119906)]minus1
1 minus 2 cos 119911119906 + 1199062119889119906)
times(minus1)120583minus1
(120583 minus 1)sin119910int
1
0
(ln V)120583minus1
1 minus 2 cos119910V + V2119889V
(44)
In particular for 120583 = 1 ] = 1
infin
sum119899=1
infin
sum119898=1
sin (119899119911) sin (119898119910)119899119898
= (sin 119911int1
0
1
(119906 minus cos 119911)2 + sin2119911119889119906)
times (sin119910int1
0
1
(V minus cos119910)2 + sin2119910119889V)
=1003816100381610038161003816100381610038161003816tanminus1 119906 minus cos 119911
sin 119911]1
0
10038161003816100381610038161003816100381610038161003816tanminus1
V minus cos119910sin119910
]
1
0
= (tanminus1 sin 1199111 minus cos 119911
)(tanminus1sin119910
1 minus cos119910)
= (120587 minus 119910
2) (
120587 minus 119911
2)
(45)
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are grateful for the very useful comments regard-ing detailed remarks which improved the presentation andthe contents of the paper Further the authors would also liketo extend their sincere appreciation to the Deanship of Scien-tific Research at King Saud University for its funding of thisresearch through the Research Group Project no RGP-VPP-117
References
[1] C J Efthimiou ldquoFinding exact values for infinite sumsrdquoMath-ematics Magazine vol 72 no 1 pp 45ndash51 1999
[2] J P Lesko andW D Smith ldquoA Laplace transform technique forevaluating infinite seriesrdquoMathematics Magazine vol 76 no 5pp 394ndash398 2003
[3] A Sofo ldquoSummation of series via Laplace transformsrdquo RevistaMatematica Complutense vol 15 no 2 pp 437ndash447 2002
[4] J J Dacunha ldquoTransition matrix and generalized matrix expo-nential via the Peano-Baker seriesrdquo Journal of Difference Equa-tions and Applications vol 11 no 15 pp 1245ndash1264 2005
6 Abstract and Applied Analysis
[5] D E Dobbs ldquoUsing infinite series and complex numbers toderive formulas involving Laplace transformsrdquo InternationalJournal of Mathematical Education in Science and Technologyvol 44 no 5 pp 752ndash761 2013
[6] J Abate and W Whitt ldquoInfinite-series representations of La-place transforms of probability density functions for numericalinversionrdquo Journal of the Operations Research Society of Japanvol 42 no 3 pp 268ndash285 1999
[7] P Cerone and A Sofo ldquoSumming series arising from integro-differential-difference equationsrdquo The Journal of AustralianMathematical Society B vol 41 no 4 pp 473ndash486 2000
[8] H Eltayeb ldquoNote on relation between double Laplace transformand double differential transformrdquo Abstract and Applied Analy-sis vol 2013 Article ID 535020 7 pages 2013
Submit your manuscripts athttpwwwhindawicom
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Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
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Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
2 Abstract and Applied Analysis
Consider the series
119876 (120572 120573 119888 119889) =
infin
sum119899=1
infin
sum119898=1
1
(119898 + 120572) (119898 + 120573)
1
(119899 + 119888) (119899 + 119889) (3)
where 120572 = 120573 and 119888 = 119889 and neither 120572 nor 120573 is a negative integerand also 119888 119889 Sums of this form arise often in problems deal-ing with Quantum Field Theory By using partial fractionswe have
119876 (120572 120573 119888 119889) =1
120572 minus 120573
infin
sum119898=1
(1
(119898 + 120573)minus
1
(119898 + 120572))
sdot1
119888 minus 119889
infin
sum119899=1
(1
(119899 + 119889)minus
1
(119899 + 119888))
(4)
By using double Laplace transform of function 119891(119905 119909) = 1
∬
infin
0
119890minus119860119905minus119861119909
119889119905 119889119909 =1
119860119861 119860 119861 gt 0 (5)
Therefore for 120572 120573 gt minus1 and 119888 119889 gt minus1 we have
119876 (120572 120573 119888 119889)
= (1
(120572 minus 120573)lim119872rarrinfin
119872
sum119898=1
int
infin
0
119890minus119898119905
(119890minus120573119905
minus 119890minus120572119905
) 119889119905)
times (1
(119888 minus 119889)lim119873rarrinfin
119873
sum119898=1
int
infin
0
119890minus119898119909
(119890minus119889119909
minus 119890minus119888119909
) 119889119909)
= ( lim119872rarrinfin
int
infin
0
(119890minus120573119905
minus 119890minus120572119905
120572 minus 120573) 119890minus119905(1 minus 119890
minus119872119905
)
1 minus 119890minus119905119889119905)
times ( lim119873rarrinfin
int
infin
0
(119890minus119889119909
minus 119890minus119888119909
119888 minus 119889) 119890minus119909
(1 minus 119890minus119873119909
)
1 minus 119890minus119909119889119909)
= (int
infin
0
(119890minus120573119905
minus 119890minus120572119905
120572 minus 120573)
119890minus119905
1 minus 119890minus119905119889119905)
times (int
infin
0
(119890minus119889119909
minus 119890minus119888119909
119888 minus 119889)
119890minus119909
1 minus 119890minus119909119889119909)
(6)
Using change of variables 119910 = 119890minus119905 and 119911 = 119890
minus119909 in the integralof (6) gives a more symmetric result
119876 (120572 120573 119888 119889) =1
(120572 minus 120573) (119888 minus 119889)∬
1
0
119910120573
minus 119910120572
1 minus 119910
119911119889
minus 119911119888
1 minus 119911119889119910 119889119911
(7)
The integral in (7) converges for 120572 = 120573 119888 = 119889 and 120572 120573 gt minus1
119888 119889 gt minus1 In case 120572 = 12 119888 = 12 and 120573 = 0 119889 = 0 (7)becomes
infin
sum119899=1
infin
sum119898=1
1
119898119899 (119898 + 12)
1
(119899 + 12)
= 4∬
1
0
1 minus radic119910
1 minus 119910
1 minus radic119911
1 minus 119911119889119910 119889119911
= 4∬
1
0
1
(1 + radic119910) (1 + radic119911)119889119910 119889119911
(8)
By using substitution method we haveinfin
sum119899=1
infin
sum119898=1
1
119898119899 (119898 + (12))
1
(119899 + (12))= (4 (1 minus ln 2))2 (9)
For another example take 120572 = 0 119888 = 0 and 120573 119889 as positiveintegers then (7) becomesinfin
sum119899=1
infin
sum119898=1
1
119898 (119898 + 120573)
1
119899 (119899 + 119889)
=1
120573119889∬
1
0
1 minus 119910120573
1 minus 119910
1 minus 119911119889
1 minus 119911119889119910 119889119911
=1
120573119889∬
1
0
(1 + 119910 + 1199102
+ sdot sdot sdot + 119910120573minus1
)
times (1 + 119911 + 1199112
+ sdot sdot sdot + 119911119889minus1
) 119889119910 119889119911
=1
120573119889(1 +
1
2+1
3+ sdot sdot sdot +
1
120573)(1 +
1
2+1
3+ sdot sdot sdot +
1
119889)
(10)Example 1 Find closed-form expressions for series of theform
infin
sum119898=2
1
(1198982 minus 1)
infin
sum119899=2
1
(1198992 minus 1) (11)
We observe from the definition of double Laplace transformthat
1
