The Mellin transform

The Mellin transform is an integral transform named after the finnish math-ematician Hjalmar Mellin (1854-1933). Its most remarkable property is theexistence of a direct mapping between asymptotic expansions of a functionnear zero and infinity and the set of singularities of the transform in thecomplex plane. While developing the theory, we will also see some proper-ties of the most famous function defined via the Mellin transform: the EulerGamma function.

A natural way to introduce this integral transform is via its connectionwith the older and more famous Fourier transform. Let f : R C be anintegrable function, its Fourier transform is:

F(f)(y) =Rf(x)e2piixydx.

The idea is to shift the Fourier transform via the topological isomor-phism (R,+) (R, ) induced by the exponential map. Let us first considerf(x) := (f exp)(x), and then compute its Fourier transform via the changeof variables u = ex and s = 2piiy:

F(f)(y) =Rf(ex)e2piixydx =

R+f(u)us1du. (1)

Looking to the right hand side of the last equation we recognize that thereis no need to restrict to s R. Indeed, once f is integrable, the existence ofthe integral only depends on f decreasing fast enough at 0 and (wherefast enough is intended in the sense of the upcoming Observation 1). Thismotivates the following definition.

Definition 1 (Mellin Transform). Let f be an integrable1 complex valuedfunction defined over the positive real numbers. Then its Mellin transform,when it makes sense, is defined by:

M(f)(s) :=R+f(t)ts1dt, s C.

1Locally Lebesgue integrable would be enough.

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Example 1. The Euler Gamma function is defined as:

(s) :=M(et)(s) = 0

etts1dt.

The careful reader will note that Euler died (1783) quite a long timebefore Mellin was born (1854). Indeed Euler defined its Gamma functiononly for real values, and its motivation was to interpolate the factorial onnatural numbers. Therefore we suggest as an exercise of integration by partsto prove that (z + 1) = z(z) for any z C.

We can now give a first rough result about the existence of the Mellintransform.

Observation 1 (Fundamental strip). Let f(t) be a continuous function onR+ such that:

f(t) =

{O(t) for t0,O(t) for t.

Then M(f)(s) exists for any s C such that Re(s) lies in the funda-mental strip (,).Proof.

|M(f)(s)| 10

|f(t)|tRe(s)1dt+ 1

|f(t)|tRe(s)1dt

C 10

tRe(s)1+dt+ C 1

tRe(s)1+dt.

Where the first integral is convergent for Re(s) > and the second one forRe(s) < .

We notice that the observation above implies that a Mellin transform notonly exists in its fundamental strip but also that it is holomorphic there. Theapplication of thii observation to the formula defining Gamma tells us thatit is well defined and holomorphic in {s C | Re(s) > 0.

We can go further with the correspondence between the two consideredintegral transforms, we use now the formula for the inverse of the Fouriertransform to provide an heuristic argument for the formula of the inverse of

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the Mellin transform. The problem is to find, given (s) complex valued andgood enough in a sense to be specified later, the function f : R+ C suchthat (s) =M(f)(s).

The inverse formula for the Fourier transform is:

F1(g)(x) =Rg(y)e2piixydy.

Using it with f(x) = f(ex) = F1(g)(x):

f(ex) =

RF(f)(y)e2piixydy.

Applying the usual change of variables u = ex, s = 2piiy and consideringthe relation between Fourier and Mellin transform as in equation (1),

f(u) =

yRM(f)(s)usdy.

Finally, taking c (a, b) where (a, b) is the fundamental strip2 of (s),we get an inversion formula for the Mellin transform:

M1()(x) = 12pii

c+ici

(s)xsds.

This justifies the formal statement, which goes under the name of Mellininversion formula.

Theorem 1 (Mellin inversion formula). Let (s) be holomorphic in the stripa < Re(s) < b such that it tends to zero uniformly for increasing Im(s) for

any c (a, b). Moreover assume that c+ici (s)ds is absolutely convergent

for any s lying in such a strip. Then, if

f(x) =1

2pii

c+ici

(s)xsds,

we have that

(s) =M(f)(s) =R+f(t)ts1dt.

2Here we are abusing of the term fundamental strip. We can do that because,heuristically, if f(x) = M1()(x) then (s) = M(f)(s) and its fundamental strip isgiven by the asymptotics of f on 0 and .

