In asymptotic analysis and analytic number theory, one often has to deal with complex integrals over infinite contours in the complex plane, and the required techniques to do so often go beyond the standard courses of complex analysis in one variable. In particular, I am interested in the following type of argument which I don´t fully understand and which I will try to illustrate by an example (taken from Paris and Kaminski, "Asymptotics and Mellin-Barnes Integrals"):

Consider the so called Cahen-Mellin integral $$e^{-z}=\frac{1}{2\pi i}\int_{(c)}\Gamma(s)z^{-s}ds,\ \arg(z)<\frac{\pi}{2}, z\neq 0,$$ which is an integral representation of the exponential function by taking the inverse Mellin transform of the Gamma function, where the integration contour $(c)$ stands for the vertical line $\{\Re(s)=c\}$ with some $c>0$. It can be shown that the integrand has "the controlling behavior" $|z|^{-\sigma}O(|t|^{\sigma-{1\over 2}}e^{t\arg(z)-{1\over 2}\pi|t|})$ as $|t|\to\infty$, where $s=\sigma + it$. Now, aside from the obvious way to show the validity of the above integral representation, Paris and Kaminski argue that because of the aforementioned exponential decay of the integrand we are allowed to move the contour of integration over the poles of the Gamma function and use the residues of the latter to obtain the exponential series. This is precisely the argument I want to understand, so I will try to break it down into few smaller questions:

(Q1) How does the exponential decay of the integrand allow us to displace the contour of intagration over singularities? Is there a more general setup where the asymptotic behavior of the integrand allows for moving the contour of integration through and over singularities?

(Q2) After the displacement of the integration contour (still a vertical line), what kind of a residue theorem allows for considering all of the infinitely many singularities of the gamma function?

The version of the residue theorem I know uses bounded interior of a (simply) closed contour in the complex plane and only finitely many residues contained in there.

Remark 1: In order to compute the above integral in a classical way, I would take a finite line segments of the vertical line, symmetric with respect to the positive real axis, i.e. $\{Re(s)=c, -r_n\leq\Im(s)\leq r_n\}$, construct circle segments with radii $r_n$ encompassing each of the poles, with $r_n\uparrow \infty$ suitably chosen such that no poles lie on the segment contours, and then show that the integrals over the half-circles tend to zero as $n\to\infty$, thus obtaining on the one hand the integral over the infinite vertical line and on the other hand the infinite sum of the residues. However, I haven´t really checked whether the exponential decay of the integrand would suffice for the half-circle integrals to vanish in the limit.

Remark 2: I could imagine that there might be a version of the residue theorem suitably formulated for the Riemann sphere resp. $\bar{\mathbb{C}}:=\mathbb{C}\cup\{\infty\}$, where basically infinite contours from the complex plane correspond to closed ones on the sphere.

(Q3) Where could one find a more systematic treatment of integrals over infinite contours in the complex plane, including contour shifting over poles, other types of contour modifications, usage of infinitely many residues, as well as other techniques for the exact computation of such integrals? I understand that such techniques are often to be applied "individually", thus such literature would ideally contain a few good examples.

Thanks in advance for your attention and sorry if I appear to sound too confused :-), I am only trying to fill in certain "gaps" in my knowledge of complex analysis.

PS: I am not interested in numerical computations or general asymptotic expansions for contour integrals (even though in the above example the residues "expansion" appears as a special case thereof).

  • $\begingroup$ I remember the Bromwich contour integral in the theory of Laplace transforms as an example of the kind of integral you are mentioning. $\endgroup$ Feb 13, 2011 at 21:26
  • $\begingroup$ Actually, both are closely related as Mellin and Laplace transform are related :-) $\endgroup$
    – M.G.
    Feb 13, 2011 at 21:33

1 Answer 1


The passage in Paris and Kaminski's "Asymptotics and Mellin-Barnes Integrals" is available here. It seems to me that the formulation "because of this exponential decay, we are free to displace the contour of integration [...] over the poles [...] to produce the exponential series" is not intended to mean that by displacing the contour over the poles we don't change the value of the integral, but rather that because of the exponential decay, we can displace the contour of integration and the value of the integral will change only due to the poles, and we can produce the exponential series by adding up the residues since the value of the remaining integral can be shown to go to zero as we move it to (negative) infinity. So perhaps there is nothing as mysterious here as you suspected? (To justify displacing the contour of integration, we have to integrate around infinite rectangular strips between two parallels of the imaginary axis. Because of the exponential decay, the pieces at infinity don't contribute, so the difference between the integrals along the two parallels of the imaginary axis must be given by the residues of the poles that lie within the strip.)

  • $\begingroup$ Danke for your answer! That explains everything. It seems I have misinterpreted or assumed too much out of the context. And btw, the rectangular solution is of course much cooler than my circular segment idea as it becomes right away evident that the integral vanishes on the two parallels of the imaginary axes due to the exponential decay. In the end it turned out to be the standard complex analysis stuff :-) $\endgroup$
    – M.G.
    Feb 14, 2011 at 13:37

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