CURRICULUM VITAE НАСТАВНИКА САРАДНИКА · 2010-12-14 · 5....

CURRICULUM VITAE НАСТАВНИКА - САРАДНИКА

А) ОСНОВНИ ПОДАЦИ

1. ИМЕ: Мирјана

2. ПРЕЗИМЕ: Видановић

3. ГОДИНА РОЂЕЊА: 1956.

4. ТЕЛЕФОН НА ПОСЛУ: 018-529-740

5. ЕЛЕКТРОНСКА ПОШТА: [email protected]

6. ОБРАЗОВАЊЕ:

Институција Природно-математички факултет Универзитета у Новом Саду (Департман за математику и информатику)

Датум 11.07.2003. Диплома; звање Доктор математичких наука

Институција Филозофски факултет Универзитета у Нишу (Група за математику)

Датум 1987. Диплома; звање Магистар математичких наука

Институција Филозофски факултет Универзитета у Нишу (Група за математику)

Датум 1980. Диплома; звање Дипломирани математичар

7. УСАВРШАВАЊА И ОБУКЕ (ИЗБОР ДО 5 НАЈЗНАЧАЈНИЈИХ)

СЕМИНАР БРОЈ САТИ ОРГАНИЗАТОР ВРЕМЕ И МЕСТО 8. ЗНАЊЕ СТРАНИХ ЈЕЗИКА (навести језике које говорите, читате и/или пишете):

Енглески и немачки (говорим, читам и пишем), руски (читам).

9. ЧЛАНСТВО У ПРОФЕСИОНАЛНИМ ОРГАНИЗАЦИЈАМА: - 10. ПРОФЕСИОНАЛНО ИСКУСТВО:

ВРЕМЕ од марта 1980. до 31.08.1981.

МЕСТО Ниш

ОРГАНИЗАЦИЈА Eлектротехничка школа ''Никола Тесла''

РАДНО МЕСТО Професор математике

ВРЕМЕ od 01.09.1981. до 02.02.2004.

МЕСТО Ниш

ОРГАНИЗАЦИЈА Факултет заштите на раду

РАДНО МЕСТО Асистент-приправник и асистент

2

10.1. УЧЕШЋЕ НА НАУЧНО-ИСТРАЖИВАЧКИМ И ДРУГИМ ПРОЈЕКТИМА : ВРЕМЕ 2002.-2003. МЕСТО Ниш ФИНАНСИРАЊЕ Министарство за науку Републике Србије НАЗИВ ПРОЈЕКТА Истраживање и развој опреме и софтвера за реинжењеринг

мониторинга, дијагностике, управљања и безбедности рада у подземним рудницима угља, бр. 0248, област ТР

ПОЗИЦИЈА Истраживач

ВРЕМЕ 2006.-2010. МЕСТО Ниш ФИНАНСИРАЊЕ Министарство за науку Републике Србије НАЗИВ ПРОЈЕКТА Теорија оператора, стохастичка анализа и примене, бр. 144003, област

Математика ПОЗИЦИЈА Истраживач

11. РАДОВИ (ИЗБОР ОД 20 НАЈЗНАЧАЈНИЈИХ)

РАДОВИ У ИСТАКНУТИМ МЕЂУНАРОДНИМ ЧАСОПИСИМА :

1. M.S. Stanković, M.V. Vidanović, S.B. Tričković: On the Summation of Series Involving Bessel or Struve functions, J. Math. Anal. Appl., 247 (2000) 15-26. M22 - 5

2. S.B. Tričković, M.V. Vidanović, M.S. Stanković: On the summation of trigonometric series, Int. Trans. Spec. Func. 19, No 6 (2008) 441-452. M22 - 5

3. S.B. Tričković, M.V. Vidanović, M.S. Stanković: On trigonometric series over integrals involving Bessel or Struve functions, Int. Trans. Spec. Func. 20, No 11 (2009) 821-834. M22 - 5

РАДОВИ У МЕЂУНАРОДНИМ ЧАСОПИСИМА :

4. M.S. Stanković, S.B. Tričković, M.V. Vidanović: Series over the product of Bessel and trigonometric functions,

Int. Trans. Spec. Func. 11, No 3 (2001) 281-290. M23 - 3 5. M.S. Stanković, M.V. Vidanović, S.B. Tričković: Some series over the product of two trigonometric functions and series involving Bessel functions, Z. Anal. Anw. 20, No 1 (2001) 235-246. M23 - 3

ВРЕМЕ 02.02.2004. до 02.02.2009.

МЕСТО Ниш


РАДНО МЕСТО Доцент (ужа научна област Математика и математичко моделирање)

ВРЕМЕ 02.02.2009. до данас

МЕСТО Ниш


РАДНО МЕСТО Ванредни професор (ужа научна област Математика и информатика)

3

6. M.S. Stanković, M.V. Vidanović, S.B. Tričković: Closed form expressions for some series over certain trigonometric integrals, Appl. Anal. 80, No 1-2 (2001) 53-64. M23 - 3 7. M.S. Stanković, M.V. Vidanović, S.B. Tričković: Summation of some trigonometric and Schlömilch series, J. Comp. Anal. Appl. 5, No 3 (2003) 313-331. M23 - 3 8. S.B. Tričković, M.V. Vidanović, M.S. Stanković: On the summation of series over the product of Bessel functions, Int. Trans. Spec. Func. 17, No 10 (2006) 749-758. M23 - 3 9. S.B. Tričković, M.S. Stanković, M.V. Vidanović: On the summation of series in terms of Bessel functions, Z. Anal. Anw. 25 (2006) 393-406. M23 - 3 10. S.B. Tričković, M.V. Vidanović, M.S. Stanković: Series involving the product of a trigonometric integral and a trigonometric function, Int. Trans. Spec. Func. 18, No 10 (2007) 751-763. M23 - 3 11. M.V. Vidanović, S.B. Tričković, M.S. Stanković: Summation of series over Bourget functions, Canad. Math. Bull. Vol. 51 (4), (2008) 627-636. M23 - 3 РАДОВИ САОПШТЕНИ НА СКУПУ МЕЂУНАРОДНОГ ЗНАЧАЈА ШТАМПАНИ У ЦЕЛИНИ :

12. M.S. Stanković, D.M. Petković, M.V. Đurić: Closed form expressions for some series involving Bessel functions of the first kind, Numerical Methods and Approximation Theory, Electronic Faculty of Niš (1987) 379-389. М33 - 1 РАДОВИ САОПШТЕНИ НА СКУПУ МЕЂУНАРОДНОГ ЗНАЧАЈА ШТАМПАНИ У ИЗВОДУ :

13. G. Milovanović, M.S. Stanković, M.V. Vidanović: Closed form expressions for some series over certain trigonometric integrals, International Memorial Conference "D.S.Mitrinović, Niš, 1996, June 20-22. M34 - 0,5 14. M.V. Vidanović, M.S. Stanković, S.B. Tričković, P.M. Rajković:

Trigonometric integral transforms and summation of some series in terms of Rieman zeta function, Symposium on Wavelets and Integral Transforms, Novi Sad, april 10-12, 2002. M34 - 0,5

РАДОВИ У ВОДЕЋИМ ЧАСОПИСИМА НАЦИОНАЛНОГ ЗНАЧАЈА :

15. M.S. Stanković, M.V. Đurić, D.M. Petković: Closed form expressions for some series over product of two trigonometric functions, Facta Universitatis, Ser. Math. Inform. 9 (1994) 69-82. M51 - 2 16. S.B. Tričković, M.S. Stanković, M.V. Vidanović, V.N. Aleksić:

Integral transforms and summation of some Schlömilch series, Matematički Vesnik 54 (2002), 211-218. M51 - 2

РАДОВИ У ЧАСОПИСИМА НАЦИОНАЛНОГ ЗНАЧАЈА :

17. Ž. Pantić, D.M. Petković, M.V. Đurić: Some series involving Bessel functions of the first kind, Zbornik radova GF, Niš, 6(1985) 369-372. M52 - 1,5 18. Ž. Pantić, D.M. Petković, M.V. Đurić: O ubrzavanju konvergencije nekih klasa redova,

4

Zbornik radova GF, Niš, 6(1985) 359-367. M52 - 1,5 19. Ž. Pantić, D.M. Petković, M.V. Đurić: Some series involving spherical Bessel functions, Zbornik radova GF, Niš, 6(1985) M52 - 1,5 20. M.S. Stanković, M.V. Đurić: Weighted weak Drazin inverses, Zbornik radova Filozofskog fakulteta u Nišu, Serija Matematika, 1 (11) (1987) 89-94. M52 - 1,5

ОБЈАВЉЕНЕ ПУБЛИКАЦИЈЕ :

1. V.N.Aleksić, M.V.Vidanović, M.S. Stanković, Matematika : elementi teorije i zadaci sa rešenjima. Deo 1. Fakultet zaštite na radu, Niš, 2006.

2. M.V.Vidanović, V.N.Aleksić, M.S. Stanković, Matematika : elementi teorije i zadaci sa rešenjima. Deo 2. Fakultet zaštite na radu, Niš, 2009.

This article was downloaded by:[Tričković, Slobodan B.]On: 3 October 2007Access Details: [subscription number 782305840]Publisher: Taylor & FrancisInforma Ltd Registered in England and Wales Registered Number: 1072954Registered office: Mortimer House, 37-41 Mortimer Street, London W1T 3JH, UK

Integral Transforms and SpecialFunctionsPublication details, including instructions for authors and subscription information:http://www.informaworld.com/smpp/title~content=t713643686

Series involving the product of a trigonometric integraland a trigonometric functionSlobodan B. Tričković a; Mirjana V. Vidanović b; Miomir S. Stanković b

a Faculty of Civil Engineering, Department of Mathematics, University of Ni , Ni ,Serbiab Faculty of Environmental Engineering, Department of Mathematics, University ofNi , Ni , Serbia

Online Publication Date: 01 January 2007To cite this Article: Tričković, Slobodan B., Vidanović, Mirjana V. and Stanković,Miomir S. (2007) 'Series involving the product of a trigonometric integral and a

trigonometric function', Integral Transforms and Special Functions, 18:10, 751 - 763To link to this article: DOI: 10.1080/10652460701446458URL: http://dx.doi.org/10.1080/10652460701446458

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Integral Transforms and Special FunctionsVol. 18, No. 10, October 2007, 751–763

Series involving the product of a trigonometric integraland a trigonometric function

SLOBODAN B. TRICKOVIC*†, MIRJANA V. VIDANOVIC‡and MIOMIR S. STANKOVIC‡

†University of Niš, Faculty of Civil Engineering, Department of Mathematics, AleksandraMedvedeva 14, 18000 Niš, Serbia

‡University of Niš, Faculty of Environmental Engineering, Department of Mathematics,Carnojevica 10a, 18000 Niš, Serbia

(Received 8 December 2006)

This paper is concerned with the summation of series (1). To find the sum of the series (1) we firstderive formulas for the summation of series whose general term contains a product of two trigonometricfunctions. These series are expressed in terms of Riemann’s zeta, Catalan’s beta function or Dirichletfunctions eta and lambda, and in certain cases, thoroughly investigated here, they can be brought inclosed form, meaning that the infinite series are represented by finite sums.

Keywords: Riemann’s zeta; Catalan’s beta function; Dirichlet; Bessel functions

Mathematics Subject Classification 1991: Primary: 33C10; Secondary: 11M06, 65B10

1. Introduction

In this paper we shall find the sum of the series

I T ,fα =

∞∑n=1

(s)n−1T ((an − b)x)

(an − b)αf ((an − b)z), α ∈ R

+, (1)

where a = {12

}b = {

01

}, s = 1 or −1, f is sin or cos, and T denotes trigonometric integral

Sφ or Cφ defined by

Sφ(x) =∫ 1

0φ(y) sin xy dy, Cφ(x) =

∫ 1

0φ(y) cos xy dy. (2)

To obtain the sum of the series (1) we do not have to calculate integrals T ((an − b)x)

previously. At first, we assume that φ is integrable. Yet, in order to extend the class of

*Corresponding author. Email: [email protected]

Integral Transforms and Special FunctionsISSN 1065-2469 print/ISSN 1476-8291 online © 2007 Taylor & Francis

http://www.tandf.co.uk/journalsDOI: 10.1080/10652460701446458

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752 S. B. Trickovic et al.

Table 1.

a b s c F for

1 0 1 1 ζ 0 < x < 2π

−1 0 η −π < x < π

2 1 1 1/2 λ 0 < x < π

−1 0 β −(π/2) < x < (π/2)

summable series, we admit that φ is differentiable on (0, 1), but not necessarily boundedin the neighborhood of 0 or 1, or not integrable on (0, 1), however, such that there existsat least one of the integrals (2). We further require that the functions ykφ(y) (k ∈ N) areintegrable on (0, 1) as well.

Obtaining sums of the series (1) relies on the summation of some trigonometric series.Making use of the method for finding the formula (6) from our article [1] (see Theorem 1,p. 396), we can derive the other particular cases as well, writing all of them, in terms ofRiemann’s ζ or Catalan’s β function or Dirichlet functions η and λ, in the form of a singleformula, i.e.,

∞∑n=1

(s)n−1f ((an − b)x)

(an − b)α= cπ

2(α)f (πα/2)xα−1 +

∞∑i=0

(−1)iF (α − 2i − δ)

(2i + δ)! x2i+δ, (3)

where α > 0, a = {12

}b = {

01

}, s = 1 or −1, and f = {

sincos

}δ = {

10

}. The values for F

and c are in the table 1, where ζ is Riemann’s zeta function ζ(z) = ∑∞k=1 k−z, η and λ are

Dirichlet functions η(z) = ∑∞k=1(−1)k−1k−z = (1 − 21−z)ζ(z), λ(z) = ∑∞

k=0(2k + 1)−z =(1 − 2−z)ζ(z), and β is Catalan’s function β(z) = ∑∞

k=0(−1)k(2k + 1)−z.

Remark 1 We note that the functions ζ, η, λ are analytic in the whole complexplane except for z = 1, where they have a pole. The integral representation β(z) =(1/(z))

∫ ∞0 (xz−1ex/(e2x + 1)) dx of Catalan’s function defines an analytical func-

tion for Re z ≥ 1, but also it satisfies the functional equation β(z) = (π/2)z−1(1 −z) cos(πz/2)β(1 − z) extending beta to the left side of the complex plane Re z < 1.

Remark 2 After multiplying by (2(x/2)ν)/((1/2)(ν + (1/2))) the integrals (2) withφ(y) = (1 − y2)ν−1/2, and substituting y = cos θ , we obtain

ϕν(z) = 2(z/2)ν

(1/2)(ν + (1/2))

∫ π/2

0sin2ν θ g(z cos θ) dθ, (4)

where Re ν > −(1/2), ϕν = {Jν

Hν

}g = { cos

sin }, Jν and Hν are the Bessel and Struve functionsrespectively of the first kind and order ν. The integrals Sφ(x) and Cφ(x) in equation (2) can beconsidered as generalizations of Bessel and Struve functions, respectively, since their integralrepresentations (4) are obtained when the particular function φ(y) = (1 − y2)ν−1/2 is chosen.Just because of this, in this paper we generalize some of our results from [2].