(1199012 minus 1) (1199042 minus 1)= ∬
infin
0
119890minus119904119905minus119901119909 sinh (119905) sinh (119909) 119889119905 119889119909 (12)
The hyperbolic sine of 119905 and 119909 can be expressed exponentiallyas follows
sinh 119905 = 119890119905
2minus119890minus119905
2 sinh119909 =
119890119909
2minus119890minus119909
2 (13)
Then (12) can be written as1
(1199012 minus 1) (1199042 minus 1)
= (1
2int
infin
0
119890119905
119890minus119904119905
119889119905 minus1
2int
infin
0
119890minus119905
119890minus119904119905
119889119905)
times (1
2int
infin
0
119890119909
119890minus119901119909
119889119909 minus1
2int
infin
0
119890minus119909
119890minus119901119909
119889119909)
(14)
By using the general summation for (14) this becomesinfin
sum119904=2
1
(1199042 minus 1)
infin
sum119901=2
1
(1199012 minus 1)
= (1
2int
infin
0
119890119905
infin
sum119904=2
(119890minus119905
)119904
119889119905 minus1
2int
infin
0
119890minus119905
infin
sum119904=2
(119890minus119905
)119904
119889119905)
times (1
2int
infin
0
119890119909
infin
sum119904=2
(119890minus119909
)119901
119889119909 minus1
2int
infin
0
119890minus119909
infin
sum119904=2
(119890minus119909
)119901
119889119909)
(15)
Abstract and Applied Analysis 3
Let us make substitution of variables 119910 = 119890minus119909 and 119911 = 119890
minus119905 wehaveinfin
sum119904=2
1
(1199042 minus 1)
infin
sum119901=2
1
(1199012 minus 1)
= (1
2int
0
1
1
119911
minus1199112
119889119911
(1 minus 119911) 119911minus1
2int
0
1
minus1199112
119911 119889119911
(1 minus 119911) 119911)
times (1
2int
0
1
1
119910
minus1199102
119889119910
(1 minus 119910) 119910minus1
2int
0
1
minus1199102
119910119889119910
(1 minus 119910) 119910)
(16)
Simplification of the above equation yieldsinfin
sum119904=2
1
(1199042 minus 1)
infin
sum119901=2
1
(1199012 minus 1)
= (1
2int
1
0
(1 + 119911) 119889119911)(1
2int
1
0
(1 + 119910) 119889119910) =9
16
(17)
In the next example we use the same method
Example 2 As a variation let us choose a general term withanalog function of 119904 and 119901 A typical function of this typemight be
119865 (119898 119899) =16119898119899
(1198982 minus 1)2
(1198992 minus 1)2 (18)
It is desired to evaluate this series from 119898 = 2 119899 = 2 to 119898 =
infin 119899 = infin by using table of Laplace transform we have4119901
(1199012 minus 1)2
4119904
(1199042 minus 1)2
= 4∬
infin
0
119890minus119904119905minus119901119909
119909119905 sinh (119909) sinh (119905) 119889119905 119889119909
(19)
When the two numbers 120572 120573 and 119888 119889 differ by an integer119896 119903 that is 120572 = 120573 + 119896 and 119888 = 119889 + 119903 respectively then thesum of (3) can be easily calculated from (4)
119876 (120572 120572 minus 119896 119888 119888 minus 119903) =1
119896119903
minus119896
sum
119895=1
minus119903
sum
119894=1
(1
119895 + 120572)(
1
119894 + 119888) (20)
The previous equation can be checked from (7) as follows
119876 (120572 120572 minus 119896 119888 119888 minus 119903) =1
119896119903∬
1
0
119910119896
minus 1
119910 minus 1119910120572119911119903
minus 1
119911 minus 1119911119888
119889119910119889119911
=1
119896119903∬
1
0
minus119896minus1
sum
ℎ=1
minus119903minus1
sum
119897=1
119910ℎ+120572
119911119897+119888
119889119910119889119911
=1
119896
119896minus1
sum
ℎ=1
(119910ℎ+120572+1
ℎ + 120572 + 1
100381610038161003816100381610038161003816100381610038161003816
1
0
1
119903
119903minus1
sum
119897=1
(119911119897+119888+1
119897 + 119888 + 1
100381610038161003816100381610038161003816100381610038161003816
1
0
(21)
Then
119876 (120572 120572 minus 119896 119888 119888 minus 119903) =1
119896119903
minus119896
sum
119895=1
minus119903
sum
119894=1
(1
119895 + 120572)(
1
119894 + 119888) (22)
We now generalize Efthirnioursquos technique to the series ofthe form
infin
sum119899=1
infin
sum119898=1
119876119899119875119899119906119898V119898 (23)
where it is convenient to write that only
119876119899= int
infin
0
119890minus119899119905
119891 (119905) 119889119905 119906119898= int
infin
0
119890minus119898119909
119891 (119909) 119889119909 (24)
are Laplace transform as follows
infin
sum119899=1
119876119899119875119899
infin
sum119898=1
119906119898V119898
= int
infin
0
119891 (119905) (
infin
sum119899=1
119875119899119890minus119899119905
)119889119905int
infin
0
119891 (119909)(
infin
sum119898=1
V119898119890minus119898119909
)119889119909
(25)
For example consider
infin
sum119899=1
infin
sum119898=1
119903119899
120572119899 + 120573
119896119898
119886119898 + 119887 (26)
where 119903 isin [minus1 1) 119896 isin [minus1 1) 120572 119886 gt 0 120573 119887 ge 0By using Laplace transform
∬
infin
0
119890minus119899119905minus119898119909
(119890minus(120573120572)119905
119890minus(119887119886)119909
)119889119909119889119905
120572119886 (27)
where 119903 = minus 1 119896 = minus 1 the partial sum of
infin
sum119899=1
infin
sum119898=1
119903119899
119896119898
119890minus119899119905minus119898119909
(119890minus(120573120572)119905
119890minus(119887119886)119909
)1
120572119886(28)
is dominated above by
infin
sum119899=1
1003816100381610038161003816100381610038161003816119903119899
119890minus119899119905
(1
120572119890minus(120573120572)119905
)1003816100381610038161003816100381610038161003816
infin
sum119898=1
1003816100381610038161003816100381610038161003816119896119898
119890minus119898119909
(1
119886119890minus(119887119886)119909
)1003816100381610038161003816100381610038161003816
=1
120572119890minus(120573120572)119905
(|119903| 119890minus119905
1 minus |119903| 119890minus119905)1
119886119890minus(119887119886)119909
times (|119896| 119890minus119909
1 minus |119896| 119890minus119909)
(29)
where
int
infin
0
1
120572119890minus(120573120572)119905
(|119903| 119890minus119905
1 minus |119903| 119890minus119905)119889119905int
infin
0
1
119886119890minus(119887119886)119909
(|119896| 119890minus119909
1 minus |119896| 119890minus119909)119889119909
le1
120572119886int
infin
0
(|119903| 119890minus119905
1 minus |119903| 119890minus119905)119889119905
times int
infin
0
(|119896| 119890minus119909
1 minus |119896| 119890minus119909)119889119909 lt infin
(30)
4 Abstract and Applied Analysis
We apply the Lebesgue dominated convergence theoremto have
infin
sum119899=1
infin
sum119898=1
119903119899
120572119899 + 120573
119896119898
119886119898 + 119887
=
infin
sum119899=1
int
infin
0
119903119899
119890minus119899119905
(1
120572119890minus(120573120572)119905
)119889119905
infin
sum119898=1
int
infin
0
119896119898
119890minus119898119909
times (1
119886119890minus(119887119886)119909
)119889119909