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Conversely suppose f(x) to be continuous on R+, and assume that the inte-gral:

(s) =

R+f(t)ts1ds

is absolutely convergent for a < Re(s) < b. Then:

f(x) =M1(M(f))(x).We make a list of some elementary properties of the Mellin transform that

describe its behavior corresponding to some basic analytic manipulations.

Proposition 1. For the s C for which M(f)(s) is well defined:1. M(thf(t))(s) =M(f(t))(s+ h), h C.2. M(f(t))(s) = sM(f(t))(s), R+.3. M(iI if(it))(s) =M(f(t))(s) (iI isi ), I finite.4. M(f(th))(s) = 1|h|M(f(t))( sh), h R.

5. M(f(t)log(t))(s) = ddsM(f(t))(s).

6. M( ddtf(t))(s) = (s 1)M(f(t))(s 1).

7. M(t ddtf(t))(s) = sM(f(t))(s).

8. M( t0f(x)dx)(s) = 1

sM(f(t))(s+ 1).

Proof. Properties 1., 2., 3., 4. and 5. are straightforward, recalling for 5. thatddxtx = txlog(t). While 6., 7. and 8. follow integrating by parts.

Finally we come to the relation between asymptotic expansions of f(t)at 0 and and singularities ofM(f)(s) on C. This relation is traditionallystated in two results. In the direct mapping Theorem we deduce resultson the singularities of M(f)(s) from the asymptotics of f(t), in the reversemapping Theorem we do the opposite. Since the nature of the results is quitetechnical we state only the former one. The latter is, for example, Theorem9.2 of [?].

Theorem 2 (Direct mapping Theorem). Let f(t) be continuous with Mellintransform M(f)(s) defined in the strip (a, b).

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1. Assume that f(t) admits the following asymptotic expansion for t0:

f(t) =ki=1

cini!tailog(t)ni +O0(t

M).

Where for any i = 1, ..., k N: ci C, ni N and M < ai awith M Z. Then M(f)(s) has a meromorphic continuation to thestrip (M, b). Furthermore, in a neighborhood of s = aj, it can beexpanded to:

M(f)(s) =ki=1ai=aj

((1)nici

(s+ aj)ni+1+Oai(1)

).3

2. Now assume that f(t) admits the asymptotic expansion as before fort (with differents k,ci,ni,ai and M). Then M(f)(s) has a mero-morphic expansion in the strip (a,M). And in a neighborhood ofs = aj it can be expanded to:

M(f)(s) = ki=1ai=aj

((1)nici

(s+ aj)ni+1+Oai(1)

).

Proof. Since the proof of the second statement is the same of the first oneexcept for minor differences, we will prove only the first statement. Let usintroduce

g(t) := f(t)ki=1

cini!tailog(t)ni ,

which is O(tM) by assumption. Then:

M(f)(s) = 10

g(t)ts1dt+ 10

ki=1

cini!tailog(t)nits1dt+

1

f(t)ts1dt.

Since g(t) is O(tM) for t0 the first integral is defined for s > M . Thethird integral is well defined because s < b and M(f)(s) is defined in the

3We remark that the sum is taken over all the i = 1, ..., k such that ai = aj .

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strip (a, b). In specific in the neighborhoods of the singularities ai they canbe regarded as constant functions.

Observe that, for any 1 < r R and n N, integration by parts gives: 10

trlog(t)ndt =(1)nn!

(r + 1)n+1.

Both the sides of the equation are analytic functions in the half plane{z C | Re(z) > 1}. The left hand side is not defined outside of this halfplane, but the right hand side is, and it is a well defined meromorphic functionon the whole complex plane. Then this equation gives us the required analyticcontinuation if we impose it to be valid for any r C.

So the analytic continuation of the second integral can be explicitly com-puted, and the result is:

ki=1

(1)nici(s+ ai)ni+1

.

Since everytime ai 6= aj the term (1)nici(s+ai)ni+1 can be regarded as a constantin a neighborhood of s = aj, the meromorphic expansion is proved.Exercise 1. Apply this Theorem to the Gamma function to deduce that it(or properly its analytic continuation) has simple poles at any non positive

integer. And the correspondent residues are Res(,n) = (1)nn!

.

Exercise 2. Copy the proof of the theorem to find the functions {fn(t)}nNsuch that (s) =M(fn)(s) in (n,n+ 1).Exercise 3. Classify the poles of the Digamma function: 0(s) :=

dds

log((s)).

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