2. Preliminaries

In order to find a sum of the series (1) we set one of the integrals (2) instead of T (x), so

that in the sequel we use denotations T (x) ={

Sφ

Cφ

}g = {

sincos

}, meaning that g is one of the

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Series of trigonometric integral and trigonometric function 753

trigonometric functions in equation (2), which depends on the choice of Sφ or Cφ . Afterwards,we shall prove that we are allowed to interchange summation and integration, i.e.,

I T ,fα =

∞∑n=1

(s)n−1f ((an − b)z)

(an − b)α

∫ 1

0φ(y)g((an − b)xy) dy

=∫ 1

0

( ∞∑n=1

(s)n−1f ((an − b)z)g((an − b)xy)

(an − b)α

)φ(y) dy (α ∈ R

+),

(5)

relying on uniform convergence of the right-hand series with respect toy by virtue of Dirichlet’stest, which says that (see [3]) the series

∑∞n=0 an(y)bn(y) is uniformly convergent in D, if

the partial sums of∑∞

n=0 an(y) are uniformly bounded in D, and the sequence bn(y), beingmonotonic for every fixed y, uniformly converges to 0. We emphasize that we treat the aboveright-hand series as a function of y ∈ (0, 1), regarding x and z as variable parameters.

LEMMA 1 Let φ(y) be integrable. Then the right-hand series converges uniformly and thereholds equation (5).

Proof We make use of an elementary trigonometric identity, whereby representing the prod-uct of f and g, as the sum of two trigonometric functions sin or cos. Now this series is splitup into two series of the type

∞∑n=1

(s)n−1τ((an − b)(z ± xy))

(an − b)α(α ∈ R

+), (6)

where τ = sin or τ = cos. Let us suppose first that a = 1, b = 0. If s = 1, then there holds∣∣∣∣∣n∑

k=1

τ(k(z ± xy))

∣∣∣∣∣ ≤ 1

sin((z ± xy)/2)≤ 1

sin ε

for 0 < z ± xy < 2π , because sin((z ± xy)/2) ≥ sin ε > 0 for each ε > 0 satisfying ε ≤((z ± xy)/2) ≤ π − ε.

Similarly, if s = −1, we would have∣∣∣∣∣n∑

k=1

(−1)k−1τ(k(z ± xy))

∣∣∣∣∣ ≤ 1

cos((z ± xy)/2)≤ 1

sin ε

for −π < z ± xy < π , because cos((z ± xy)/2) ≥ sin ε > 0 for each ε > 0 satisfying−(π/2) + ε ≤ ((z ± xy)/2) ≤ (π/2) − ε.

We now suppose a = 2, b = 1. If s = 1, then∣∣∣∣∣n∑

k=1

τ((2k − 1)(z ± xy))

∣∣∣∣∣ ≤ 1

sin(z ± xy)≤ 1

sin ε

for 0 < z ± xy < π , because sin(z ± xy) ≥ sin ε > 0 for each ε > 0 satisfying ε ≤ z ± xy ≤π − ε. Finally, if s = −1, then∣∣∣∣∣

n∑k=1

(−1)k−1τ((2k − 1)(z ± xy))

∣∣∣∣∣ ≤ 1

cos(z ± xy)≤ 1

sin ε

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for −(π/2) < z ± xy < (π/2), because cos(z ± xy) ≥ sin ε > 0 for each ε > 0 satisfying−(π/2) + ε ≤ z ± xy ≤ (π/2) − ε.

So in all these cases, partial sums are uniformly bounded with respect to y ∈ (0, 1), andon this basis, we determine values of x and z giving rise to this, thus finding boundaries forconvergence regions of the series (1) in each of four considered cases. They are in table 2.First, we have immediately −|x| ≤ ±xy ≤ |x|, and to obtain the condition 0 < z ± xy < 2π ,it is necessary to take |x| < z < 2π − |x|, from where there follows |x| < π . In the secondcase, to satisfy −π < z ± xy < π , it is necessary to take |x| − π < z < π − |x|, and we haveagain |x| < π . In the third, 0 < z ± xy < π , it is necessary to take |x| < z < π − |x|, so thatwe easily find |x| < (π/2). Finally, for −(π/2) < z ± xy < (π/2), it is necessary to take|x| − (π/2) < z < (π/2) − |x| giving again |x| < (π/2).

On the other hand, the sequence 1/((an − b)α) obviously monotonically tends to 0 forα > 0, and uniformly converges to 0 with respect to y. Together with the above, this provesan uniform convergence of the right-hand series in equation (5) with respect to y ∈ (0, 1). Sothe interchange of integration and summation in equation (5) is permitted, and accordingly,there holds equation (5). �

LEMMA 2 Suppose that, for a differentiable on (0, 1) and unbounded in the neighborhood of0 or 1 function φ, the integral

∫ 10 φ(y)dy does not converge. Let there exist at least one of the

integrals T ((an − b)x) (T is Sφ or Cφ defined by equation (2)), so that |T ((an − b)x)| ≤Mn(x), and for each corresponding x from table 2, the sequence (Mn(x))/((an − b)α), α > 0,monotonically tends to 0. Then there holds (5).

Proof Because of the assumption that φ is differentiable function, we know that it is contin-uous on each closed interval within (0, 1), and, as it is not bounded in the neighborhood of 0or 1, without loss of generality, we can consider [δ, 1 − δ], 0 < δ < 1. Continuous functionon a closed interval is bounded, and referring again to Dirichlet’s test, we prove uniform con-vergence of the right-hand series in equation (5) with respect to y, but this time on [δ, 1 − δ],so that there holds

∞∑n=1

(s)n−1f ((an − b)z)

(an − b)α

∫ 1−δ

δ

φ(y)g((an − b)xy) dy

=∫ 1−δ

δ

( ∞∑n=1

(s)n−1f ((an − b)z)g((an − b)xy)

(an − b)α

)φ(y) dy (α ∈ R

+).

(7)

We regard the left-hand series as a function of δ with variable parameters z and x. In viewof the conditions, by virtue of Dirichlet’s test, the left-hand series in equation (7) converges

Table 2.

a b s c F Convergence regions

1 0 1 1 ζ {(x, z) : −π < x < π, |x| < z < 2π − |x|}1 0 −1 0 η {(x, z) : −π < x < π, |x| − π < z < π − |x|}2 1 1 1/2 λ {(x, z) : −(π/2) < x < (π/2), |x| < z < π − |x|}2 1 −1 0 β {(x, z) : −(π/2) < x < (π/2), |x| − (π/2) < z < (π/2) − |x|}

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uniformly with respect to δ on (0, 1). Hence, we have

limδ→0+

∞∑n=1

(s)n−1f ((an − b)z)

(an − b)α

∫ 1−δ

δ

φ(y)g((an − b)xy) dy

=∞∑

n=1

(s)n−1f ((an − b)z)

(an − b)α

∫ 1

0φ(y)g((an − b)xy) dy,

meaning that the right-hand integral in equation (7) converges, so there holds equation (5). �

3. Series over the product of trigonometric functions

We have seen that in finding the summation formula for equation (1), the key role plays theseries including the product of two trigonometric functions. That is why we are going toinvestigate them thoroughly. Now, by applying equation (3) to both series in each of the aboveparticular cases, we obtain the following general formula

Sf,gα =

∞∑n=1

(s)n−1g((an − b)xy) f ((an − b)z)

(an − b)α

= cπ(−1)δ(δ−d)

4(α)h(πα/2)

((z + xy)α−1 + (−1)δ(z − xy)α−1

)

+∞∑i=0

(−1)δ(δ−d)+iF (α − 2i − d)

(2i + d)!i∑

j=0

(2i + d

2j + δ

)z2i−2j+d−δ(xy)2j+δ, (8)

where g = {sincos

}δ = {

10

}, and d =

{0 f =g1 f �=g , and also there holds h =

{cos f =gsin f �=g . All the other

relevant parameters are in table 2.

3.1 Limiting values of equation (8)

When on the right-hand side of equation (8) appears h = sin and α = 2m or h = cos andα = 2m − 1, where m ∈ N, one should take limit. Let us consider a particular case of theformula (8), taking a = 1, b = 0, s = 1, which implies c = 1, F = ζ . If g = cos and f = sinthen δ = 0, h = sin, d = 1, so we have

∞∑n=1

sin nz cos nxy

nα= π

(z + xy)α−1 + (z − xy)α−1

4(α) sin(πα/2)

+∞∑i=0

(−1)iζ(α − 2i − 1)

(2i + 1)!i∑

j=0

(2i + 1

2j

)z2i−2j+1(xy)2j .

As we have said, we take limit

limα→2m

[π

(z + xy)α−1 + (z − xy)α−1

4(α) sin(πα/2)

+m−1∑i=0

(−1)iζ(α − 2i − 1)

(2i + 1)!i∑

j=0

(2i + 1

2j

)z2i−2j+1(xy)2j

]= G2m(x, y, z),

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where we have found

G2m(x, y, z) = (−1)m

2(2m − 1)!((z + xy)2m−1(log(z + xy) − ψ(2m) − γ )

+ (z − xy)2m−1(log(z − xy) − ψ(2m) − γ ))

+m−1∑k=1

(−1)k−1ζ(2m − 2k + 1)

(2k − 1)!k−1∑j=0

(2k − 1

2j

)z2k−2j−1(xy)2j ,

where γ is Euler’s constant and ψ is the digamma function, ψ(s) = (′(s)/(s)), whoserelation to the harmonic numbers Hn = ∑n

j=1(1/j) is ψ(n) = Hn−1 − γ , with ψ(1) = −γ =′(1). Finally,

∞∑n=1

sin nz cos nxy

n2m= G2m(x, y, z) +

∞∑i=m

(−1)iζ(2m − 2i − 1)

(2i + 1)!

×i∑

j=0

(2i + 1

2j

)z2i−2j+1(xy)2j .

Example 1 If we additionally choose m = 1, we obtain

∞∑n=1

sin nz cos nxy

n2= z − 1

2

(z log(z2 − (xy)2) + xy log

z + xy

z − xy

)

+∞∑i=1

(−1)iζ(1 − 2i)

(2i + 1)!i∑

j=0

(2i + 1

2j

)z2i−2j+1(xy)2j .

In the same way, we can come to a similar result for any of particular cases if h = sin andα = 2m. Also, we repeat the same procedure, if h = cos and α = 2m − 1.

Example 2 If we choose the same parameters as above, but take f = cos instead of f = sinand again m = 1, then h = cos and α = 1, so we first have to find the limiting value

limα→1

(π

(z + xy)α−1 + (z − xy)α−1

4(α) sin(πα/2)+ zζ(α)

)= −1

2log(z2 − (xy)2),

so that we have

∞∑n=1

cos nxy cos nz

n= −1

2log(z2 − (xy)2) +

∞∑i=1

(−1)iζ(1 − 2i)

(2i)!i∑

j=0

(2i

2j

)z2i−2j (xy)2j .

However, based on the Fourier expansion of the function −(1/2) log(2 sin(t/2)) for 0 <

t < 2π , there holds −(1/2) log(2 sin(t/2)) = ∑∞n=1(1/n) cos nt , and we can easily find

∞∑n=1

cos nxy cos nz

n= −1

2log

(4 sin

z − xy

2sin

z + xy

2

),

where 0 < z ± xy < 2π . Relying on this result, we can conclude

∞∑i=1

(−1)iζ(1 − 2i)

(2i)!i∑

j=0

(2i

2j

)z2i−2j (xy)2j = log

√(z − xy)(z + xy)

4 sin((z − xy)/2) sin((z + xy)/2),

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which is a new summation formula for the series involving Riemann’s zeta function. Further,if we let xy → z, and make use of the combinatorial identity

∑ij=0

(2i

2j

) = 22i−1, the abovesummation formula will be reduced to

∞∑i=1

(−4z2)iζ(1 − 2i)

(2i)! = log

(z

sin z

).

3.2 Closed form cases and some applications of equation (8)

If α − d = 2m and F = ζ, η, λ or α − d = 2m − 1 and F = β (m ∈ N), the sum of the serieson the right-hand side of equation (8) consists of a finite number of terms because the functionsζ, η, λ vanish at even negative integers, and the function β vanishes at odd negative integers.

We shall denote α = 2m + d − ε, where ε ={

0, F = ζ,η,λ1, F = β . So, for this choice of parameters,

the formula (8) is brought into a so-called closed form

Sf,g

2m+d−ε =∞∑

n=1

(s)n−1g((an − b)xy) f ((an − b)z)

(an − b)2m+d−ε

= cπ

2

m∑j=0

(2m + d − ε − 1

2j + δ

)(−1)δ(δ−d)z2m−2j+d−δ−ε−1(xy)2j+δ

(2m + d − ε − 1)! h(mπ + ((d − ε)π)/2)

+m∑

i=0

(−1)δ(δ−d)+iF (2m − 2i − ε)

(2i + d)!i∑

j=0

(2i + d

2j + δ

)z2i−2j+d−δ(xy)2j+δ, (9)

where g = {sincos

}δ = {

10

}, and independently of that d =

{0 f =g1 f �=g , h =

{cos f =gsin f �=g . The other

relevant parameters are in table 2. The formula (9) comprises some particular results from[4], but it is more suitable for immediately obtaining sums of some infinite series as well. Forexample, in [2], §5.4.15, entry 8, we find

∞∑n=1

(−1)n−1

n3sin nxy cos nz = xy

12

(π2 − (xy)2 − 3z2

).

It holds if |xy ± z| ≤ π , which is the condition for uniform convergence of the left-handside series in equation (8) with a = 1, b = 0 and s = −1. After referring to table 2 and theconditions following equation (9), we obtain the above formula, placing in equation (9):g = sin, f = cos, m = 1, d = 1, ε = 0, c = 0, F = η, δ = 1, h = sin.

3.3 Some applications

In [5] the solution of the boundary value problem

U ′′t t = a2U ′′

xx, U(x, 0) = 4hx(L − x)

L2, U ′

t (x, 0) = 0, 0 ≤ x ≤ L, t ≥ 0,

is given by the infinite series

U(x, t) = 32h

π3

∞∑n=1

cos((π(2n − 1)at)/L) sin(((2n − 1)πx)/L)

(2n − 1)3.

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However, using equation (9), where we take m = 1, a = 2, b = 1, s = 1, c = (1/2) andF = λ, then δ = 0, d = 1 and ε = 0, and denoting (atπ/L) = xy, (xπ/L) = z, we obtainits solution in closed form

U(x, t) = 4h

L2(xL − x2 − a2t2), 0 ≤ at

L≤ 1

2,

∣∣∣∣at

L

∣∣∣∣ ≤ x

L≤ 1 −

∣∣∣∣at

L

∣∣∣∣ .Also (see [5]), the solution of the boundary value problem

U ′′t t = U ′′

xx + x(x − L), U(x, 0) = U ′t (x, 0)

= U(0, t) = U(L, t) = 0, 0 ≤ x ≤ L, t ≥ 0,

is

U(x, t) = 8L4

π5

∞∑n=1

cos(πt (2n − 1)/L) sin(πx(2n − 1)/L)

(2n − 1)5− x

12(x3 − 2x2L + L3).