= int
infin
0
(1
120572119890minus(120573120572)119905
)
infin
sum119899=1
119903119899
119890minus119899119905
119889119905int
infin
0
(1
119886119890minus(119887119886)119909
)
times
infin
sum119898=1
119896119898
119890minus119898119909
119889119909
=1
120572119886int
infin
0
119890minus(120573120572)119905
(119903119890minus119905
1 minus 119903119890minus119905)119889119905int
infin
0
119890minus(119887119886)119909
times (119896119890minus119909
1 minus 119896119890minus119909)119889119909
=1
120572119886∬
1
0
119903119906120573120572
1 minus 119903119906
119896V119887119886
1 minus 119896V119889119906 119889V
(31)
Now let 119886 = 1 119887 = 0 and 120572 = 1 120573 = 0 yield
infin
sum119899=1
infin
sum119898=1
119903119899
119899
119896119898
119898= ∬
1
0
119903
1 minus 119903119906
119896
1 minus 119896V119889119906 119889V
= ln( 1
1 minus 119903) ln( 1
1 minus 119896)
(32)
and when 119903 = minus1 119896 = minus1 119886 = 1 120572 = 1 and 119887 = 12 120573 = 12we have
infin
sum119899=1
infin
sum119898=1
(minus1)119899
119899 + (12)
(minus1)119898
119898 + (12)= ∬
1
0
11990612
1 + 119906
V12
1 + V119889119906 119889V
= (120587
2minus 2)2
(33)
Consider one more example of this technique
119876 (120572 120573 119886 119887) =
infin
sum119899=1
infin
sum119898=1
119896119898
(119898 + 119886) (119898 + 119887)
119903119899
(119899 + 120572) (119899 + 120573)
= (
infin
sum119899=1
119903119899
int
infin
0
119890minus119899119905
(119890minus120572119905
minus 119890minus120573119905
120573 minus 120572)119889119905)
times (
infin
sum119898=1
119896119898
int
infin
0
119890minus119898119909
(119890minus119886119909
minus 119890minus119887119909
119887 minus 119886)119889119909)
= (int
infin
0
(119890minus120572119905
minus 119890minus120573119905
120573 minus 120572)
infin
sum119899=1
119903119899
119890minus119899119905
119889119905)
times (int
infin
0
(119890minus119886119909
minus 119890minus119887119909
119887 minus 119886)
infin
sum119898=1
119896119898
119890minus119898119909
119889119909)
= (1
120573 minus 120572int
infin
0
(119890minus120572119905
minus 119890minus120573119905
) (119903119890minus119905
1 minus 119903119890minus119905)119889119905)
times (1
119887 minus 119886int
infin
0
(119890minus119886119909
minus 119890minus119887119909
) (119896119890minus119909
1 minus 119896119890minus119909)119889119909)
=119903119896
(120573 minus 120572) (119887 minus 119886)∬
1
0
119906120572
minus 119906120573
1 minus 119903119906
V119886 minus V119887
1 minus 119896V119889119906 119889V
(34)
Now let 119886 = 1 119887 = 0 and 120572 = 1 120573 = 0 and 119903 119896 isin (minus1 1)then (34) becomesinfin
sum119899=1
infin
sum119898=1
119896119898
119898(119898 + 1)
119903119899
119899 (119899 + 1)
= 119903119896∬
1
0
1 minus 119906
1 minus 119903119906
1 minus V1 minus 119896V
119889119906 119889V
= (1 + (1 minus 119903
119903) ln (1 minus 119903)) (1 + (
1 minus 119896
119896) ln (1 minus 119896))
(35)
To apply the method to trigonometric series we need tobe able to handle series of the form
119878 =
infin
sum119899=1
infin
sum119898=1
sin (119898119910) sin (119899119911) 119890minus119898119909minus119899119905
119862 =
infin
sum119899=1
infin
sum119898=1
cos (119898119910) cos (119899119911) 119890minus119898119909minus119899119905
(36)
In particular we assume 119910 and 119911 are real number and 119909 gt
0 and 119905 gt 0 whereinfin
sum119899=1
infin
sum119898=1
sin (119898119910) sin (119899119911) 119890minus119898119909minus119899119905
=119890minus119909 sin119910
1 minus 2 cos119910119890minus119909 + 119890minus2119909
119890minus119905 sin 119911
1 minus 2 cos 119911119890minus119905 + 119890minus2119905
infin
sum119899=1
infin
sum119898=1
cos (119898119910) cos (119899119911) 119890minus119898119909minus119899119905
=119890minus119909
(cos119910 minus 119890minus119909
)
1 minus 2 cos119910119890minus119909 + 119890minus2119909
119890minus119905
(cos 119911 minus 119890minus119905
)
1 minus 2 cos 119911119890minus119905 + 119890minus2119905
(37)
We start by the seriesinfin
sum119899=1
infin
sum119898=1
cos (119899119911) cos (119898119910)119899]119898120583
(38)
where ] 120583 isin N and 0 lt 119910 119911 lt 2120587 if 120583 ] = 1 or 0 le 119910 119911 le 2120587
if 120583 ] gt 1 Using
1
119899]119898120583=
1
(] minus 1) (120583 minus 1)∬
infin
0
119890minus119898119909minus119899119905
119905]minus1
119909120583minus1
119889119905 119889119909 (39)
Abstract and Applied Analysis 5
we can write (38) asinfin
sum119899=1
infin
sum119898=1
cos (119899119911) cos (119898119910)119899]119898120583
= (1
(] minus 1)
infin
sum119899=1
cos (119899119911) intinfin
0
119890minus119899119905
119905]minus1
119889119905)
times (1
(120583 minus 1)
infin
sum119898=1
cos (119898119910)intinfin
0
119890minus119898119909
119909120583minus1
119889119909)
= (1
(] minus 1)int
infin
0
(
infin
sum119899=1
cos (119899119911) 119890minus119899119905) 119905]minus1
119889119905)
times (1
(120583 minus 1)int
infin
0
(
infin
sum119898=1
cos (119898119910) 119890minus119898119909)119909120583minus1
119889119909)
= (1
(] minus 1)int
infin
0
119890minus119905
(cos 119911 minus 119890minus119905
)
1 minus 2 cos 119911119890minus119905 + 119890minus2119905119905]minus1
119889119905)
times1
(120583 minus 1)int
infin
0
119890minus119909
(cos119910 minus 119890minus119909
)
1 minus 2 cos119910119890minus119909 + 119890minus2119909119909120583minus1
119889119909
(40)
With the change of variables 119906 = 119890minus119905 and V = 119890
minus119909 (40)becomes
infin
sum119899=1
infin
sum119898=1
cos (119899119911) cos (119898119910)119899]119898120583
= ((minus1)
]minus1
(] minus 1)int
1
0
(cos 119911 minus 119906)
1 minus 2 cos 119911119906 + 1199062(ln 119906)]minus1119889119906)
times(minus1)120583minus1
(120583 minus 1)int
1
0
(cos119910 minus V)1 minus 2 cos119910V + V2
(ln V)120583minus1119889V
(41)
In the special case let 120583 = 1 ] = 1 we haveinfin
sum119899=1
infin
sum119898=1
cos (119899119911) cos (119898119910)119899119898
= (minus1
2int
1
0
119889 (1 minus 2 cos 119911119906 + 1199062
)
1 minus 2 cos 119911119906 + 1199062)
times (minus1
2int
1
0
119889 (1 minus 2 cos119910V + V2)
1 minus 2 cos119910V + V2)
= (minus1
2[ln (1 minus 2 cos 119911119906 + 119906
2
)]1
0
)
times (minus1
2[ln (1 minus 2 cos119910V + V2)]
1
0
)
= ln(2 sin 119911
2) ln(2 sin
119910
2)
(42)
We use the same steps for the seriesinfin
sum119899=1
infin
sum119898=1
sin (119899119911) sin (119898119910)119899]119898120583
(43)
we arrive at the integral representation
infin
sum119899=1
infin
sum119898=1