Applying equation (9) with m = 2, using the same parameters as above, but denoting(tπ/L) = xy and (xπ/L) = z, this solution in closed form becomes

U(x, t) = 1

2x2t2 − 1

2Lxt2 + 1

12t4, 0 ≤ t ≤ L

2, t ≤ x ≤ L − t.

4. Sum of the series (1)

Now we make use of equation (8), which is, as we have shown, the formula for findingsum of the right-hand side series of equation (5). We develop in equation (8) the binomials(z ± xy)α−1 into binomial series, and after a rearrangement we actually obtain the summationformula of the left-hand side series in equation (5), which is actually the summation formulafor the series (1)

I T ,fα = cπ(−1)δ(δ−d)

4(α)h(πα/2)

∞∑k=1

(α − 1

2k − δ

)zα−1−2k+δx2k−δ

∫ 1

0y2k−δφ(y)dy

+∞∑i=0

(−1)δ(δ−d)+iF (α − 2i − d)

(2i + d)!i∑

j=0

(2i + d

2j + δ

)z2i−2j+d−δx2j+δ

∫ 1

0y2j+δφ(y)dy,

(10)

where T (x) ={

Sφ(x)

Cφ(x)

}g = {

sincos

}δ = {1

0

}, h =

{cos, f =gsin, f �=g and d =

{0, f =g1, f �=g . The other rele-

vant parameters are in table 2.

4.1 Limiting values

If h = sin and α = 2m or h = cos and α = 2m − 1 (m ∈ N), we encounter a singularity in thefirst term of equation (10). That is why the limit should be taken. We shall demonstrate it for

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the following choice of parameters: a = 1, b = 0, s = 1, c = 1, F = ζ , g = sin, f = cos,δ = 1, h = sin and d = 1, and, apart from this, φ(y) = y−1, so that equation (10) becomes

ISφ,cosα =

∞∑n=1

cos nz

nα

∫ 1

0

sin nxy

ydy = π

2(α) sin(πα/2)

∞∑k=1

(α − 1

2k − 1

)zα−2kx2k−1

2k − 1

+∞∑i=0

(−1)iζ(α − 2i − 1)

(2i + 1)!i∑

j=0

(2i + 1

2j + 1

)z2i−2j x2j+1

2j + 1.

Having chosen φ(y) = 1/y, the integral∫ 1

0 (dy/y) does not converge, but there exists the

integral∫ 1

0 (sin nxy/y)dy = Si(nx), where Si(z) = ∫ z

0 (sin t/t)dt is the integral sine, whichcould not be elementarily calculated, but we know it is bounded by π/2, so Lemma 2 holds.Irrespective of this, without even trying to find its value, and simply by applying equation(10), we obtain the above sum. Yet, because of h = sin, there still remains to take the limit forα = 2m, m ∈ N, i.e.,

limα→2m

[π

2(α) sin(πα/2)

m∑k=1

(α − 1

2k − 1

)zα−2kx2k−1

2k − 1

+m−1∑i=0

(−1)iζ(α − 2i − 1)

(2i + 1)!i∑

j=0

(2i + 1

2j + 1

)z2i−2j x2j+1

2j + 1

⎤⎦ = G2m(x, z),

and we find

G2m(x, z) =m−1∑k=1

(−1)k−1ζ(2m − 2k + 1)

(2k − 1)!k−1∑j=0

(2k − 1

2j + 1

)z2k−2j−2x2j+1

2j + 1+ (−1)m

(2m − 1)!

×( m−1∑

k=1

log z − ψ(2m − 2k + 1) − γ

2k − 1

(2m − 1

2k − 1

)x2k−1z2m−2k + x2m−1 log z

2m − 1

)

as well as

limα→2m

π

2(α) sin(πα/2)

∞∑k=m+1

(α − 1

2k − 1

)zα−2kx2k−1

2k − 1

= (−1)m−1z2m−1

(2m)!∞∑

k=m+1

(x/z)2k−1(2k−12m

)(2k − 1)

,

so that we get

∞∑n=1

cos nz

n2m

∫ 1

0

sin nxy

ydy = G2m(x, z) + (−1)m−1z2m−1

(2m)!∞∑

k=m+1

(x/z)2k−1(2k−12m

)(2k − 1)

+∞∑

i=m

(−1)iζ(2m − 2i − 1)

(2i + 1)!i∑

j=0

(2i + 1

2j + 1

)z2i−2j x2j+1

2j + 1.

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For instance,

∞∑n=1

cos nz

n4

∫ 1

0

sin nxy

ydy = −11x3

108− 19 x z2

18+

(x2 z

4+ 11z3

36

)arcth

x

z

+ x5

24z2�

(x2

z2, 2,

5

2

)+

(x3

36+ x z2

4

)log

(1 − x2

z2

)

+(

x3

18+ x z2

2

)log z + x ζ(3)

+∞∑i=2

(−1)iζ(3 − 2i)

(2i + 1)!i∑

j=0

(2i + 1

2j + 1

)z2i−2j x2j+1

2j + 1,

where � is the Lerch transcendent function defined by the series �(z, s, α) = ∑∞n=0(z

n/(n +α)s), and |x| < π , |x| < z < 2π − |x| (see table 2).

In the same way, we can come to a similar result for any of particular cases.

4.2 Closed form cases

They ensue if α − d = 2m and F = ζ, η, λ or α − d = 2m − 1 and F = β (m ∈ N). The otherrelevant parameters and convergence regions are in table 2. When we choose the function φ inequation (10), we additionally have to calculate both integrals having in fact a similar structure.

We have already seen that the formula (9) contains all closed form cases for the productof trigonometric functions. So, using the formula (9) we bring the sum of the series (1) intoclosed form:

IT ,f

2m+d−ε =∞∑

n=1

(s)n−1T ((an − b)x)

(an − b)2m+d−εf ((an − b)z)

= cπ

2

m∑j=0

(2m + d − ε − 1

2j + δ

)(−1)δ(δ−d)z2m−2j+d−δ−ε−1x2j+δ

(2m + d − ε − 1)! h(mπ + ((d − ε)π)/2)

×∫ 1

0y2j+δφ(y)dy +

m∑i=0

(−1)δ(δ−d)+iF (2m − 2i − ε)

(2i + d)!

×i∑

j=0

(2i + d

2j + δ

)z2i−2j+d−δx2j+δ

∫ 1

0y2j+δφ(y)dy, (11)

where T (x) ={

Sφ(x)

Cφ(x)

}g = {

sincos

}δ = {

10


{0 f =g1 f �=g , h ={

cos f =gsin f �=g . For F = ζ, η, λ there holds ε = 0, but for F = β it is ε = 1. The other relevant

parameters are in table 2.

Example 3 First we take a = 1, b = 0 and s = 1 in equation (11), there follows c = 1and F = ζ (see table 2), then ε = 0. We choose φ(y) = ctgy. For g = cos the integral∫ 1

0 ctgy cos nxy dy does not exist, so there must be g = sin, which means δ = 1. If we take

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f = cos, then h = sin and d = 1 because f �= g. Let m = 1. For 0 < |x| < π , we have∣∣∣x ∫ 1

0ctgy sin nxy dy

∣∣∣ =∣∣∣ ∫ x

0ctg(t/x) sin nt dt

∣∣∣ �∫ x

0

∣∣∣ctg(t/x) sin nt

∣∣∣ dt

�∫ π

0

∣∣∣ctg(t/π) sin nt

∣∣∣ dt.

Also, |tg(t/π)| > (|t |/π) implies |ctg(t/π)| < (π/|t |), for |t | < π . Additionally,limt→0+ ctg(t/π) sin nt = nπ . The function |ctg(t/π) sin nt | is non-negative, fast oscillatory,and its maximal value between two consecutive zeros constantly decreases. There follows∫ π

0

∣∣∣ctgt

πsin nt

∣∣∣ dt =∫ 1/n

0

∣∣∣ctgt

πsin nt

∣∣∣ dt +∫ π

1/n

∣∣∣ctgt

π

∣∣∣ ·∣∣∣ sin nt

∣∣∣ dt

<

∫ 1/n

0nπ dt + π

∫ π

1/n

dt

t= π(1 + log nπ),

which means that∣∣∣ ∫ 1

0 ctgy sin nxy dy

∣∣∣ < (π(1 + log nπ)/|x|) = Mn(x), and we can easily

see that for α > 0 and each x, 0 < |x| < π , the sequence (Mn(x)/nα) → 0 monotonically,so that we may apply Lemma 2. Also, we find∫ 1

0y2j+1ctgy dy =

{log(2 sin 1) + (1/2)Cl2(2), j = 0

log(2 sin 1) + (3/2)Cl2(2) + (3/2)Cl3(2) − (3/4)Cl4(2), j = 1

where on the right-hand side are Clausen functions defined by (see [1])

Cl2ν(x) =∞∑

n=1

sin nx

n2ν, Cl2ν−1(x) =

∞∑n=1

cos nx

n2ν−1, ν ∈ N.

Making use of equation (9), we obtain

ISφ,cos1 =

∞∑n=1

cos nz

n3

∫ 1

0ctgy sin nxy dy

= x

12(2π2 − 6πz + 3z2)

(log(2 sin 1) + 1

2Cl2(2)

)

+ x3

12

(log(2 sin 1) + 3

2Cl2(2) + 3

2Cl3(2) − 3

4Cl4(2)

),

where |x| < π and |x| < z < 2π − |x| (see table 2).

Example 4 Let a = 1, b = 0 and s = −1 in equation (11), implying c = 0, F = η and ε = 0.For g = cos there must be δ = 0. Further, we take f = cos, there follows h = cos, d = 0because f = g. Let m = 2. We choose φ(y) = (1 − y2)−1/2. It is unbounded about 1, butintegrable on (0, 1), thus satisfying conditions of Lemma 1. So applying equation (11), wefind

ICφ,cos4 =

∞∑n=1

(−1)n−1 cos nz

n4

∫ 1

0

cos nxy√1 − y2

dy

= 7π5

1440− π3

96(x2 + 2z2) + π

32

(x4

8+ x2z2 + z4

3

),

where |x| < π and |x| − π < z < π − |x| (see table 2).

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Remark 3 We note that this particular result can be obtained in a different way. Namely,substituting y = cos θ , then multiplying with 2/π , we deal with Bessel functions (see equation(4)), and the above series becomes

∞∑n=1

(−1)n−1J0(nx)

n4cos nz,

which is the series over the product of Bessel and trigonometric function, and its sum can beobtained by means of the formula on the page 289 in [2].

Example 5 Further, we take a = 2, b = 1 and s = 1 in equation (11). In table 2, we readc = (1/2) and F = λ. So ε = 0. If we choose g = cos, f = sin, then we have δ = 0, d = 1and h = sin. Let m = 1 and φ(y) = log(sin y). It is unbounded in the neighborhood of 0,however

∫ 10 log(sin y) dy = − log 2 − (1/2)Cl2(2), and Lemma 1 holds. Applying equation

(11), we obtain

ISφ,sin1 =

∞∑n=1

sin((2n − 1)z)

(2n − 1)3

∫ 1

0log(sin y) cos((2n − 1)xy) dy

= π

96

(6z(z − π)(Cl2(2) + log 4) + x2(6Cl2(2) + 6Cl3(2) − 3Cl4(2) + log 16)

),

where |x| < (π/2) and |x| − (π/2) < z < (π/2) − |x| (see table 2).

Example 6 Finally, we take in equation (11): a = 2, b = 1, s = −1, c = 0, F = β, g =sin, f = sin, d = 0, δ = 1, h = cos, ε = 1 and m = 2, and choose φ(y) = log y. The inte-gral

∫ 10 log y dy converges, φ meets requirements of Lemma 1, and making use of equation

(11), the infinite series is brought in closed form

ISφ,sin5 =

∞∑n=1

(−1)n−1 sin((2n − 1)z)

(2n − 1)5

∫ 1

0log y sin((2n − 1)xy) dy

= πxz

32

(π2

4− x2

12− z2

3

),

where |x| < (π/2) and |x| − (π/2) < z < (π/2) − |x| (see table 2).This series can take another form. At first, by integrating by parts, we come to the rela-

tion∫ 1

0 log y sin((2n − 1)xy) dy = (1/((2n − 1)x))∫ 1

0 ((cos((2n − 1)xy) − 1/y) dy. After-wards, we prove

∫ p

0 ((1 − e−t )/t)dt − ∫ ∞p

(e−t /t) dt = log p + γ . On this basis, we find

Ci(p) + Cin(p) = log p + γ , where Ci(p) = − ∫ ∞p

(cos t/t)dt is the integral cosine and

Cin(p) = ∫ p

0 ((1 − cos t)/t)dt its related function. Hence, there follows

∫ 1

0log y sin((2n − 1)xy) dy

= 1

(2n − 1)x

(Ci((2n − 1)x) − γ − log((2n − 1)x)

)(x > 0),

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and we obtain the sum of a new series

∞∑n=1

(−1)n−1 sin((2n − 1)z)

(2n − 1)6x

(Ci((2n − 1)x) − γ − log((2n − 1)x)

)

= πxz

32

(π2

4− x2

12− z2

3

).

References[1] Trickovic, S.B., Vidanovic, M.V. and Stankovic, M.S., 2006, On the summation of series in terms of Bessel

functions. Zeitschrift für Analysis und ihre Anwendungen, 25, 393–406.[2] Stankovic, M.S., Trickovic, S.B. and Vidanovic, M.V., 2001, Series over the product of Bessel and trigonometric

functions. Integral Transform and Special Function, 11(3), 281–290.[3] Knopp, K., 1990, Theory and Application of Infinite Series (New York: Dover Publications).[4] Prudnikov, A.P., Brychkov, Y.A. and Marichev, O.I., 1986, Integrals and Series, Elementary Functions, Vol. 1

(New York: Gordon and Breach Science Publishers).[5] E.A. Vukolov, A.V. Efimov and others, 1984, Collection of Problems in Mathematics (Moscow: Nauka) (in

Russian)

This article was downloaded by:[Tričković, Slobodan B.]On: 2 June 2008Access Details: [subscription number 793618343]Publisher: Taylor & FrancisInforma Ltd Registered in England and Wales Registered Number: 1072954Registered office: Mortimer House, 37-41 Mortimer Street, London W1T 3JH, UK

Integral Transforms and SpecialFunctionsPublication details, including instructions for authors and subscription information:http://www.informaworld.com/smpp/title~content=t713643686

On the summation of trigonometric seriesSlobodan B. Tričković a; Mirjana V. Vidanović b; Miomir S. Stanković b

a Department of Mathematics, Faculty of Civil Engineering, University of Ni , Ni ,Serbiab Department of Mathematics, Faculty of Environmental Engineering, University ofNi , Ni , Serbia

Online Publication Date: 01 January 2008

To cite this Article: Tričković, Slobodan B., Vidanović, Mirjana V. and Stanković,Miomir S. (2008) 'On the summation of trigonometric series', Integral Transforms

and Special Functions, 19:6, 441 — 452

To link to this article: DOI: 10.1080/10652460801936689URL: http://dx.doi.org/10.1080/10652460801936689



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http://dx.doi.org/10.1080/10652460801936689


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Integral Transforms and Special FunctionsVol. 19, No. 6, June 2008, 441–452

On the summation of trigonometric series

Slobodan B. Trickovica*, Mirjana V. Vidanovicb and Miomir S. Stankovicb

aDepartment of Mathematics, Faculty of Civil Engineering, University of Niš, Niš, Serbia; bDepartment ofMathematics, Faculty of Environmental Engineering, University of Niš, Niš, Serbia

(Received 28 September 2007; final version)

We deal with the series∞∑

n=1

(s)n−1 f ((an − b)x)

(an − b)α, α ∈ R

+,

and express it as a power series in terms of Riemann’s ζ or Catalan’s β function or Dirichlet functions η

and λ. Also, closed form cases as well as those when it is necessary to take limit have been thoroughlyanalyzed. Some applications such as convergence acceleration are considered too.

Keywords: Riemann’s ζ and Catalan’s β function; Dirichlet η and λ functions

2000 Mathematics Subject Classification: Primary: 33C10; Secondary: 11M06, 65B10

1. Introduction and preliminaries

There have been many particular cases (see, for example, [2] and [7]) of the trigonometric seriesof the type

∞∑n=1

(s)n−1 f ((an − b)x)

(an − b)α, α ∈ R

+, (1)

where s is 1 or −1, f = sin or f = cos and a = { 12 }b = { 0

1 }. We first derive a general formulafor finding the sum of Equation (1), meaning that we consider α to be a positive real parameterwith the exclusion of positive integers. Afterwards we regard α as a positive integer, whereuponthere appears a necessity to distinguish between the cases when infinite series (1) reduces to afinite sum, and those when limiting value must be taken. Apart from all known particular cases,the general formula yields new ones. We start with the following

LEMMA 1 Series (1) is uniformly convergent for α > 0. Convergence regions are given inTable 1, where ζ is Riemann’s zeta function ζ(z) = ∑∞

k=1 k−z, η and λ are Dirichlet functionsη(z) = ∑∞

k=1(−1)k−1k−z = (1 − 21−z)ζ(z), λ(z) = ∑∞k=0(2k + 1)−z = (1 − 2−z)ζ(z), and β is

Catalan’s beta function β(z) = ∑∞k=0(−1)k(2k + 1)−z.


ISSN 1065-2469 print/ISSN 1476-8291 online© 2008 Taylor & FrancisDOI: 10.1080/10652460801936689http://www.informaworld.com

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442 S.B. Trickovic et al.

Table 1. Parameters and convergence region

a b s c F f δ r α ∈ Convergence region

1 0 1 1 ζ sin 1 2m − 1 R+ \ 2N 0 < x < 2π

cos 0 2m R+ \ 2N − 1

−1 0 η sin 1 2m − 1 R+ \ 2N −π < x < π

cos 0 2m R+ \ 2N − 1

2 1 1 1/2 λ sin 1 2m − 1 R+ \ 2N 0 < x < π

cos 0 2m R+ \ 2N − 1

−1 0 β sin 1 2m R+ \ 2N − 1 −π/2 < x < π/2

cos 0 2m − 1 R+ \ 2N

Remark 1 We note that ζ, η, λ are analytic in the whole complex plane except for z = 1, wherethey have a pole, whereas Catalan’s beta function β(z) = ∑∞

k=0(−1)k(2k + 1)−z satisfies thefunctional equation β(z) = (π/2)z−1(1 − z) cos(πz/2)β(1 − z) extending the beta function tothe left side of the complex plane Re z < 1.

Proof We shall make use of Dirichlet’s test saying that (see [3]) the series∑∞

n=0 an(x)bn(x) isuniformly convergent in D, if the partial sums of

∑∞n=0 an(x) are uniformly bounded in D, and

the sequence bn(x), being monotonic for every fixed x, uniformly converges to 0.Let us suppose first that a = 1, b = 0. If s = 1, then there holds

∣∣∣∣n∑

k=1

f (kx)

∣∣∣∣ ≤ 1

sin(x/2)≤ 1

sin ε

for 0 < x < 2π , because sin(x/2) ≥ sin ε > 0 for each ε > 0 satisfying ε ≤ x/2 ≤ π − ε,meaning that the partial sums are uniformly bounded with respect to x ∈ (0, 2π). Similarly,if s = −1, we would have ∣∣∣∣

n∑k=1

(−1)k−1f (kx)

∣∣∣∣ ≤ 1

cos(x/2)≤ 1

sin ε

for−π < x < π , because cos(x/2) ≥ sin ε > 0 for each ε > 0 satisfying−(π/2) + ε ≤ (x/2) ≤(π/2) − ε. Here the partial sums are uniformly bounded with respect to x ∈ (−π, π).

We now suppose a = 2, b = 1. If s = 1, then

∣∣∣∣n∑

k=1

f ((2k − 1)x)

∣∣∣∣ ≤ 1

sin x≤ 1

sin ε

for 0 < x < π , because sin x ≥ sin ε > 0 for each ε > 0 satisfying ε ≤ x ≤ π − ε. The partialsums are uniformly bounded with respect to x ∈ (0, π). Finally if s = −1, then

∣∣∣∣n∑

k=1

(−1)k−1f ((2k − 1)x)

∣∣∣∣ ≤ 1

cos x≤ 1

sin ε

for −(π/2) < x < (π/2), because cos x ≥ sin ε > 0 for each ε > 0 satisfying −(π/2) + ε ≤x ≤ (π/2) − ε, and the partial sums are uniformly bounded with respect to x ∈ (−(π/2), (π/2)).Since in each of the particular cases the sequence 1/(an − b)α tends to 0 for α > 0, by virtue ofDirichlet’s test all the series in question uniformly converge, whereby the proof is complete. �

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Integral Transforms and Special Functions 443

2. The general formula

THEOREM 1 For the values of α and the other relevant parameters from Table 1, there holds

∞∑n=1

(s)n−1f ((an − b)x)

(an − b)α= cπxα−1

2(α)f (πα/2)+

∞∑k=0

(−1)kF (α − 2k − δ)

(2k + δ)! x2k+δ. (2)

Proof We make use of the polylogarithm Liα(z) defined by the series (see [4]), and with thefollowing integral representation

Liα(z) =∞∑

n=1

zn

nα= 1

(α)

∫ ∞

0

tα−1

et /z − 1dt,

where the right-hand side integral converges for z ∈ C \ {z | z ∈ R, z ≥ 1}, and it is referred toas Bose’s integral. So for α > 0, we have

∞∑n=1

sin nx

nα= i

2

∞∑n=1

e−inx − einx

nα= i

2

(Liα(e−ix) − Liα(eix)

),

∞∑n=1

cos nx

nα= 1

2

∞∑n=1

e−inx + einx

nα= 1

2

(Liα(e−ix) + Liα(eix)

) (3)

and

∞∑n=1

(−1)n−1 sin nx

nα= i

2

(Liα(−eix) − Liα(−e−ix)

),

∞∑n=1

(−1)n−1 cos nx

nα= −1

2

(Liα(−e−ix) + Liα(−eix)

).

(4)

Also, we have

∞∑n=1

sin(2n − 1)x

(2n − 1)α= i

2

((Liα(e−ix) − 1

2αLiα(e−2ix)

)−

(Liα(eix) − 1

2αLiα(e2ix)

)),

∞∑n=1

cos(2n − 1)x

(2n − 1)α= 1

2

((Liα(e−ix) − 1

2αLiα(e−2ix)

)+

(Liα(eix) − 1

2αLiα(e2ix)

)) (5)

and

∞∑n=1

(−1)n−1 sin(2n − 1)x

(2n − 1)α= 1

4

(Liα(−ieix) − Liα(−ie−ix) − Liα(ieix) + Liα(ie−ix)

),

∞∑n=1

(−1)n−1 cos(2n − 1)x

(2n − 1)α= i

4

(Liα(−ieix) + Liα(−ie−ix) − Liα(ieix) − Liα(ie−ix)

).

(6)

Here, we note that in order to obtain the right-hand sides of Equation (4) and (6), we have takenadvantage of the representation (−1)n = cos nπ , and then, by means of elementary trigonometricidentities, we split up the product into the sum of two trigonometric functions.

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We shall now consider the Mellin transform of the polylogarithm in the form of Bose’s integral.The Mellin transform of a function f and the inverse transform of a function ϕ are (see [6])

M(f (x)) =∫ ∞

0xu−1f (x)dx, M−1(ϕ(u)) = 1

2πi

∫ c+i∞

c−i∞x−uϕ(u)du.

This integral transform is closely connected to the theory of Dirichlet series and is often used innumber theory and the theory of asymptotic expansions. Also, it is closely related to the Laplaceand Fourier transform as well as to the theory of the gamma function and allied special functions.So we find

M(Liα(p e−x)) =∫ ∞

0xu−1Liα(p e−x)dx = 1

(α)

∫ ∞

0

∫ ∞

0

tα−1xu−1

et+x/p − 1dtdx.

The change of variables x = ab, t = a(1 − b) allows the integrals to be separated

M(Liα(p e−x)) = 1

(α)

∫ 1

0bu−1(1 − b)α−1db

∫ ∞

0

aα+u−1

ea/p − 1da = (u)Liα+u(p). (7)

For p = 1, because Liα+u(1) = ζ(α + u), through the inverse Mellin transform, we have

Liα(e−x) = 1

2πi

∫ c+i∞

c−i∞(u)ζ(α + u)x−udu,

where c is a constant to the right of the poles of the integrand. The path of integration maybe converted into a closed contour, and the poles of the integrand are those of (u) at u =0, −1, −2, . . . , and of ζ(α + u) at u = 1 − α. Summing the residues yields a representation ofthe polylogarithm as a power series

Liα(eμ) = (−μ)α−1(1 − α) +∞∑

k=0

ζ(α − k)

k! μk, |μ| < 2π, α �= 1, 2, 3, . . . (8)

about μ = 0. Further, following Equation (3), we have

i

2(Liα(e−μ) − Liα(eμ)) = i

2(μα−1 − (−μ)α−1)(1 − α) − i

∞∑k=0

ζ(α − 2k − 1)

(2k + 1)! μ2k+1,

1

2(Liα(eμ) + Liα(e−μ)) = 1

2((−μ)α−1 + μα−1)(1 − α) +

∞∑k=0

ζ(α − 2k)

(2k)! μ2k,

(9)

where we replace μ with ix, 0 < x < 2π . For a positive real non-integer α, the gamma functionis finite, and so in view of

(± i)α−1 = e± i(α−1)π/2 = cosπ

2(α − 1) ± i sin

π

2(α − 1), (10)

we calculate

i

2(μα−1 − (−μ)α−1)(1 − α) = iπxα−1

2(α) sin πα(iα−1 − (−i)α−1) = πxα−1

2(α) sin(π/2)α,

1

2((−μ)α−1 + μα−1)(1 − α) = πxα−1

2(α) sin πα((−i)α−1 + iα−1) = πxα−1

2(α) cos(π/2)α,

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which, in conjunction with Equation (9) gives the sums of the series in Equation (3), i.e.

∞∑n=1

sin nx

nα= πxα−1

2(α) sin(π/2)α+

∞∑k=0

(−1)kζ(α − 2k − 1)

(2k + 1)! x2k+1,

∞∑n=1

cos nx

nα= πxα−1

2(α) cos(π/2)α+

∞∑k=0

(−1)kζ(α − 2k)

(2k)! x2k.

We now consider Equation (4). Because of Equation (7), we find M(Liα(−e−x)) =−(u)η(α + u), since Liα+u(−1) = −η(α + u). Through the inverse Mellin transform andconversion of path of integration into a closed contour, after summing the residues at u =0, −1, −2, . . . and u = 1 − α (which is 0), we have

Liα(−eμ) = −∞∑

k=0

η(α − k)

k! μk, |μ| < 2π, α �= 1, 2, 3, . . . ,

whence we find

i

2(Liα(−e−μ) − Liα(−eμ)) = i

∞∑k=0

η(α − 2k − 1)

(2k + 1)! μ2k+1,

−1

2(Liα(−eμ) + Liα(−e−μ)) =

∞∑k=0

η(α − 2k)

(2k)! μ2k,

(11)

and for μ = ix (−π < x < π), we obtain the sums of the series in Equation (4), i.e.

∞∑n=1

(−1)n−1 sin nx

nα=

∞∑k=0

(−1)kη(α − 2k − 1)

(2k + 1)! x2k+1,

∞∑n=1

(−1)n−1 cos nx

nα=

∞∑k=0

(−1)kη(α − 2k)

(2k)! x2k.

As regards Equation (5), judging by its structure, we have to repeat the procedure as that forobtaining Equation (8), i.e. once for Liα(e±μ) (|μ| < 2π ), and once again for Liα(e±2μ) (|μ| < π ),so that we have

i

2

((Liα(e−μ) − 1

2αLiα(e−2μ)

)−

(Liα(eμ) − 1

2αLiα(e2μ)

))

= i

4(μα−1 − (−μ)α−1)(1 − α) − i

∞∑k=0

λ(α − 2k − 1)

(2k + 1)! μ2k+1,

1

2

((Liα(e−μ) − 1

2αLiα(e−2μ)

)+

(Liα(eμ) − 1

2αLiα(e2μ)

))

= 1

4(μα−1 + (−μ)α−1)(1 − α) +

∞∑k=0

λ(α − 2k)

(2k)! μ2k, |μ| < π, α �= 1, 2, 3, . . . ,

(12)

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where we have taken into account λ(z) = (1 − 2−z)ζ(z). Thus, for μ = ix (0 < x < π), weobtain the sums of the series in Equation (5), i.e.

∞∑n=1

sin(2n − 1)x

(2n − 1)α= πxα−1

4(α) sin(π/2)α+

∞∑k=0

(−1)kλ(α − 2k − 1)

(2k + 1)! x2k+1,

∞∑n=1

cos(2n − 1)x

(2n − 1)α= πxα−1

4(α) cos(π/2)α+

∞∑k=0

(−1)kλ(α − 2k)

(2k)! x2k.