sin (119899119911) sin (119898119910)119899]119898120583
= ((minus1)
]minus1
(] minus 1)sin 119911int
1
0
(ln 119906)]minus1
1 minus 2 cos 119911119906 + 1199062119889119906)
times(minus1)120583minus1
(120583 minus 1)sin119910int
1
0
(ln V)120583minus1
1 minus 2 cos119910V + V2119889V
(44)
In particular for 120583 = 1 ] = 1
infin
sum119899=1
infin
sum119898=1
sin (119899119911) sin (119898119910)119899119898
= (sin 119911int1
0
1
(119906 minus cos 119911)2 + sin2119911119889119906)
times (sin119910int1
0
1
(V minus cos119910)2 + sin2119910119889V)
=1003816100381610038161003816100381610038161003816tanminus1 119906 minus cos 119911
sin 119911]1
0
10038161003816100381610038161003816100381610038161003816tanminus1
V minus cos119910sin119910
]
1
0
= (tanminus1 sin 1199111 minus cos 119911
)(tanminus1sin119910
1 minus cos119910)
= (120587 minus 119910
2) (
120587 minus 119911
2)
(45)
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are grateful for the very useful comments regard-ing detailed remarks which improved the presentation andthe contents of the paper Further the authors would also liketo extend their sincere appreciation to the Deanship of Scien-tific Research at King Saud University for its funding of thisresearch through the Research Group Project no RGP-VPP-117
References
[1] C J Efthimiou ldquoFinding exact values for infinite sumsrdquoMath-ematics Magazine vol 72 no 1 pp 45ndash51 1999
[2] J P Lesko andW D Smith ldquoA Laplace transform technique forevaluating infinite seriesrdquoMathematics Magazine vol 76 no 5pp 394ndash398 2003
[3] A Sofo ldquoSummation of series via Laplace transformsrdquo RevistaMatematica Complutense vol 15 no 2 pp 437ndash447 2002
[4] J J Dacunha ldquoTransition matrix and generalized matrix expo-nential via the Peano-Baker seriesrdquo Journal of Difference Equa-tions and Applications vol 11 no 15 pp 1245ndash1264 2005
6 Abstract and Applied Analysis
[5] D E Dobbs ldquoUsing infinite series and complex numbers toderive formulas involving Laplace transformsrdquo InternationalJournal of Mathematical Education in Science and Technologyvol 44 no 5 pp 752ndash761 2013
[6] J Abate and W Whitt ldquoInfinite-series representations of La-place transforms of probability density functions for numericalinversionrdquo Journal of the Operations Research Society of Japanvol 42 no 3 pp 268ndash285 1999
[7] P Cerone and A Sofo ldquoSumming series arising from integro-differential-difference equationsrdquo The Journal of AustralianMathematical Society B vol 41 no 4 pp 473ndash486 2000
[8] H Eltayeb ldquoNote on relation between double Laplace transformand double differential transformrdquo Abstract and Applied Analy-sis vol 2013 Article ID 535020 7 pages 2013
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 3
Let us make substitution of variables 119910 = 119890minus119909 and 119911 = 119890
minus119905 wehaveinfin
sum119904=2
1
(1199042 minus 1)
infin
sum119901=2
1
(1199012 minus 1)
= (1
2int
0
1
1
119911
minus1199112
119889119911
(1 minus 119911) 119911minus1
2int
0
1
minus1199112
119911 119889119911
(1 minus 119911) 119911)
times (1
2int
0
1
1
119910
minus1199102
119889119910
(1 minus 119910) 119910minus1
2int
0
1
minus1199102
119910119889119910
(1 minus 119910) 119910)
(16)
Simplification of the above equation yieldsinfin
sum119904=2
1
(1199042 minus 1)
infin
sum119901=2
1
(1199012 minus 1)
= (1
2int
1
0
(1 + 119911) 119889119911)(1
2int
1
0
(1 + 119910) 119889119910) =9
16
(17)
In the next example we use the same method
Example 2 As a variation let us choose a general term withanalog function of 119904 and 119901 A typical function of this typemight be
119865 (119898 119899) =16119898119899
(1198982 minus 1)2
(1198992 minus 1)2 (18)
It is desired to evaluate this series from 119898 = 2 119899 = 2 to 119898 =
infin 119899 = infin by using table of Laplace transform we have4119901
(1199012 minus 1)2
4119904
(1199042 minus 1)2
= 4∬
infin
0
119890minus119904119905minus119901119909
119909119905 sinh (119909) sinh (119905) 119889119905 119889119909
(19)
When the two numbers 120572 120573 and 119888 119889 differ by an integer119896 119903 that is 120572 = 120573 + 119896 and 119888 = 119889 + 119903 respectively then thesum of (3) can be easily calculated from (4)
119876 (120572 120572 minus 119896 119888 119888 minus 119903) =1
119896119903
minus119896
sum
119895=1
minus119903
sum
119894=1
(1
119895 + 120572)(
1
119894 + 119888) (20)
The previous equation can be checked from (7) as follows
119876 (120572 120572 minus 119896 119888 119888 minus 119903) =1
119896119903∬
1
0
119910119896
minus 1
119910 minus 1119910120572119911119903
minus 1
119911 minus 1119911119888
119889119910119889119911
=1
119896119903∬
1
0
minus119896minus1
sum
ℎ=1
minus119903minus1
sum
119897=1
119910ℎ+120572
119911119897+119888
119889119910119889119911
=1
119896
119896minus1
sum
ℎ=1
(119910ℎ+120572+1
ℎ + 120572 + 1
100381610038161003816100381610038161003816100381610038161003816
1
0
1
119903
119903minus1
sum
119897=1
(119911119897+119888+1
119897 + 119888 + 1
100381610038161003816100381610038161003816100381610038161003816
1
0
(21)
Then
119876 (120572 120572 minus 119896 119888 119888 minus 119903) =1
119896119903
minus119896
sum
119895=1
minus119903
sum
119894=1
(1
119895 + 120572)(
1
119894 + 119888) (22)
We now generalize Efthirnioursquos technique to the series ofthe form
infin
sum119899=1
infin
sum119898=1
119876119899119875119899119906119898V119898 (23)
where it is convenient to write that only
119876119899= int
infin
0
119890minus119899119905
119891 (119905) 119889119905 119906119898= int
infin
0
119890minus119898119909
119891 (119909) 119889119909 (24)
are Laplace transform as follows
infin
sum119899=1
119876119899119875119899
infin
sum119898=1
119906119898V119898
= int
infin
0
119891 (119905) (
infin
sum119899=1
119875119899119890minus119899119905
)119889119905int
infin
0
119891 (119909)(
infin
sum119898=1
V119898119890minus119898119909
)119889119909
(25)
For example consider
infin
sum119899=1
infin
sum119898=1
119903119899
120572119899 + 120573
119896119898
119886119898 + 119887 (26)
where 119903 isin [minus1 1) 119896 isin [minus1 1) 120572 119886 gt 0 120573 119887 ge 0By using Laplace transform
∬
infin
0
119890minus119899119905minus119898119909
(119890minus(120573120572)119905
119890minus(119887119886)119909
)119889119909119889119905
120572119886 (27)
where 119903 = minus 1 119896 = minus 1 the partial sum of
infin
sum119899=1
infin
sum119898=1
119903119899
119896119898
119890minus119899119905minus119898119909
(119890minus(120573120572)119905
119890minus(119887119886)119909
)1
120572119886(28)
is dominated above by
infin
sum119899=1
1003816100381610038161003816100381610038161003816119903119899
119890minus119899119905
(1
120572119890minus(120573120572)119905
)1003816100381610038161003816100381610038161003816
infin
sum119898=1
1003816100381610038161003816100381610038161003816119896119898
119890minus119898119909
(1
119886119890minus(119887119886)119909
)1003816100381610038161003816100381610038161003816
=1
120572119890minus(120573120572)119905
(|119903| 119890minus119905
1 minus |119903| 119890minus119905)1
119886119890minus(119887119886)119909
times (|119896| 119890minus119909
1 minus |119896| 119890minus119909)
(29)
where
int
infin
0
1
120572119890minus(120573120572)119905
(|119903| 119890minus119905
1 minus |119903| 119890minus119905)119889119905int
infin
0
1
119886119890minus(119887119886)119909
(|119896| 119890minus119909
1 minus |119896| 119890minus119909)119889119909
le1
120572119886int
infin
0
(|119903| 119890minus119905
1 minus |119903| 119890minus119905)119889119905
times int
infin
0
(|119896| 119890minus119909
1 minus |119896| 119890minus119909)119889119909 lt infin
(30)
4 Abstract and Applied Analysis
We apply the Lebesgue dominated convergence theoremto have
infin
sum119899=1
infin
sum119898=1
119903119899
120572119899 + 120573
119896119898
119886119898 + 119887
=
infin
sum119899=1
int
infin
0
119903119899
119890minus119899119905
(1
120572119890minus(120573120572)119905
)119889119905
infin
sum119898=1
int
infin
0
119896119898
119890minus119898119909
times (1
119886119890minus(119887119886)119909
)119889119909
= int
infin
0
(1
120572119890minus(120573120572)119905
)
infin
sum119899=1
119903119899
119890minus119899119905
119889119905int
infin
0
(1
119886119890minus(119887119886)119909
)
times
infin
sum119898=1
119896119898
119890minus119898119909
119889119909
=1
120572119886int
infin
0
119890minus(120573120572)119905
(119903119890minus119905
1 minus 119903119890minus119905)119889119905int
infin
0
119890minus(119887119886)119909
times (119896119890minus119909
1 minus 119896119890minus119909)119889119909
=1
120572119886∬
1
0
119903119906120573120572
1 minus 119903119906
119896V119887119886
1 minus 119896V119889119906 119889V
(31)
Now let 119886 = 1 119887 = 0 and 120572 = 1 120573 = 0 yield
infin
sum119899=1
infin
sum119898=1
119903119899
119899
119896119898
119898= ∬
1
0
119903
1 minus 119903119906
119896
1 minus 119896V119889119906 119889V
= ln( 1
1 minus 119903) ln( 1
1 minus 119896)
(32)
and when 119903 = minus1 119896 = minus1 119886 = 1 120572 = 1 and 119887 = 12 120573 = 12we have
infin
sum119899=1
infin
sum119898=1
(minus1)119899
119899 + (12)
(minus1)119898
119898 + (12)= ∬
1
0
11990612
1 + 119906
V12
1 + V119889119906 119889V
= (120587
2minus 2)2
(33)
Consider one more example of this technique
119876 (120572 120573 119886 119887) =
infin
sum119899=1
infin
sum119898=1
119896119898
(119898 + 119886) (119898 + 119887)
119903119899
(119899 + 120572) (119899 + 120573)
= (
infin
sum119899=1
119903119899
int
infin
0
119890minus119899119905
(119890minus120572119905
minus 119890minus120573119905
120573 minus 120572)119889119905)
times (
infin
sum119898=1
119896119898
int
infin
0
119890minus119898119909
(119890minus119886119909
minus 119890minus119887119909
119887 minus 119886)119889119909)
= (int
infin
0
(119890minus120572119905
minus 119890minus120573119905
120573 minus 120572)
infin
sum119899=1
119903119899
119890minus119899119905
119889119905)
times (int
infin
0
(119890minus119886119909
minus 119890minus119887119909
119887 minus 119886)
infin
sum119898=1
119896119898
119890minus119898119909
119889119909)
= (1
120573 minus 120572int
infin
0
(119890minus120572119905
minus 119890minus120573119905
) (119903119890minus119905
1 minus 119903119890minus119905)119889119905)
times (1
119887 minus 119886int
infin
0
(119890minus119886119909
minus 119890minus119887119909
) (119896119890minus119909
1 minus 119896119890minus119909)119889119909)
=119903119896
(120573 minus 120572) (119887 minus 119886)∬
1
0
119906120572
minus 119906120573
1 minus 119903119906
V119886 minus V119887
1 minus 119896V119889119906 119889V
(34)
Now let 119886 = 1 119887 = 0 and 120572 = 1 120573 = 0 and 119903 119896 isin (minus1 1)then (34) becomesinfin
sum119899=1
infin
sum119898=1
119896119898
119898(119898 + 1)
119903119899
119899 (119899 + 1)
= 119903119896∬
1
0
1 minus 119906
1 minus 119903119906
1 minus V1 minus 119896V
119889119906 119889V
= (1 + (1 minus 119903
119903) ln (1 minus 119903)) (1 + (
1 minus 119896
119896) ln (1 minus 119896))
(35)
To apply the method to trigonometric series we need tobe able to handle series of the form
119878 =
infin
sum119899=1
infin
sum119898=1
sin (119898119910) sin (119899119911) 119890minus119898119909minus119899119905
119862 =
infin
sum119899=1
infin
sum119898=1
cos (119898119910) cos (119899119911) 119890minus119898119909minus119899119905
(36)
In particular we assume 119910 and 119911 are real number and 119909 gt
0 and 119905 gt 0 whereinfin
sum119899=1
infin
sum119898=1
sin (119898119910) sin (119899119911) 119890minus119898119909minus119899119905
=119890minus119909 sin119910
1 minus 2 cos119910119890minus119909 + 119890minus2119909
119890minus119905 sin 119911
1 minus 2 cos 119911119890minus119905 + 119890minus2119905
infin
sum119899=1
infin
sum119898=1
cos (119898119910) cos (119899119911) 119890minus119898119909minus119899119905
=119890minus119909
(cos119910 minus 119890minus119909
)
1 minus 2 cos119910119890minus119909 + 119890minus2119909
119890minus119905
(cos 119911 minus 119890minus119905
)
1 minus 2 cos 119911119890minus119905 + 119890minus2119905
(37)
We start by the seriesinfin
sum119899=1
infin
sum119898=1
cos (119899119911) cos (119898119910)119899]119898120583
(38)
where ] 120583 isin N and 0 lt 119910 119911 lt 2120587 if 120583 ] = 1 or 0 le 119910 119911 le 2120587
if 120583 ] gt 1 Using
1
119899]119898120583=
1
(] minus 1) (120583 minus 1)∬
infin
0
119890minus119898119909minus119899119905
119905]minus1
119909120583minus1
119889119905 119889119909 (39)
Abstract and Applied Analysis 5
we can write (38) asinfin
sum119899=1
infin
sum119898=1
cos (119899119911) cos (119898119910)119899]119898120583
= (1
(] minus 1)
infin
sum119899=1
cos (119899119911) intinfin
0
119890minus119899119905
119905]minus1
119889119905)
times (1
(120583 minus 1)
infin
sum119898=1
cos (119898119910)intinfin
0
119890minus119898119909
119909120583minus1
119889119909)
= (1
(] minus 1)int
infin
0
(
infin
sum119899=1
cos (119899119911) 119890minus119899119905) 119905]minus1
119889119905)
times (1
(120583 minus 1)int
infin
0
(
infin
sum119898=1
cos (119898119910) 119890minus119898119909)119909120583minus1
119889119909)