Finally, in the case of Equation (6), we first calculate Liα+u(±i) = 2−(α+u)η(α + u) ± iβ(α +u) and find M(Liα(±ie−x)) = (u)

(2−(α+u)η(α + u) ± iβ(α + u)

). Applying again the inverse

Mellin transform and converting the path of integration into closed contour, we evaluate theresidues at u = 0, −1, −2, . . . and u = 1 − α (which is 0), so that after their summation, arearrangement and simplification, we obtain

1

4

(Liα(−ieμ) − Liα(−ie−μ) − Liα(ieμ) + Liα(ie−μ)

) = −i

∞∑k=0

β(α − 2k − 1)

(2k + 1)! μ2k+1,

i

4

(Liα(−ieμ) + Liα(−ie−μ) − Liα(ieμ) − Liα(ie−μ)

) =∞∑

k=0

β(α − 2k)

(2k)! μ2k,

(13)

where |μ| < π/2, α �= 1, 2, 3, . . . . For μ = ix(−π/2 < x < π/2), we find the sums of the seriesin Equation (6), i.e.

∞∑n=1

(−1)n−1 sin(2n − 1)x

(2n − 1)α=

∞∑k=0

(−1)kβ(α − 2k − 1)

(2k + 1)! x2k+1,

∞∑n=1

(−1)n−1 cos(2n − 1)x

(2n − 1)α=

∞∑k=0

(−1)kβ(α − 2k)

(2k)! x2k.

Gathering all these results, we conclude that Equation (2) holds. �

3. Closed form cases

We shall express now, for certain values of parameters, series (1) as polynomials, saying then thatwe have brought the infinite series in closed form.

THEOREM 2 For the values of r and the other relevant parameters from Table 1, there holds

∞∑n=1

(s)n−1 f (an − b)x)

(an − b)r= (−1)[r/2]cπxr−1

2(r − 1)! +[r/2]∑k=0

(−1)kF (r − 2k − δ)

(2k + δ)! x2k+δ. (14)

Proof First of all, by setting z = 2n + 1 in the functional equation for the Riemann zeta function(see [1])

ζ(1 − z) = 2ζ(z)(z)

(2π)zcos

zπ

2,

we find ζ(−2n) = 0, n ∈ N. Taking into account the relations of the Riemann zeta functions toDirichlet functions η and λ (see Lemma 1), we easily notice that they also vanish at even negative

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integers. Moreover, λ(0) = 0. It is the other way round with the Catalan function β, i.e. it vanishesat odd negative integers. In order to prove this, we need the Hurwitz zeta function ζ(z, a) initiallydefined for σ > 1 (z = σ + iτ ) by the series

ζ(z, a) =+∞∑k=0

1

(k + a)z,

where a is a fixed real number, 0 < a ≤ 1. The function β(z) can now be represented as follows

β(z) =+∞∑k=0

(−1)k

(2k + 1)z= 4−z

(ζ

(z,

1

4

)− ζ

(z,

3

4

)),

whence we have

β(−(2n − 1)) = 42n−1

(ζ

(−(2n − 1),

1

4

)− ζ

(−(2n − 1),

3

4

)), n ∈ N.

Considering that for a non-negative integern, there holds (see [1]) ζ(−n, a) = −Bn+1(a)/(n + 1),where Bn(x) are Bernoulli polynomials defined by relations

text

et − 1=

+∞∑n=0

Bn(x)tn

n! , B0(x) = 1,

we first replace x with 1 − x, and have

+∞∑n=0

Bn(1 − x)tn

n! = te(1−x)t

et − 1= (−t)e(−t)x

e−t − 1=

+∞∑n=0

Bn(x)(−1)n tn

n! ,

whence we obtain Bn(1 − x) = (−1)n Bn(x) implying B2n(3/4) = B2n(1/4), so that we calculate

β(−(2n − 1)) = 42n−1

(−B2n(1/4)

2n+ B2n(3/4)

2n

)= 0.

It is known that for r ∈ N, the value (1 − r) of the gamma function becomes infinite, andwe cannot place α = r immediately in Equation (9). In order to get a finite value we take, on theright-hand side sum in Equation (9), the first m − 1 terms if r = 2m − 1, and the first m terms ifr = 2m. Then we take limits

limα→2m−1

(i

2(μα−1 − (−μ)α−1)(1 − α) − i

m−1∑k=0

ζ(α − 2k − 1)

(2k + 1)! μ2k+1

)

= −ilog(−μ) − log μ

2(2m − 2)! μ2m−2 − i

m−1∑k=0

ζ(2m − 2k − 2)

(2k + 1)! μ2k+1,

limα→2m

(1

2((−μ)α−1 + μα−1)(1 − α) +

m∑k=0

ζ(α − 2k)

(2k)! μ2k

)

= − log(−μ) − log μ

2(2m − 1)! μ2m−1 +m∑

k=0

ζ(2m − 2k)

(2k)! μ2k.

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By virtue of ζ(−2n) = 0 (n ∈ N), the remainders of the series in Equation (9) vanish, so placingμ = ix, 0 < x < 2π , we obtain

∞∑n=1

sin nx

n2m−1= (−1)m−1πx2m−2

2(2m − 2)! +m−1∑k=0

(−1)kζ(2m − 2k − 2)

(2k + 1)! x2k+1,

∞∑n=1

cos nx

n2m= (−1)mπx2m−1

2(2m − 1)! +m∑

k=0

(−1)kζ(2m − 2k)

(2k)! x2k (m ∈ N),

(15)

which is formula (14) for s = 1, a = 1, b = 0, c = 1, F = ζ , f = {sincos

}, r = {2m−1

2m

}.

When dealing with Equation (11), we do not have a problem with the singularity of the gammafunction any more, so that we may replace there α with r ∈ N, and considering that η(−2n) = 0(n ∈ N), we obtain

∞∑n=1

(−1)n−1 sin nx

n2m−1=

m−1∑k=0

(−1)kη(2m − 2k − 2)

(2k + 1)! x2k+1,

∞∑n=1

(−1)n−1 cos nx

n2m=

m∑k=0

(−1)kη(2m − 2k)

(2k)! x2k (m ∈ N),

(16)

which is formula (14) for the choice of parameters s = −1, a = 1, b = 0, c = 0, F = η, f ={sincos

}r = {2m−1

2m

}.

In the case of Equation (12), there appears again the singularity of (1 − r) at r ∈ N, so quitesimilarly as above we take limits, and after placing μ = ix, 0 < x < π , because λ(−2n) = 0(n ∈ N), we obtain

∞∑n=1

sin(2n − 1)x

(2n − 1)2m−1= (−1)m−1πx2m−2

4(2m − 2)! +m−1∑k=0

(−1)kλ(2m − 2k − 2)

(2k + 1)! x2k+1,

∞∑n=1

cos(2n − 1)x

(2n − 1)2m= (−1)mπx2m−1

4(2m − 1)! +m∑

k=0

(−1)kλ(2m − 2k)

(2k)! x2k (m ∈ N),

(17)

which is formula (14) for the choice of s = 1, a = 2, b = 1, c = 1/2, F = λ, f = {sincos

}r ={2m−1

2m

}.

Finally, in Equation (13), similarly as for Equation (11), we do not have to deal with thesingularity of the gamma function and replacing α with r ∈ N, considering that β(−2n + 1) = 0(n ∈ N), we obtain

∞∑n=1

(−1)n−1 sin(2n − 1)x

(2n − 1)2m=

m∑k=0

(−1)kβ(2m − 2k − 1)

(2k + 1)! x2k+1,

∞∑n=1

(−1)n−1 cos(2n − 1)x

(2n − 1)2m−1=

m−1∑k=0

(−1)kβ(2m − 2k)

(2k)! x2k (m ∈ N),

(18)

which is formula (14) for the choice of parameters s = −1, a = 2, b = 1, c = 0, F = β, f ={sincos

}r = { 2m

2m−1

}, wherby we complete the proof. �

We note that because of λ(0) = 0 the upper bounds in Equation (17) are actually m − 2 andm − 1 and m − 1 for the first sum in Equation (18) since β(−1) = 0, but purely for the sake offitting them into a general formula (14), we retain m − 1 and m in Equation (17), and m for thefirst sum in Equation (18), without formally changing values.

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3.1. Some applications

Now we are going to present significant and important applications of our closed formformula (14).

3.1.1. Integral transforms

Applying some of the integral transforms, one can obtain various series in closed form. Forinstance, if we take f = cos, δ = 0 in Equation (14), then apply the Laplace transform, we have

∞∑n=1

(s)n−1p

(an − b)r(p2 + (an − b)2)= (−1)[r/2] cπ

2p r+

[r/2]∑i=0

(−1)iF (r − 2i)

p 2i+1.

Further, if we set s = 1, a = 1, b = 0, then F = ζ, c = 1 must be taken, and we obtain

∞∑n=1

p

nr(p2 + n2)= (−1)[r/2] π

2p r+

[r/2]∑i=0

(−1)iζ(r − 2i)

p2i+1. (19)

Now we apply the inverse Mellin transform to this series, knowing that (see [6, p. 166, 2.16 andp. 167, 2.25])

M−1

(z

z2 + n2

)=

{cos(n log x) x < 1

0 x > 1,M−1

(1

zν

)=

⎧⎨⎩

(log(1/x))ν−1

(ν)x < 1

0 x > 1,

coming to the sum of a trigonometric series

∞∑n=1

cos(n log x)

nr= (−1)[r/2]π

2(r − 1)!(

log1

x

)r−1

+[r/2]∑i=0

(−1)iζ(r − 2i)

(2i)!(

log1

x

)2i

, x < 1.

However, if we want to apply the Bessel instead of Mellin transform to series (19), we first referto (see [5, p. 36, 4.23 and p. 33, 4.6])

B

(xν+1/2

(a2 + x2)μ

)= aν−μ+1 yμ−1/2

2μ−1 (μ)Kν−μ+1(ay)

and

B(xμ) = 2μ+1/2 ((ν/2) + (μ/2) + (3/4))

yμ+1 ((ν/2) − (μ/2) + (1/4)),

where Kν is Hankel’s function. Thus we obtain the sum of one of the Schlömilch series

∞∑n=1

K1/2(ny)

nr−(1/2)= (−1)[r/2]πyr−(3/2)(1 − (r/2))

2r+1/2((r + 1)/2)+

[r/2]∑i=0

(−1)iζ(r − 2i)((1/2) − i)

i! 22i+1/2y2i−1/2.

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3.1.2. Convergence acceleration

On the basis of Equation (14), the convergence acceleration of trigonometric series is obtained.As an example, we consider the series

T =∞∑

n=1

n

n2 + 1sin nx. (20)

For each natural number M we prove, by the method of mathematical induction, that there holds

n

n2 + 1= n

n2· 1

1 + (1/n2)= 1

n

(1 − 1

n2+ 1

n4− · · ·

)

=M∑

m=1

(−1)m−1

n2m−1+ (−1)M

n2M−1(n2 + 1).

(21)

Replacing Equation (21) in Equation (20), we have

T =M∑

m=1

(−1)m−1T sin2m−1 +

∞∑n=1

(−1)M sin nx

(n2 + 1)n2M−1,

where T sin2m−1 denotes first of the closed form formulas (15), i.e.

T sin2m−1 =

∞∑n=1

sin nx

n2m−1= (−1)m−1πx2m−2

2(2m − 2)! +m−1∑k=0

(−1)kζ(2m − 2k − 2)

(2k + 1)! x2k+1.

So, the greater M , the faster convergence of the remaining series, and accordingly we have thefaster convergence of the series T . Here is shown how many terms of the remaining series oughtto be taken for the given M and accuracy ε

ε 10−1 10−2 10−5 10−8

M = 1 3 8 224 7072M = 3 2 2 6 16M = 10 1 2 2 3

4. Limiting values

THEOREM 3 For the values of α complementary to those in Table 1, there holds

∞∑n=1

(s)n−1f ((an − b)x)

(an − b)α= �α(x) +

∞∑k=[(α−1)/2]+1

(−1)kF (α − 2k − δ)

(2k + δ)! x2k+δ, (22)

where

�α(x) = (−1)[(α−1)/2] c(α − 1)! (ψ(α) + γ − log x) xα−1 +

[(α−1)/2]∑k=1

(−1)[(α−1)/2]−kF (2k + 1)

(α − 2k − 1)! xα−2k−1

and if c = 1, then F = ζ or if c = 1/2, then F = λ.

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Proof We have seen that if we let α tend to 2m − 1 and to 2m, respectively, in Equation (9), theright-hand side sums truncate because the zeta function vanish at even negative integers, and aslimiting values, we obtain polynomials (see Equation (15)). However, if we let α tend to 2m andto 2m − 1, respectively in Equation (9), we find

limα→2m

(i

2(μα−1 − (−μ)α−1)(1 − α) − i

m−1∑k=0

ζ(α − 2k − 1)

(2k + 1)! μ2k+1

)

= i

(2m − 1)!(

ψ(2m) + γ − log(−μ) + log μ

2

)μ2m−1 − i

m−1∑k=1

ζ(2k + 1)

(2m − 2k − 1)! μ2m−2k−1

limα→2m−1

(1

2((−μ)α−1 + μα−1)(1 − α) +

m−1∑k=0

ζ(α − 2k)

(2k)! μ2k

)

= 1

(2m − 2)!(ψ(2m − 1) + γ − log(−μ) + log μ

2

)μ2m−2 +

m−1∑k=1

ζ(2k + 1)

(2m − 2k − 2)! μ2m−2k−2,

where ψ is the digamma function. Setting μ = ix, 0 < x < 2π , we obtain

∞∑n=1

sin nx

n2m= (−1)m−1

(2m − 1)! (ψ(2m) + γ − log x) x2m−1

+m−1∑k=1

(−1)m−1−kζ(2k + 1)

(2m − 2k − 1)! x2m−2k−1 +∞∑

k=m

(−1)kζ(2m − 2k − 1)

(2k + 1)! x2k+1,

∞∑n=1

cos nx

n2m−1= (−1)m−1

(2m − 2)! (ψ(2m − 1) + γ − log x) x2m−2

+m−1∑k=1

(−1)m−1−kζ(2k + 1)

(2m − 2k − 2)! x2m−2k−2 +∞∑

k=m

(−1)kζ(2m − 2k − 1)

(2k)! x2k,

(23)

which is Equation (22) for s = 1, a = 1, b = 0, c = 1, F = ζ , f = {sincos

}α = { 2m

2m−1

}δ = {1

0

}.

In the case of Equation (12), bearing in mind λ(z) = (1 − 2−z)ζ(z), and applying the sameprocedure as above yields

∞∑n=1

sin(2n − 1)x

(2n − 1)2m= (−1)m−1

2(2m − 1)! (ψ(2m) + γ − log x) x2m−1

+m−1∑k=1

(−1)m−1−kλ(2k + 1)

(2m − 2k − 1)! x2m−2k−1 +∞∑

k=m

(−1)kλ(2m − 2k − 1)

(2k + 1)! x2k+1,

∞∑n=1

cos(2n − 1)x

(2n − 1)2m−1= (−1)m−1

2(2m − 2)! (ψ(2m − 1) + γ − log x) x2m−2

+m−1∑k=1

(−1)m−1−kλ(2k + 1)

(2m − 2k − 2)! x2m−2k−2 +∞∑

k=m

(−1)kλ(2m − 2k − 1)

(2k)! x2k,

which is Equation (22) for s = 1, a = 2, b = 1, c = 1/2, F = λ, f = {sincos

}α = { 2m

2m−1

}δ = {1

0

}.

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Remark 2 We note that formulas (23) can be obtained by letting in Equation (2) α tend to 2m iff = sin or to 2m − 1 if f = cos (m ∈ N), since we encounter division by zero. Apart from this,the left-hand side series are known as Clausen functions defined by (see [4])

Cl2ν(x) =∞∑

n=1

sin nx

n2ν, Cl2ν−1(x) =

∞∑n=1

cos nx

n2ν−1, ν ∈ N,

and, as a by-product, we have managed to express each of them as the sum of a polynomial anda power series in terms of Riemann’s ζ function, that is

Cl2ν(x) = (−1)ν−1

(2ν − 1)! (ψ(2ν) + γ − log x) x2ν−1 +ν−1∑k=1

(−1)ν−1−kζ(2k + 1)

(2ν − 2k − 1)! x2ν−2k−1

+∞∑

k=ν

(−1)kζ(2ν − 2k − 1)

(2k + 1)! x2k+1, ν ∈ N,

Cl2ν−1(x) = (−1)ν−1

(2ν − 2)! (ψ(2ν − 1) + γ − log x) x2ν−2 +ν−1∑k=1

(−1)ν−1−kζ(2k + 1)

(2ν − 2k − 2)! x2ν−2k−2

+∞∑

k=ν

(−1)kζ(2ν − 2k − 1)

(2k)! x2k, ν ∈ N.

References

[1] T.M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, New York, 1976.[2] I.S. Gradshteyn and I.M. Ryzhik, Tables of Integrals, Series and Products,A. Jeffrey and D. Zwillinger, eds.,Academic

Press, New York, 2000.[3] K. Knopp, Theory and Application of Infinite Series, Dover, New York, 1990.[4] L. Lewin, Polylogarithms and Associated Functions, North Holland, New York, 1975.[5] F. Oberhettinger, Tables of Bessel Transform, Springer-Verlag, 1972.[6] ———, Tables of Mellin Transform, Springer-Verlag/Academic Press, 1974.[7] A.P. Prudnikov, Y.A. Brychkov, and O.I. Marichev, Integrals and Series, Elementary Functions, Vol. 1 Gordon and

Breach Science Publishers, New York, 1986.


This article was downloaded by:On: 23 January 2010Access details: Access Details: Free AccessPublisher Taylor & FrancisInforma Ltd Registered in England and Wales Registered Number: 1072954 Registered office: Mortimer House, 37-41 Mortimer Street, London W1T 3JH, UK

Integral Transforms and Special FunctionsPublication details, including instructions for authors and subscription information:http://www.informaworld.com/smpp/title~content=t713643686

On trigonometric series over integrals involving Bessel or Struve functionsSlobodan B. Tričković a; Mirjana V. Vidanović b; Miomir S. Stanković b

a Faculty of Civil Engineering, Department of Mathematics, University of Niš, Niš, Serbia b Faculty ofEnvironmental Engineering, Department of Mathematics, University of Niš, Niš, Serbia

To cite this Article Tričković, Slobodan B., Vidanović, Mirjana V. and Stanković, Miomir S.(2009) 'On trigonometric seriesover integrals involving Bessel or Struve functions', Integral Transforms and Special Functions, 20: 11, 821 — 834To link to this Article: DOI: 10.1080/10652460902868062URL: http://dx.doi.org/10.1080/10652460902868062


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http://dx.doi.org/10.1080/10652460902868062


Integral Transforms and Special FunctionsVol. 20, No. 11, November 2009, 821–834

On trigonometric series over integrals involving Besselor Struve functions

Slobodan B. Trickovica*, Mirjana V. Vidanovicb and Miomir S. Stankovicb

aFaculty of Civil Engineering, Department of Mathematics, University of Niš, Aleksandra Medvedeva 14,18000 Niš, Serbia; bFaculty of Environmental Engineering, Department of Mathematics, University of Niš,

Carnojevica 10a, 18000 Niš, Serbia

(Received 1 September 2008 )

To find formulas for the summation of trigonometric series over integrals involving Bessel or Struvefunctions, we rely on trigonometric series involving Bessel or Struve functions, which are in turn obtainedby using summation formulas for series over the product of two trigonometric functions. All these sums areexpressed either as power series in terms of Riemann’s ζ or Catalan’s β function or Dirichlet functions η

and λ, or, in certain cases, they are brought in so called closed form, which means that the infinite seriesare represented by finite sums. Important limiting values cases are considered too.

Keywords: Riemann’s; Catalan’s function; Bessel; Struve and Dirichlet functions

MSC (2000): Primary: 33C10; Secondary: 11M06; 65B10

1. Introduction

In this article, we deal with finding the sum of the series

ID,fα =

∞∑n=1

(s)n−1Dν((an − b)x)

(an − b)αf ((an − b)z), (1)

where a = {12

}b = {

01

}, s = 1 or − 1, α ∈ R

+, f = sin or f = cos, and, Dν(x) denotes anintegral Bν, φ(x) or Sν, φ , defined by

Bν,φ(x) =∫ 1

0Jν(xy)φ(y)dy, Sν,φ(x) =

∫ 1

0Hν(xy)φ(y)dy, (2)


ISSN 1065-2469 print/ISSN 1476-8291 online© 2009 Taylor & FrancisDOI: 10.1080/10652460902868062http://www.informaworld.com

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Table 1. Parameters and convergence regions.

a b s c F for

1 0 1 1 ζ 0 < x < 2π

− 1 0 η − π < x < π

2 1 11

2λ 0 < x < π

− 1 0 β −π

2< x <

π

2

where Jν and Hν are Bessel or Struve functions of the first kind and order ν ∈ R. To obtainthe sum of the series (1) we do not have to previously calculate integrals Dν((an − b)x), whichare not necessarily found elementarily. However, if we are able to do so, the series (1) will takea different form, leading possibly to a new class of summation formulas. At first, we assumethat φ is integrable. Yet, in order to extend the class of summable series, we admit that φ isdifferentiable on (0, 1), but not bounded in the neighbourhood of 0 or 1, or even not integrable on(0, 1), however, such that there exists at least one of the integrals (2). We further require that thefunctions ykφ(y) (k ∈ N) are integrable on (0, 1) as well.

Obtaining the sum of the series (1) relies on the summation of some trigonometric series(see [6]) in terms of Riemann’s ζ or Catalan’s β function or Dirichlet functions η and λ, in theform of a single formula, i.e.

∞∑n=1

(s)n−1f ((an − b)x)

(an − b)α= cπ

2(α)f (πα/2)xα−1 +

∞∑i=0

(−1)iF (α − 2i − δ)

(2i + δ)! x2i+δ, (3)

where α > 0, a = { 12

}b = { 0

1

}, s = 1 or − 1, and f = {sin

cos

}δ = {1

0

}. The values for F and c are in

the Table 1, where ζ is Riemann’s zeta function ζ(z) = ∑∞k=1 k−z, η and λ are Dirichlet functions

η(z) = ∑∞k=1(−1)k−1k−z = (1 − 21−z)ζ(z), λ(z) = ∑∞

k=0(2k + 1)−z = (1 − 2−z)ζ(z) and β isCatalan’s function β(z) = ∑∞

k=0(−1)k(2k + 1)−z.We note that the functions ζ , η, λ are analytic in the whole complex plane except for z = 1,

where they have a pole. The integral representation β(z) = 1/(z)∫ ∞

0 (xz−1ex)/(e2x + 1) dx ofCatalan’s function defines an analytical function for Rz ≥ 1; but, also it satisfies the functionalequation β(z) = (π/2)z−1(1 − z) cos (πz/2) β(1 − z) extending beta to the left side of thecomplex plane Rz < 1.

2. Preliminaries

We place the integral D(x) in Equation (1), and check whether the interchange of summation andintegration

ID,fα =

∞∑n=1

(s)n−1f ((an − b)z)

(an − b)α

∫ 1

0ϕν((an − b)xy) φ(y)dy

=∫ 1

0

( ∞∑n=1

(s)n−1ϕν((an − b)xy)

(an − b)αf ((an − b)z)

)φ(y)dy

(4)



may take place. Here is α > 0, a = { 12

}b = { 0

1

}, s = 1 or − 1, f = sin or cos, and ϕν is a Bessel

Jν or Struve function Hν . In order to justify Equation (4), we prove uniform convergence of theseries

Sϕ,fα =

∞∑n=1

(s)n−1ϕν((an − b)xy)f ((an − b)z)

(an − b)α(5)

with respect to y ∈ (0, 1). For this purpose, we use an integral representation of the Bessel or Struvefunction [1]

ϕν(t) = 2(t/2)ν

(1/2)(ν + 1/2)

∫ π/2

0sin2ν θg(t cos θ) dθ, (6)

where ν > − 1/2, ϕν = {Jν

Hν

}g = {cos

sin

}. After replacing ϕν in Equation (5) with the right-hand

side integral of Equation (6), we shall first prove that we may interchange integration andsummation, i.e.

Sϕ,fα = 2(xy/2)ν√

π(ν + (1/2))

∞∑n=1

(s)n−1f ((an − b)z)

(an − b)α−ν

∫ π/2

0sin2ν θ g((an − b)xy cos θ) dθ

= 2(xy/2)ν√π(ν + (1/2))

∫ π/2

0sin2ν θ

( ∞∑n=1

(s)n−1f ((an − b)z)

(an − b)α−νg((an − b)xy cos θ)

)dθ,

(7)

by showing uniform convergence of the right-hand series with respect to (y, θ ) ∈ (0, 1) × [0, π /2]on the basis of Dirichlet’s test, which says that (see [3]) the series

∑∞n=0 an(y)bn(y) is uniformly

convergent in D, if the partial sums of∑∞

n=0 an(y) are uniformly bounded in D and the sequencebn(y), being monotonic for every fixed y, uniformly converges to 0.

Lemma 1 The series (5) converges uniformly with respect to y on (0, 1).

Proof In order to prove this, we treat the right-hand series of Equation (7) as a function ofy ∈ (0, 1), regarding x, z and θ as variable parameters. By making use of an elementary trigono-metric identity, the product of f and g is represented as a sum of two trigonometric functions sinor cos. Consequently, the series in question is split up into two series of the type

∞∑n=1

(s)n−1τ((an − b)(z ± xy cos θ))

(an − b)α−ν, (8)

where τ = sin or τ = cos. Let us suppose first that a = 1, b = 0. If s = 1, then there holds∣∣∣∣∣n∑

k=1

τ(k(z ± xy cos θ))

∣∣∣∣∣ ≤ 1

sin(z ± xy cos θ)/2≤ 1

sin ε

for 0 < z ± xy cos θ < 2π , because sin (z ± xy cos θ)/2 ≥ sin ε > 0 for each ε > 0 satisfyingε ≤ (z ± xy cos θ)/2 ≤ π − ε. Similarly, if s = − 1, we would have∣∣∣∣∣

n∑k=1

(−1)k−1τ(k(z ± xy cos θ))

∣∣∣∣∣ ≤ 1

cos(z ± xy cos θ)/2≤ 1

sin ε

for −π < z ± xy cos θ < π , because cos (z ± xy cos θ)/2 ≥ sin ε > 0 for each ε > 0 satisfying−π/2 + ε ≤ (z ± xy cos θ)/2 ≤ π/2 − ε.



We now suppose a = 2, b = 1. If s = 1, then

∣∣∣∣∣n∑

k=1

τ((2k − 1)(z ± xy cos θ))

∣∣∣∣∣ ≤ 1

sin(z ± xy cos θ)≤ 1

sin ε

for 0 < z ± xy cos θ < π , because sin(z ± xy cos θ) ≥ sin ε > 0 for each ε > 0 satisfying ε ≤z ± xy cos θ ≤ π − ε. Finally if s = − 1, then

∣∣∣∣∣n∑

k=1

(−1)k−1τ((2k − 1)(z ± xy cos θ))

∣∣∣∣∣ ≤ 1

cos(z ± xy cos θ)≤ 1

sin ε

for −π/2 < z ± xy cos θ < π/2, because cos(z ± xy cos θ) ≥ sin ε > 0 for each ε > 0 satis-fying −π/2 + ε ≤ z ± xy cos θ ≤ π/2 − ε.

On the one hand, all the partial sums are uniformly bounded with respect to(y, θ ) ∈ [0, 1] × [0, π /2], and considering that 0 ≤ y cos θ ≤ 1, we are able to determine bound-aries for x and z, i.e. the convergence regions for the series (1). They are in Table 2. For instance,in the first case, from 0 ≤ y cos θ ≤ 1, we have immediately −|x| ≤ ±xy cos θ ≤ |x|, and tocome to the condition 0 < z ± xy cos θ < 2π , it is necessary to take |x|< z < 2π −|x|; hence,there follows |x|<π . In a similar way the rest of the convergence regions are determined. Notethat the boundaries for x are the same as for xy cos θ , because 0 ≤ y cos θ ≤ 1.

On the other hand, it is obvious that in each case the sequence 1/(an − b)α−ν monotonically tendsto 0 for α >ν, which proves uniform convergence of the right-hand series in Equation (7) withrespect to (y, θ ) ∈ (0, 1) × [0, π /2]. So the interchange of integration and summation in Equation (7)is permitted. There still remains to prove that the left-hand series in Equation (7) uniformlyconverges with respect to y ∈ (0, 1). Namely, relying on Equation (7) and uniform convergence ofthe right-hand series in Equation (7) with respect to (y, θ ) ∈ (0, 1) × [0, π /2], we state that for anarbitrary ε > 0, there exists k0, so that k ≥ k0 implies

∣∣∣∣∣ 2(ν + 1)√π(ν + (1/2))

∞∑n=k+1

(s)n−1f ((an − b)z)

(an − b)α−ν

∫ π/2

0sin2ν θ g((an − b)xy cos θ) dθ

∣∣∣∣∣=

∣∣∣∣∣∫ π/2

0sin2ν θ

(2(ν + 1)√

π(ν + (1/2))

∞∑n=k+1

(s)n−1f ((an − b)z)

(an − b)α−νg((an − b) xy cos θ)

)dθ

∣∣∣∣∣≤

∫ π/2

0sin2ν θ

∣∣∣∣∣ 2(ν + 1)√π(ν + (1/2))

∞∑n=k+1

(s)n−1f ((an − b)z)

(an − b)α−νg((an − b)xy cos θ)

∣∣∣∣∣ dθ < ε.