= (1
(] minus 1)int
infin
0
119890minus119905
(cos 119911 minus 119890minus119905
)
1 minus 2 cos 119911119890minus119905 + 119890minus2119905119905]minus1
119889119905)
times1
(120583 minus 1)int
infin
0
119890minus119909
(cos119910 minus 119890minus119909
)
1 minus 2 cos119910119890minus119909 + 119890minus2119909119909120583minus1
119889119909
(40)
With the change of variables 119906 = 119890minus119905 and V = 119890
minus119909 (40)becomes
infin
sum119899=1
infin
sum119898=1
cos (119899119911) cos (119898119910)119899]119898120583
= ((minus1)
]minus1
(] minus 1)int
1
0
(cos 119911 minus 119906)
1 minus 2 cos 119911119906 + 1199062(ln 119906)]minus1119889119906)
times(minus1)120583minus1
(120583 minus 1)int
1
0
(cos119910 minus V)1 minus 2 cos119910V + V2
(ln V)120583minus1119889V
(41)
In the special case let 120583 = 1 ] = 1 we haveinfin
sum119899=1
infin
sum119898=1
cos (119899119911) cos (119898119910)119899119898
= (minus1
2int
1
0
119889 (1 minus 2 cos 119911119906 + 1199062
)
1 minus 2 cos 119911119906 + 1199062)
times (minus1
2int
1
0
119889 (1 minus 2 cos119910V + V2)
1 minus 2 cos119910V + V2)
= (minus1
2[ln (1 minus 2 cos 119911119906 + 119906
2
)]1
0
)
times (minus1
2[ln (1 minus 2 cos119910V + V2)]
1
0
)
= ln(2 sin 119911
2) ln(2 sin
119910
2)
(42)
We use the same steps for the seriesinfin
sum119899=1
infin
sum119898=1
sin (119899119911) sin (119898119910)119899]119898120583
(43)
we arrive at the integral representation
infin
sum119899=1
infin
sum119898=1
sin (119899119911) sin (119898119910)119899]119898120583
= ((minus1)
]minus1
(] minus 1)sin 119911int
1
0
(ln 119906)]minus1
1 minus 2 cos 119911119906 + 1199062119889119906)
times(minus1)120583minus1
(120583 minus 1)sin119910int
1
0
(ln V)120583minus1
1 minus 2 cos119910V + V2119889V
(44)
In particular for 120583 = 1 ] = 1
infin
sum119899=1
infin
sum119898=1
sin (119899119911) sin (119898119910)119899119898
= (sin 119911int1
0
1
(119906 minus cos 119911)2 + sin2119911119889119906)
times (sin119910int1
0
1
(V minus cos119910)2 + sin2119910119889V)
=1003816100381610038161003816100381610038161003816tanminus1 119906 minus cos 119911
sin 119911]1
0
10038161003816100381610038161003816100381610038161003816tanminus1
V minus cos119910sin119910
]
1
0
= (tanminus1 sin 1199111 minus cos 119911
)(tanminus1sin119910
1 minus cos119910)
= (120587 minus 119910
2) (
120587 minus 119911
2)
(45)
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are grateful for the very useful comments regard-ing detailed remarks which improved the presentation andthe contents of the paper Further the authors would also liketo extend their sincere appreciation to the Deanship of Scien-tific Research at King Saud University for its funding of thisresearch through the Research Group Project no RGP-VPP-117
References
[1] C J Efthimiou ldquoFinding exact values for infinite sumsrdquoMath-ematics Magazine vol 72 no 1 pp 45ndash51 1999
[2] J P Lesko andW D Smith ldquoA Laplace transform technique forevaluating infinite seriesrdquoMathematics Magazine vol 76 no 5pp 394ndash398 2003
[3] A Sofo ldquoSummation of series via Laplace transformsrdquo RevistaMatematica Complutense vol 15 no 2 pp 437ndash447 2002
[4] J J Dacunha ldquoTransition matrix and generalized matrix expo-nential via the Peano-Baker seriesrdquo Journal of Difference Equa-tions and Applications vol 11 no 15 pp 1245ndash1264 2005
6 Abstract and Applied Analysis
[5] D E Dobbs ldquoUsing infinite series and complex numbers toderive formulas involving Laplace transformsrdquo InternationalJournal of Mathematical Education in Science and Technologyvol 44 no 5 pp 752ndash761 2013
[6] J Abate and W Whitt ldquoInfinite-series representations of La-place transforms of probability density functions for numericalinversionrdquo Journal of the Operations Research Society of Japanvol 42 no 3 pp 268ndash285 1999
[7] P Cerone and A Sofo ldquoSumming series arising from integro-differential-difference equationsrdquo The Journal of AustralianMathematical Society B vol 41 no 4 pp 473ndash486 2000
[8] H Eltayeb ldquoNote on relation between double Laplace transformand double differential transformrdquo Abstract and Applied Analy-sis vol 2013 Article ID 535020 7 pages 2013
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
4 Abstract and Applied Analysis
We apply the Lebesgue dominated convergence theoremto have
infin
sum119899=1
infin
sum119898=1
119903119899
120572119899 + 120573
119896119898
119886119898 + 119887
=
infin
sum119899=1
int
infin
0
119903119899
119890minus119899119905
(1
120572119890minus(120573120572)119905
)119889119905
infin
sum119898=1
int
infin
0
119896119898
119890minus119898119909
times (1
119886119890minus(119887119886)119909
)119889119909
= int
infin
0
(1
120572119890minus(120573120572)119905
)
infin
sum119899=1
119903119899
119890minus119899119905
119889119905int
infin
0
(1
119886119890minus(119887119886)119909
)
times
infin
sum119898=1
119896119898
119890minus119898119909
119889119909
=1
120572119886int
infin
0
119890minus(120573120572)119905
(119903119890minus119905
1 minus 119903119890minus119905)119889119905int
infin
0
119890minus(119887119886)119909
times (119896119890minus119909
1 minus 119896119890minus119909)119889119909
=1
120572119886∬
1
0
119903119906120573120572
1 minus 119903119906
119896V119887119886
1 minus 119896V119889119906 119889V
(31)
Now let 119886 = 1 119887 = 0 and 120572 = 1 120573 = 0 yield
infin
sum119899=1
infin
sum119898=1
119903119899
119899
119896119898
119898= ∬
1
0
119903
1 minus 119903119906
119896
1 minus 119896V119889119906 119889V
= ln( 1
1 minus 119903) ln( 1
1 minus 119896)
(32)
and when 119903 = minus1 119896 = minus1 119886 = 1 120572 = 1 and 119887 = 12 120573 = 12we have
infin
sum119899=1
infin
sum119898=1
(minus1)119899
119899 + (12)
(minus1)119898
119898 + (12)= ∬
1
0
11990612
1 + 119906
V12
1 + V119889119906 119889V
= (120587
2minus 2)2
(33)
Consider one more example of this technique
119876 (120572 120573 119886 119887) =
infin
sum119899=1
infin
sum119898=1
119896119898
(119898 + 119886) (119898 + 119887)
119903119899
(119899 + 120572) (119899 + 120573)
= (
infin
sum119899=1
119903119899
int
infin
0
119890minus119899119905
(119890minus120572119905
minus 119890minus120573119905
120573 minus 120572)119889119905)
times (
infin
sum119898=1
119896119898
int
infin
0
119890minus119898119909
(119890minus119886119909
minus 119890minus119887119909
119887 minus 119886)119889119909)
= (int
infin
0
(119890minus120572119905
minus 119890minus120573119905
120573 minus 120572)
infin
sum119899=1
119903119899
119890minus119899119905
119889119905)
times (int
infin
0
(119890minus119886119909