(9)

Table 2. Parameters and convergence regions.

a b s c F Convergence region

1 0 1 1 ζ K1 = {(x, z) � − π < x < π , |x| < z < 2π − |x|}1 0 − 1 0 η K2 = {(x, z) � − π < x < π , |x| − π < z < π − |x|}

2 1 11

2λ K3 = {(x, z) | −π

2< x <

π

2, |x| < z < π − |x|}

2 1 − 1 0 β K4 = {(x, z) | −π

2< x <

π

2, |x| − π

2< z <

π

2− |x|}



Hence, we conclude that the convergence speed of the first series in Equation (9) does not dependon y, meaning that the left-hand side series in Equation (7) uniformly converges with respect toy ∈ (0, 1), and so does the series (5). �

Lemma 2 Let φ(y) be integrable. Then there holds Equation (4).

Proof Let us denote by

Sϕ,f

α,k =k∑

n=1

(s)k−1ϕν((an − b)xy)f ((an − b)z)

(an − b)α(10)

the kth partial sum of the series (5). We have

∞∑n=1

(s)n−1f ((an − b)z)

(an − b)α

∫ 1

0ϕν((an − b)xy)φ(y) dy

= limk→∞

k∑n=1

(s)n−1f ((an − b)z)

(an − b)α

∫ 1

0ϕν((an − b)xy)φ(y) dy

= limk→∞

∫ 1

0

(k∑

n=1


(an − b)α

)φ(y) dy

=∫ 1

0

( ∞∑n=1


(an − b)α

)φ(y) dy.

The last passage is permitted because both the sequence (10) uniformly converges with respectto y on (0, 1) (Lemma 1) and

∫ 10 φ(y)dy exists. �

Lemma 3 Suppose that, for a differentiable on (0, 1) and unbounded in the neighbourhood of0 or 1 function φ, the integral

∫ 10 φ(y)dy does not converge. Let there exist at least one of the

integrals D((an − b)x) (D is Bφ or Sφ defined by Equation (2)), so that |D((an − b)x)|≤ Mn(x),and for each corresponding x from Table 2, the sequence Mn(x)/(an − b)α , α > 0, monotonicallytends to 0. Then there holds Equation (4).

Proof Because of the assumption that φ is a differentiable function, we know that it is continuouson each closed interval within (0, 1), and, as it is not bounded in the neighbourhood of 0 or 1,without loss of generality, we can consider [δ, 1 − δ], 0 <δ < 1. Continuous function on a closedinterval is bounded, so referring again to Dirichlet’s test, we prove uniform convergence of theseries (5) with respect to y, but this time on [δ, 1 − δ], so that there holds

∞∑n=1

(s)n−1f ((an − b)z)

(an − b)α

∫ 1−δ

δ

ϕν((an − b)xy) φ(y) dy

=∫ 1−δ

δ

( ∞∑n=1

(s)n−1f ((an − b)z)ϕν((an − b)xy))

(an − b)α

)φ(y) dy (α > 0). (11)

We regard the left-hand side series as a function of δ with variable parameters z and x. In view ofthe conditions, by virtue of Dirichlet’s test, the left-hand side series in Equation (11) converges



uniformly with respect to δ on (0, 1). Hence, we have

limδ→0+

∞∑n=1

(s)n−1f ((an − b)z)

(an − b)α

∫ 1−δ

δ

ϕν((an − b)xy) φ(y)) dy

=∞∑

n=1

(s)n−1f ((an − b)z)

(an − b)α

∫ 1

0ϕν((an − b)xy) φ(y)) dy,

meaning that the right-hand side integral in Equation (11) converges, so there holdsEquation (4). �

3. Sum of the series over the product of a Bessel and a trigonometric function

The series involving the product of sine or cosine play a key role in finding the summation formulafor the series (1). That is why we investigated them thoroughly in [5].After representing the productof g and f as the sum of two trigonometric functions sin or cos, and applying Equation (3) to bothof series in each of the particular cases, in [5] we obtained a general formula

T f,gα =

∞∑n=1

(s)n−1g((an − b)xy cos θ) f ((an − b)z)

(an − b)α−ν

= cπ(−1)δ(δ−d)

4(α − ν)h(π(α − ν)/2)((z + xy cos θ)α−ν−1 + (−1)δ(z − xy cos θ)α−ν−1) (12)

+∞∑i=0

(−1)δ(δ−d)+iF (α − ν − 2i − d)

(2i + d)!i∑

j=0

(2i + d

2j + δ

)z2i−2j+d−δ(xy cos θ)2j+δ,

where g = {sincos

}δ = {1

0

}and d =

{0 f =g1 f �=g , h =

{cos f =gsin f �=g . All the other relevant parameters are

in Table 2.When on the right-hand side of Equation (12) appears h = sin and α − ν = 2m or h = cos and

α − ν = 2m − 1, where m ∈ N, one should take limit. However, if α − ν − d = 2m and F = ζ , η, λor α − ν − d = 2m − 1 and F =β (m ∈ N), the sum of the series on the right-hand side ofEquation (12) consists of a finite number of terms because of the vanishing functions ζ , η, λat even negative integers, and the function β at odd negative integers. So, for this choice ofparameters, the formula (12) is brought into so called closed form (see [5])

Tf,g

2m+d+ε =∞∑

n=1

(s)n−1g((an − b)xy cos θ) f ((an − b)z)

(an − b)2m+d+ε

= cπ

2

m∑j=0

(2m + d + ε − 1

2j + δ

)(−1)δ(δ−d)z2m−2j+d−δ+ε−1(xy cos θ)2j+δ

(2m + d + ε − 1)! h(mπ + (d + ε)π/2)(13)

+m∑

i=0

(−1)δ(δ−d)+iF (2m − 2i + ε)

(2i + d)!i∑

j=0

(2i + d

2j + δ

)z2i−2j+d−δ(xy cos θ)2j+δ,

where g = {sincos

}δ = {1

0


{0 f =g1 f �=g , h =

{cos f =gsin f �=g . For F = ζ , η, λ,

there holds ε = 0, but for F =β it is ε = 1. The other relevant parameters are in Table 2. The



formula (13) comprises some particular results from [4], but it is more suitable for immediateobtaining sums of some infinite series as well.

Now, we find the sum (5) by virtue of the summation formula (12), replacing it in Equation (7)and developing the binomials (z ± xy cos θ)α−ν−1 into binomial series. After a rearrangementwe obtain

Sϕ,fα = (xy/2)ν

Gν

⎛⎝ cπ(−1)δ(δ−d)

2(α − ν)h(π(α − ν)/2)

∞∑j=0

(α − ν − 1

2j + δ

)zα−ν−1−2j−δ(xy)2j+δI2ν,2j+δ

+∞∑i=0

(−1)δ(δ−d)+iF (α − ν − 2i − d)

(2i + d)!i∑

j=0

(2i + d

2j + δ

)z2i−2j+d−δ(xy)2j+δI2ν,2j+δ

⎞⎠ ,

(14)

where, for the sake of simplicity, Gν = √π(ν + 1/2) and I2ν,2j+δ = ∫ π/2

0 sin2ν θ cos2j+δ θ dθ .Introducing sin θ = t in the last integral, we have

I2ν,2j+δ = 1

2

∫ 1

0(t2)(2ν+1)/2−1(1 − t2)(2j+δ+1)/2−1 d(t2) = 1

2B

(ν + 1

2, j + δ + 1

2

), (15)

and come to the required summation formula for the series (5)

Sϕ,fα = (−1)δ(δ−d)c

√π(xy/2)ν

2(α − ν)h(π(α − ν)/2)

∞∑j=0

(α − ν − 1

2j + δ

)zα−ν−2j−1−δ(xy)2j+δGj

+ (xy/2)ν√π

∞∑i=0

(−1)δ(δ−d)+iF (α − ν − 2i − d)

(2i + d)!i∑

j=0

(2i + d

2j + δ

)z2i−2j+d−δ(xy)2j+δGj ,

(16)

where α >ν > − 1/2, ϕν = {Jν

Hν

}g = {cos

sin

}δ = {0

1

}; independently of that d =

{0 f =g1 f �=g and

h ={

cos f =gsin f �=g . The other relevant parameters are given in Table 2, and for the sake of brevity, we

have introduced Gj =( j + (δ + 1)/2)/( j + ν + 1 + δ/2).

3.1. Limiting values

We shall now consider some important particular cases of the formula (16). If h = sin andα − ν = 2m or h = cos and α − ν = 2m − 1, m ∈ N division by zero is not defined, and wehave to take limit. After choosing a = 1, b = 0, s = 1, there must be c = 1, F = ζ . Also if we setϕν = Jν, g = cos, δ = 0, f = cos, then d = 0, h = cos, and we have

SJ,cosα =

√π(xy/2)νzα−ν−1

2(α − ν) cos(π(α − ν)/2)

∞∑j=0

(α − ν − 1

2j

)(xy/z)2j(j + 1/2)

(j + ν + 1)

+ (xy/2)ν√π

∞∑i=0

(−z2)iζ(α − ν − 2i)

(2i)!i∑

j=0

(xy/z)2j(2i

2j

)(j + 1/2)

(j + ν + 1).



In order to take limit, we first denote σ =α − ν and replace α with σ + ν, then we find

�2m−1,ν(x, y, z) = limσ→2m−1

⎡⎣ √

π(xy/2)νzσ−1

2(σ) cos(πσ/2)

m−1∑j=0

(σ − 1

2j

)(xy/z)2j(j + 1/2)

(j + ν + 1)

+ (xy/2)ν√π

m−1∑i=0

(−z2)iζ(σ − 2i)

(2i)!i∑

j=0

(xy/z)2j(2i

2j

)(j + 1/2)

(j + ν + 1)

⎤⎦

= (−z2)m−1(xy/2)ν

(2m − 2)!√π

m−1∑k=0

(xy/z)2k(2m−2

2k

)(ψ(2m − 2k − 1)

+γ − log z)(k + 1/2)

(ν + k + 1)

+ (xy/2)ν√π

m−2∑k=0

(−z2)kζ(2m − 2k − 1)

(2k)!k∑

j=0

(xy/z)2j(2k

2j

)(j + 1/2)

(j + ν + 1),

and

R2m−1,ν(x, y, z) = limσ→2m−1

√π(xy/2)νzσ−1

2(σ) cos(πσ/2)

∞∑j=m

(σ − 1

2j

)(xy/z)2j(j + 1/2)

(j + ν + 1)

= − (−z2)m−1(xy/2)ν

(2m − 1)!√π

∞∑j=m

(xy/z)2j(j + 1/2)( 2j

2m−1

)(ν + j + 1)

.

Finally, we obtain

SJ,cos2m−1+ν =

∞∑n=1

Jν(nxy)

n2m−1+νcos nz = �2m−1,ν(x, y, z) + R2m−1,ν(x, y, z)

+ (xy/2)ν√π

∞∑i=m

(−z2)iζ(2m − 1 − 2i)

(2i)!i∑

j=0

(xy/z)2j(2i

2j

)(j + 1/2)

(j + ν + 1),

which holds for (x, z) ∈ K1 (see Table 2 on the page 4).

Example 1 Let m = 3. Then we have

∞∑n=1

Jν(nxy)

n7+νcos nz = (xy/2)ν√

π

(− 49 z6

14400 (ν + 1)− 25 (xy)2 z4

1152 (ν + 2)− 3 (xy)4 z2

128 (ν + 3)

+(

z6

720 (ν + 1)+ (xy)2 z4

96 (ν + 2)+ (xy)4 z2

64 (ν + 3)+ (xy)6

384 (ν + 4)

)log z

+(

z4

24 (ν + 1)+ (xy)2 z2

8 (ν + 2)+ (xy)4

32 (ν + 3)

)ζ(3) −

(z2

2 (ν + 1)

+ (xy)2

4 (ν + 2)

)ζ(5) + ζ(7)

(ν + 1)+ z6

7!∞∑

j=4

(xy/z)2j(j + 1/2)(2j

7

)(ν + j + 1)

+∞∑i=4

(−1)iζ(7 − 2i)

(2i)!i∑

j=0

(xy)2j z2i−2j(2i

2j

)(j + 1/2)

(j + ν + 1)

⎞⎠ .



3.2. Closed form cases

The summation formula (16) takes a closed form in certain cases. Namely, the series on theright-hand side truncates because of the vanishing of F functions, i.e. α − ν − d = 2m if F = ζ , η, λand α − ν − d = 2m − 1 if F =β (m ∈ N), so that we write α = ν + 2m + d − ε, and obtain

∞∑n=1


(an − b)ν+2m+d−ε

= (−1)δ(δ−d)c√

π(xy/2)ν

2(2m + d − ε)h(mπ + π(d − ε)/2)

×m∑

j=0

(2m + d − ε − 1

2j + δ

)z2m+d−ε−2j−δ−1(xy)2j+δGj

+ (xy/2)ν√π

m∑i=0

(−1)δ(δ−d)+iF (2m − 2i − ε)

(2i + d)!

×i∑

j=0

(2i + d

2j + δ

)z2i−2j+d−δ(xy)2j+δGj , (17)

where ϕν = {Jν

Hν

}g = {cos

sin

}δ = {0

1

}, d =

{0 f =g

1 f �=g, h =

{cos f =gsin f �=g , ε = 0 if F = ζ , η, λ and

ε = 1 if F =β. The parameters a, b, s, c, F and convergence regions are read from Table 2.Particular closed form cases found in the literature (see Example 6), can be obtained fromEquation (17).

Example 2 Consider the formula (74.1.19) in [2]∞∑

n=1

Jν(nx)

nν+2cos nx = 1

3(ν + 2)2−ν−3xν[3x2 + (ν + 1)(6x2 − 12πx + 4π2)],

where 0 < x <π , Rν > − 3/2. The same result can be obtained by means of Equation (17) forz = x, y = 1, Rν > − 1/2, taking ϕν = Jν , a = 1, b = 0, s = 1, c = 1, F = ζ , ε = 0, δ = 0, d = 0,f = cos,h = cos, m = 1,α = ν + 2. Similarly, the sums (74.1.20), (74.1.21) and (74.1.22) from [2]are contained in Equation (17).

Example 3 By means of the formula (17) we can find the sums for the series that are not knownin the literature. If we consider only finite sums, there are no results for the series with α − ν ≥ 3,α − ν ∈ N0, whereas Equation (17) contains these cases too. For example, for the choice ofparameters α − ν = 3, ϕ = J, f = sin, a = 2, b = 1, s = 1, from the formula (17) one obtains

∞∑n=1

Jν((2n − 1)x)

(2n − 1)ν+3sin(2n − 1)z = (π − z)π

2ν+3(ν + 1)xνz − π

2ν+4(ν + 2)xν+2,

for ν > − 1/2, where the convergence region is K3.

Example 4 If we now choose α − ν = 4, ϕ = J, f = cos, a = 1, b = 0, s = −1,Equation (17) becomes

∞∑n=1

(−1)n−1Jν(nx)

nν+4cos nz = (7π4 − 30z2π2 + 15z4)xν

45 · 2ν+4(ν + 1)+ (3z2 − π2)xν+2

3 · 2ν+4(ν + 2)+ xν+4

2ν+6(ν + 3),

for ν > − 1/2. The convergence region is K2.



Example 5 If we finally choose parameters: α = 5, ν = 2, ϕ = H, f = sin, a = 2, b = 1,s = − 1, we obtain a sum

∞∑n=1

(−1)n−1H2((2n − 1)x)

(2n − 1)5sin(2n − 1)z = 1

30x3z,

valid in the convergence region K4. Note that this result coming out of the formula (17) cannotbe found in the literature for α − ν = 3.

4. Sum of the series (1)

Now, we make use of Equation (16), which is, as we have shown, the formula for finding sum ofthe series (5), i.e. the right-hand series of Equation (4). Thus we obtain the summation formulaof the left-hand side series in Equation (4), which is actually the summation formula for theseries (1):

∞∑n=1


(an − b)αf ((an − b)z)

= (−1)δ(δ−d)c√

π(x/2)ν

2(α − ν)h(π(α − ν)/2)

∞∑j=0

(α − ν − 1

2j + δ

)zα−ν−2j−1−δx2j+δIν+2j+δ

+ (x/2)ν√π

∞∑i=0

(−1)δ(δ−d)+iF (α − ν − 2i − d)

(2i + d)!i∑

j=0

(2i + d

2j + δ

)z2i−2j+d−δx2j+δIν+2j+δ,

(18)

where Dν = {Bν,φ

Sν,φ

}g = {cos

sin

}δ = {0

1

}, d =

{0 f =g

1 f �=gand h =

{cos f =gsin f �=g , and for the sake of sim-

plicity and brevity, we have denoted Iν+2j+δ = Gj

∫ 10 φ(y)yν+2j+δ dy. The parameters a, b, s, c, F

as well as convergence regions (which we determined earlier) are read from Table 2.

4.1. Limiting value cases

Very important particular cases of the formula (18) ensue if h = sin and α − ν = 2m or h = cosand α − ν = 2m − 1, m ∈ N, when the first term of Equation (18) has zero as a divisor, so we haveto deal with a limiting value. We denote σ =α − ν and replace α with σ + ν in Equation (18).Afterwards, choosing, for instance, a = 2, b = 1, s = 1, c = 1/2, F =λ, ϕν =Hν , g = sin, δ = 1,f = cos, d = 1, h = sin, and finally φ(y) = y−1, we have

ISφ,cosσ+ν =

√π(x/2)ν

4(σ) sin(πσ/2)

∞∑j=0

(σ − 1

2j + 1

)x2j+1zσ−2j−2j !

(ν + 2j + 1)(j + ν + 3/2)

+ (x/2)ν√π

∞∑i=0

(−1)iλ(σ − 2i − 1)

(2i + 1)!i∑

j=0

(2i + 1

2j + 1

)x2j+1z2i−2j j !

(ν + 2j + 1)(j + ν + 3/2),



so that we find

�2m,ν(x, z) = limσ→2m

⎡⎣ √

π(x/2)ν

4(σ) sin(πσ/2)

m−1∑j=0

(σ − 1

2j + 1

)x2j+1zσ−2j−2j !

(ν + 2j + 1)(j + ν + 3/2)

+ (x/2)ν√π

m−1∑i=0

(−1)iλ(σ − 2i − 1)

(2i + 1)!i∑

j=0

(2i + 1

2j + 1

)x2j+1z2i−2j j !

(ν + 2j + 1)(j + ν + 3/2)

⎤⎦

= (x/2)ν√π

[(−1)m

2

m−1∑i=0

(log(z/2) − ψ(2m − 2i − 1) − γ )x2i+1z2m−2i−2i!(ν + 2i + 1)(2m − 2i − 2)!(2i + 1)!(ν + i + 3/2)

+m−2∑i=0

(−1)iλ(2m − 2i − 1)

(2i + 1)!i∑

j=0

(2i + 1

2j + 1

)x2j+1z2i−2j j !

(ν + 2j + 1)(ν + j + 3/2)

⎤⎦,

and

R2m,ν(x, z) = limσ→2m

√π(x/2)ν

4(σ) sin(πσ/2)

∞∑j=m

(σ − 1

2j + 1

)x2j+1zσ−2j−2j !

(ν + 2j + 1)(j + ν + 3/2)

= (−1)m(x/2)ν

2(2m)!√π

∞∑j=m

x2j+1z2m−2j−2j !(2j+12m

)(ν + 2j + 1)(ν + j + 3/2)

.

Finally, the series involving the product of a Struve integral and cosine is as follows:

ISφ,cos2m+ν =

∞∑n=1

cos(2n − 1)z

(2n − 1)2m+ν

∫ 1

0

Hν((2n − 1)xy)

ydy = �2m,ν(x, z) + R2m,ν(x, z)

+ (x/2)ν√π

∞∑i=m

(−1)iλ(2m − 2i − 1)

(2i + 1)!i∑

j=0

(2i + 1

2j + 1

)x2j+1z2i−2j j !

(ν + 2j + 1)(ν + j + 3/2)

where (x, z) ∈ K3 (see Table 2 on the page 4).

Example 6 If we choose m = 2, then we obtain

∞∑n=1

cos(2n − 1)z

(2n − 1)4+ν

∫ 1

0

Hν((2n − 1)xy)

ydy

= (x/2)ν√π

[x((log(z/2) − (3/2))z2 + (7/2)ζ(3))

4(ν + 1)(ν + 3/2)

+ 1

48

∞∑j=2

x2j+1z2−2j j !(2j+14

)(ν + 2j + 1)(ν + j + 3/2)

+ x3 log(z/2)

12(ν + 3)(ν + 5/2)

+∞∑i=2

(−1)iλ(3 − 2i)

(2i + 1)!i∑

k=0

(2i + 1

2k + 1

)x2k+1z2i−2kk!

(ν + 2k + 1)(k + ν + 3/2)

].

Similarly, we can obtain the other particular cases, without having previously to calculate theintegral involved.



4.2. Closed form cases

The infinite series (18) is brought in closed form if F = ζ , η, λ and α − ν − d = 2m or F =β andα − ν − d = 2m − 1 (m ∈ N), so that we write α = ν + 2m + d − ε, and have

∞∑n=1


(an − b)ν+2m+d−εf ((an − b)z)

= (−1)δ(δ−d)c√

π(x/2)ν

2(2m + d − ε)h(mπ + π(d − ε)/2)

×m∑

j=0

(2m + d − ε − 1

2j + δ

)z2m+d−ε−2j−1−δx2j+δIν+2j+δ

+ (x/2)ν√π

m∑i=0

(−1)δ(δ−d)+iF (2m − 2i − ε)

(2i + d)!

×i∑

j=0

(2i + d

2j + δ

)z2i−2j+d−δx2j+δIν+2j+δ, (19)

where Dν = {Bν,φ

Sν,φ

}g = {cos

sin

}δ = {0

1

}, d =

{0 f =g

1 f �=gand h =

{cos f =gsin f �=g , ε = 0 if F = ζ , η, λ and

ε = 1 if F =β. The parameters a, b, s, c, F and convergence regions are read from Table 2.

Example 7 First we take a = 1, b = 0, s = 1 in Equation (19), there follows c = 1 and F = ζ

(see Table 2), then ε = 0. For D2 = S2, φ , there must be g = sin, which means δ = 1. If we takef = cos, then h = sin and d = 1 because f �= g. We choose φ(y) = ctg y. Let m = 1. The functionctg y is unbounded in the neighbourhood of 0 and the integral

∫ 10 ctg y dy does not converge, so

we cannot apply Lemma 1. However, for 0 < |x|<π , limy→0+ ctg yH2(nxy) = 0, where n ∈ N,the function ctg yH2(nxy) is integrable with respect to y ∈ (0, 1), and we have∣∣∣∣x

∫ 1

0ctg yH2(nxy) dy

∣∣∣∣ =∣∣∣∣∫ x

0ctg

t

xH2(nt) dt

∣∣∣∣ �∫ x

0

∣∣∣∣ctgt

xH2(nt)

∣∣∣∣ dt

�∫ π

0

∣∣∣∣ctgt

πH2(nt)

∣∣∣∣ dt.

Also, |tg (t/π)| > |t |/π implies |ctg (t/π)| < π/|t |, for |t|<π . Additionally, we find

|H2(nt)| = 2n|t |3π

∣∣∣∣1 − 3πJ1(nt)

2nt+ 3πJ2(nt)

n2t2

∣∣∣∣ <2n|t |3π

(1 + 3π

8

)

So, ∣∣∣∣ctgt

πH2(nt)

∣∣∣∣ =∣∣∣∣ctg

t

π

∣∣∣∣ · |H2(nt)| <π

|t | · 2n|t |3π

(1 + 3π

8

)= 2n

3

(1 + 3π

8

),

and there follows∫ π

0

∣∣∣∣ctgt

πH2(nt)

∣∣∣∣ dt <2n

3

(1 + 3π

8

) ∫ π

0dt = 2nπ

3

(1 + 3π

8

),

which means that∣∣∣∫ 1

0 ctg yH2(nxy) dy

∣∣∣ < nπ(8 + 3π)/12|x| = Mn(x), and we can easily see

that for each x, 0 < |x|<π , the sequence Mn(x)/n3 → 0 monotonically, so that Lemma 3 may be



applied. Making use of Equation (13), we obtain

∞∑n=1

cos nz

n5

∫ 1

0ctg yH2(nxy) dy

= x3

1156√

π((8π2 − 24πz + 12z2) log(2 sin 1) + (12π2 − 36πz + 18z2)(Cl2(2) + Cl3(2))

− (6π2 − 18πz + 9z2)Cl4(2)) + x5

4608√

π

(log(2 sin 1) + 10Cl2(2) + 20Cl3(2)

− 30(Cl4(2) + Cl5(2)) + 15Cl6(2)),

where |x|<π and |x|< z < 2π −|x| (see Table 2). On the right-hand side are Clausen functionsdefined by [1]

Cl2ν(x) =∞∑

n=1

sin nx

n2ν, Cl2ν−1(x) =

∞∑n=1

cos nx

n2ν−1, ν ∈ N.

Example 8 Let a = 1, b = 0, s = − 1 in Equation (19), implying c = 0 and F =η, ε = 0. ForD1/2 = B1/2, φ there must be g = cos and δ = 0. Further, we take f = cos implying h = cos, d = 0because f = g. Let m = 2. We choose φ(y) = (1 − y2)−1/2. It is unbounded about 1, but integrableon (0, 1), thus satisfying conditions of Lemma 2. So applying Equation (19), we find

∞∑n=1

(−1)n−1 cos nz

n9/2

∫ 1

0

J1/2(nxy)√1 − y2

dy

= π√

xπ

42(1/4)

(7π4

90− π2z2

3+ z4

6− π2x2

20+ 3x2z2

20+ 7x4

720

),

where |x|<π and |x|−π < z <π −|x| (see Table 2).

We have already said that in order to find the sum of the series (1), it is not necessary to calculatethe integrals (2). Besides, it does not have to be done elementarily.Yet, if we calculate the integralin Example 8, the above series takes a different form, giving rise to the following formula

∞∑n=1

(−1)n−1 cos nz

n9/2J 2

1/4

(nx

2

)=

√xπ

22(1/4)

(7π4

90− π2z2

3+ z4

6− π2x2

20+ 3x2z2

20+ 7x4

720

),

whereby we obtain the sum of a new series. Generally speaking, for ν > − 1, there holds∫ 1

0

Jν(nxy)√1 − y2

dy = π

2J 2

ν/2

(nx

2

),

so for this type of integrals there exists a whole class of new closed form formulas.

Example 9 Further, we take a = 2, b = 1, s = 1 in Equation (19). In Table 2, we read c = 1/2 andF =λ. So ε = 0. If we choose D1/3 = S1/3, φ , f = sin, then we have g = cos, δ = 0, d = 1, h = sin.Let m = 1 and φ(y) = log y, which is unbounded in the neighbourhood of 0, but integrable on(0, 1), so Lemma 2 holds. Applying Equation (19), we obtain

ISφ,sin1 =

∞∑n=1

sin(2n − 1)z

(2n − 1)10/3

∫ 1

0log yH1/3((2n − 1)xy) dy = 27 3

√x(1/6)

640 3√

2

(zπ + z2 + 12x2

275

),

where |x|<π /2 and |x|< z <π −|x| (see Table 2).



Example 10 Finally, let in Equation (19) be a = 2, b = 1, s = − 1, implying c = 0, F =β, ε = 1.If we choose D1 = B1, φ , there follows g = cos and δ = 0. If f = sin, then d = 1 and h = sinbecause f �= g. Further, let φ(y) = yμ, μ> − 2, and m = 1. The integral

∫ 10 yμ dy does not nec-

essarily converge for μ> − 2. However, making use of Equation (6), where we substitute u forcos t , we come, for |x|<π /2, to the following estimate

yμ|J1((2n − 1)xy)| � 2|x|π

(2n − 1)yμ+1

∣∣∣∣ sin((2n − 1)xy)

(2n − 1)xy

∣∣∣∣ < (2n − 1)yμ+1,

whereupon we find∣∣∣∣∫ 1

0J1((2n − 1)xy)yμ dy

∣∣∣∣ �∫ 1

0|J1((2n − 1)xy)|yμ dy < (2n − 1)

∫ 1

0yμ+1 dy = 2n − 1

μ + 2,

with Mn(x) = (2n − 1)/(μ+ 2), so Mn(x)/(2n − 1)3 monotonically tends to zero when n increasesto infinity, which means that Lemma 3 holds, and by applying Equation (19), we obtain

IBφ,sin3 =

∞∑n=1

(−1)n−1 sin(2n − 1)z

(2n − 1)3

∫ 1

0J1((2n − 1)xy) yμdy = πxz

8(μ + 2),

|x|<π /2 and |x|−π /2 < z <π /2 −|x| (see Table 2).

Acknowledgements

This article was supported by the Ministry of Science of the Republic of Serbia.

References

[1] M. Abramowitz and A. Stegun, Handbook of Mathematical Functions, with Formulas, Graphs and MathematicalTables, Dover Publications, New York, 1972.

[2] E.R. Hansen, A Table of Series and Products, Prentice-Hall Inc, Englewood Cliffs, N.J., 1975.[3] K. Knopp, Theory and Application of Infinite Series, Dover Publications, New York, 1990.[4] A.P. Prudnikov, Y.A. Brychkov, and O.I. Marichev, Integrals and Series, Vol. 1: Elementary Functions, Gordon and

Breach Science Publishers, New York, 1986.[5] S.B. Trickovic, M.V. Vidanovic, and M.S. Stankovic, Series involving the product of a trigonometric integral and a

trigonometric function, Integral Transforms Spec. Func. 18(10) (2007), pp. 751–763.[6] S.B. Trickovic, M.V. Vidanovic, and M.S. Stankovic, On the summation of trigonometric series, Integral Transforms

Spec. Func. 19(6) (2008), pp. 441–452.


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