minus 119890minus119887119909
119887 minus 119886)
infin
sum119898=1
119896119898
119890minus119898119909
119889119909)
= (1
120573 minus 120572int
infin
0
(119890minus120572119905
minus 119890minus120573119905
) (119903119890minus119905
1 minus 119903119890minus119905)119889119905)
times (1
119887 minus 119886int
infin
0
(119890minus119886119909
minus 119890minus119887119909
) (119896119890minus119909
1 minus 119896119890minus119909)119889119909)
=119903119896
(120573 minus 120572) (119887 minus 119886)∬
1
0
119906120572
minus 119906120573
1 minus 119903119906
V119886 minus V119887
1 minus 119896V119889119906 119889V
(34)
Now let 119886 = 1 119887 = 0 and 120572 = 1 120573 = 0 and 119903 119896 isin (minus1 1)then (34) becomesinfin
sum119899=1
infin
sum119898=1
119896119898
119898(119898 + 1)
119903119899
119899 (119899 + 1)
= 119903119896∬
1
0
1 minus 119906
1 minus 119903119906
1 minus V1 minus 119896V
119889119906 119889V
= (1 + (1 minus 119903
119903) ln (1 minus 119903)) (1 + (
1 minus 119896
119896) ln (1 minus 119896))
(35)
To apply the method to trigonometric series we need tobe able to handle series of the form
119878 =
infin
sum119899=1
infin
sum119898=1
sin (119898119910) sin (119899119911) 119890minus119898119909minus119899119905
119862 =
infin
sum119899=1
infin
sum119898=1
cos (119898119910) cos (119899119911) 119890minus119898119909minus119899119905
(36)
In particular we assume 119910 and 119911 are real number and 119909 gt
0 and 119905 gt 0 whereinfin
sum119899=1
infin
sum119898=1
sin (119898119910) sin (119899119911) 119890minus119898119909minus119899119905
=119890minus119909 sin119910
1 minus 2 cos119910119890minus119909 + 119890minus2119909
119890minus119905 sin 119911
1 minus 2 cos 119911119890minus119905 + 119890minus2119905
infin
sum119899=1
infin
sum119898=1
cos (119898119910) cos (119899119911) 119890minus119898119909minus119899119905
=119890minus119909
(cos119910 minus 119890minus119909
)
1 minus 2 cos119910119890minus119909 + 119890minus2119909
119890minus119905
(cos 119911 minus 119890minus119905
)
1 minus 2 cos 119911119890minus119905 + 119890minus2119905
(37)
We start by the seriesinfin
sum119899=1
infin
sum119898=1
cos (119899119911) cos (119898119910)119899]119898120583
(38)
where ] 120583 isin N and 0 lt 119910 119911 lt 2120587 if 120583 ] = 1 or 0 le 119910 119911 le 2120587
if 120583 ] gt 1 Using
1
119899]119898120583=
1
(] minus 1) (120583 minus 1)∬
infin
0
119890minus119898119909minus119899119905
119905]minus1
119909120583minus1
119889119905 119889119909 (39)
Abstract and Applied Analysis 5
we can write (38) asinfin
sum119899=1
infin
sum119898=1
cos (119899119911) cos (119898119910)119899]119898120583
= (1
(] minus 1)
infin
sum119899=1
cos (119899119911) intinfin
0
119890minus119899119905
119905]minus1
119889119905)
times (1
(120583 minus 1)
infin
sum119898=1
cos (119898119910)intinfin
0
119890minus119898119909
119909120583minus1
119889119909)
= (1
(] minus 1)int
infin
0
(
infin
sum119899=1
cos (119899119911) 119890minus119899119905) 119905]minus1
119889119905)
times (1
(120583 minus 1)int
infin
0
(
infin
sum119898=1
cos (119898119910) 119890minus119898119909)119909120583minus1
119889119909)
= (1
(] minus 1)int
infin
0
119890minus119905
(cos 119911 minus 119890minus119905
)
1 minus 2 cos 119911119890minus119905 + 119890minus2119905119905]minus1
119889119905)
times1
(120583 minus 1)int
infin
0
119890minus119909
(cos119910 minus 119890minus119909
)
1 minus 2 cos119910119890minus119909 + 119890minus2119909119909120583minus1
119889119909
(40)
With the change of variables 119906 = 119890minus119905 and V = 119890
minus119909 (40)becomes
infin
sum119899=1
infin
sum119898=1
cos (119899119911) cos (119898119910)119899]119898120583
= ((minus1)
]minus1
(] minus 1)int
1
0
(cos 119911 minus 119906)
1 minus 2 cos 119911119906 + 1199062(ln 119906)]minus1119889119906)
times(minus1)120583minus1
(120583 minus 1)int
1
0
(cos119910 minus V)1 minus 2 cos119910V + V2
(ln V)120583minus1119889V
(41)
In the special case let 120583 = 1 ] = 1 we haveinfin
sum119899=1
infin
sum119898=1
cos (119899119911) cos (119898119910)119899119898
= (minus1
2int
1
0
119889 (1 minus 2 cos 119911119906 + 1199062
)
1 minus 2 cos 119911119906 + 1199062)
times (minus1
2int
1
0
119889 (1 minus 2 cos119910V + V2)
1 minus 2 cos119910V + V2)
= (minus1
2[ln (1 minus 2 cos 119911119906 + 119906
2
)]1
0
)
times (minus1
2[ln (1 minus 2 cos119910V + V2)]
1
0
)
= ln(2 sin 119911
2) ln(2 sin
119910
2)
(42)
We use the same steps for the seriesinfin
sum119899=1
infin
sum119898=1
sin (119899119911) sin (119898119910)119899]119898120583
(43)
we arrive at the integral representation
infin
sum119899=1
infin
sum119898=1
sin (119899119911) sin (119898119910)119899]119898120583
= ((minus1)
]minus1
(] minus 1)sin 119911int
1
0
(ln 119906)]minus1
1 minus 2 cos 119911119906 + 1199062119889119906)
times(minus1)120583minus1
(120583 minus 1)sin119910int
1
0
(ln V)120583minus1
1 minus 2 cos119910V + V2119889V
(44)
In particular for 120583 = 1 ] = 1
infin
sum119899=1
infin
sum119898=1
sin (119899119911) sin (119898119910)119899119898
= (sin 119911int1
0
1
(119906 minus cos 119911)2 + sin2119911119889119906)
times (sin119910int1
0
1
(V minus cos119910)2 + sin2119910119889V)
=1003816100381610038161003816100381610038161003816tanminus1 119906 minus cos 119911
sin 119911]1
0
10038161003816100381610038161003816100381610038161003816tanminus1
V minus cos119910sin119910
]
1
0
= (tanminus1 sin 1199111 minus cos 119911
)(tanminus1sin119910
1 minus cos119910)
= (120587 minus 119910
2) (
120587 minus 119911
2)
(45)
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are grateful for the very useful comments regard-ing detailed remarks which improved the presentation andthe contents of the paper Further the authors would also liketo extend their sincere appreciation to the Deanship of Scien-tific Research at King Saud University for its funding of thisresearch through the Research Group Project no RGP-VPP-117
References
[1] C J Efthimiou ldquoFinding exact values for infinite sumsrdquoMath-ematics Magazine vol 72 no 1 pp 45ndash51 1999
[2] J P Lesko andW D Smith ldquoA Laplace transform technique forevaluating infinite seriesrdquoMathematics Magazine vol 76 no 5pp 394ndash398 2003
[3] A Sofo ldquoSummation of series via Laplace transformsrdquo RevistaMatematica Complutense vol 15 no 2 pp 437ndash447 2002
[4] J J Dacunha ldquoTransition matrix and generalized matrix expo-nential via the Peano-Baker seriesrdquo Journal of Difference Equa-tions and Applications vol 11 no 15 pp 1245ndash1264 2005
6 Abstract and Applied Analysis
[5] D E Dobbs ldquoUsing infinite series and complex numbers toderive formulas involving Laplace transformsrdquo InternationalJournal of Mathematical Education in Science and Technologyvol 44 no 5 pp 752ndash761 2013
[6] J Abate and W Whitt ldquoInfinite-series representations of La-place transforms of probability density functions for numericalinversionrdquo Journal of the Operations Research Society of Japanvol 42 no 3 pp 268ndash285 1999
[7] P Cerone and A Sofo ldquoSumming series arising from integro-differential-difference equationsrdquo The Journal of AustralianMathematical Society B vol 41 no 4 pp 473ndash486 2000
[8] H Eltayeb ldquoNote on relation between double Laplace transformand double differential transformrdquo Abstract and Applied Analy-sis vol 2013 Article ID 535020 7 pages 2013
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 5
we can write (38) asinfin
sum119899=1
infin
sum119898=1
cos (119899119911) cos (119898119910)119899]119898120583
= (1
(] minus 1)
infin
sum119899=1
cos (119899119911) intinfin
0
119890minus119899119905
119905]minus1
119889119905)
times (1
(120583 minus 1)
infin
sum119898=1
cos (119898119910)intinfin
0
119890minus119898119909
119909120583minus1
119889119909)
= (1
(] minus 1)int
infin
0
(
infin
sum119899=1
cos (119899119911) 119890minus119899119905) 119905]minus1
119889119905)
times (1
(120583 minus 1)int
infin
0
(
infin
sum119898=1
cos (119898119910) 119890minus119898119909)119909120583minus1
119889119909)
= (1
(] minus 1)int
infin
0
119890minus119905
(cos 119911 minus 119890minus119905
)
1 minus 2 cos 119911119890minus119905 + 119890minus2119905119905]minus1
119889119905)
times1
(120583 minus 1)int
infin
0
119890minus119909
(cos119910 minus 119890minus119909
)
1 minus 2 cos119910119890minus119909 + 119890minus2119909119909120583minus1
119889119909
(40)
With the change of variables 119906 = 119890minus119905 and V = 119890
minus119909 (40)becomes
infin
sum119899=1
infin
sum119898=1
cos (119899119911) cos (119898119910)119899]119898120583
= ((minus1)
]minus1
(] minus 1)int
1
0
(cos 119911 minus 119906)
1 minus 2 cos 119911119906 + 1199062(ln 119906)]minus1119889119906)
times(minus1)120583minus1
(120583 minus 1)int
1
0
(cos119910 minus V)1 minus 2 cos119910V + V2
(ln V)120583minus1119889V
(41)
In the special case let 120583 = 1 ] = 1 we haveinfin
sum119899=1
infin
sum119898=1
cos (119899119911) cos (119898119910)119899119898
= (minus1
2int
1
0
119889 (1 minus 2 cos 119911119906 + 1199062
)
1 minus 2 cos 119911119906 + 1199062)
times (minus1
2int
1
0
119889 (1 minus 2 cos119910V + V2)
1 minus 2 cos119910V + V2)
= (minus1
2[ln (1 minus 2 cos 119911119906 + 119906
2
)]1
0
)
times (minus1
2[ln (1 minus 2 cos119910V + V2)]
1
0
)
= ln(2 sin 119911
2) ln(2 sin
119910
2)
(42)
We use the same steps for the seriesinfin
sum119899=1
infin
sum119898=1
sin (119899119911) sin (119898119910)119899]119898120583
(43)
we arrive at the integral representation
infin
sum119899=1
infin
sum119898=1
sin (119899119911) sin (119898119910)119899]119898120583
= ((minus1)
]minus1
(] minus 1)sin 119911int
1
0
(ln 119906)]minus1
1 minus 2 cos 119911119906 + 1199062119889119906)
times(minus1)120583minus1
(120583 minus 1)sin119910int
1
0
(ln V)120583minus1
1 minus 2 cos119910V + V2119889V
(44)
In particular for 120583 = 1 ] = 1
infin
sum119899=1
infin
sum119898=1
sin (119899119911) sin (119898119910)119899119898
= (sin 119911int1
0
1
(119906 minus cos 119911)2 + sin2119911119889119906)
times (sin119910int1
0
1
(V minus cos119910)2 + sin2119910119889V)
=1003816100381610038161003816100381610038161003816tanminus1 119906 minus cos 119911
sin 119911]1
0
10038161003816100381610038161003816100381610038161003816tanminus1
V minus cos119910sin119910
]
1
0
= (tanminus1 sin 1199111 minus cos 119911
)(tanminus1sin119910
1 minus cos119910)
= (120587 minus 119910
2) (
120587 minus 119911
2)
(45)
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are grateful for the very useful comments regard-ing detailed remarks which improved the presentation andthe contents of the paper Further the authors would also liketo extend their sincere appreciation to the Deanship of Scien-tific Research at King Saud University for its funding of thisresearch through the Research Group Project no RGP-VPP-117
References
[1] C J Efthimiou ldquoFinding exact values for infinite sumsrdquoMath-ematics Magazine vol 72 no 1 pp 45ndash51 1999
[2] J P Lesko andW D Smith ldquoA Laplace transform technique forevaluating infinite seriesrdquoMathematics Magazine vol 76 no 5pp 394ndash398 2003
[3] A Sofo ldquoSummation of series via Laplace transformsrdquo RevistaMatematica Complutense vol 15 no 2 pp 437ndash447 2002
[4] J J Dacunha ldquoTransition matrix and generalized matrix expo-nential via the Peano-Baker seriesrdquo Journal of Difference Equa-tions and Applications vol 11 no 15 pp 1245ndash1264 2005
6 Abstract and Applied Analysis
[5] D E Dobbs ldquoUsing infinite series and complex numbers toderive formulas involving Laplace transformsrdquo InternationalJournal of Mathematical Education in Science and Technologyvol 44 no 5 pp 752ndash761 2013
[6] J Abate and W Whitt ldquoInfinite-series representations of La-place transforms of probability density functions for numericalinversionrdquo Journal of the Operations Research Society of Japanvol 42 no 3 pp 268ndash285 1999
[7] P Cerone and A Sofo ldquoSumming series arising from integro-differential-difference equationsrdquo The Journal of AustralianMathematical Society B vol 41 no 4 pp 473ndash486 2000
[8] H Eltayeb ldquoNote on relation between double Laplace transformand double differential transformrdquo Abstract and Applied Analy-sis vol 2013 Article ID 535020 7 pages 2013
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
6 Abstract and Applied Analysis
[5] D E Dobbs ldquoUsing infinite series and complex numbers toderive formulas involving Laplace transformsrdquo InternationalJournal of Mathematical Education in Science and Technologyvol 44 no 5 pp 752ndash761 2013
[6] J Abate and W Whitt ldquoInfinite-series representations of La-place transforms of probability density functions for numericalinversionrdquo Journal of the Operations Research Society of Japanvol 42 no 3 pp 268ndash285 1999
[7] P Cerone and A Sofo ldquoSumming series arising from integro-differential-difference equationsrdquo The Journal of AustralianMathematical Society B vol 41 no 4 pp 473ndash486 2000
[8] H Eltayeb ldquoNote on relation between double Laplace transformand double differential transformrdquo Abstract and Applied Analy-sis vol 2013 Article ID 535020 7 pages 2013
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
