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I have been studying a definite integral that I found out to be a particular (and possibly one of the simplest) case(s) of the arcane Mellin-Barnes integral. Solving this problem would lead to a breakthrough in my current research.

For any complex $z$, $e^{-z}$ can be represented by a line integral taken over a contour $C$,

$e^{-z}=\frac{1}{2\pi i}\int_{C}z^{-x}\Gamma(x)dx$

as long as the curve $C$ is chosen conveniently depending on the argument $z$. An almost complete discussion of this type of integral can be found in Paris and Kaminski's "Asymptotics and Mellin-Barnes Integrals" (pp. 89-90, Cambridge University Press, 2001).

For certain arguments, the curve $C$ is trivial. For example, if $z$ is a positive real number, then $C$ is a vertical line in the complex plane, and the integral can be expressed simply as,

$e^{-z}= \frac{1}{2\pi}\int_{-\infty}^{+\infty}\frac{\Gamma(1+i x)}{z^{1+i x}}dx$

The big challenge I am having with this type of integral though is that I am not able to figure out what the contour $C$ should be for an arbitrary non-real complex number (let's say $z=i$, or $z=1+i$), and the reference I posted doesn't have any examples. Ideally I'd like to be able to come up with an explicit parametrization of $C$ so that I can turn the Mellin-Barnes integral into a regular Riemann integral for any complex parameter (or at least some). I'm not sure if that is even possible, as evidenced by the lack of mention to such results in the aforementioned reference (or in the lengthy video tutorials that I watched on the Wolfram website).

Edit: I think I need to make the solution I'm looking for more explicit. As in the definition of line integral, and assuming $(-\infty,+\infty)$ as the domain of the curve, I'm looking for $h(x)$ such that,

$\int_{C}f(x)dx=\int_{-\infty}^{+\infty}f(h(x))*h'(x)dx$

PS Feel free to point out any inaccuracies in this description of Mellin-Barnes integrals, I am by no means an expert.

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    $\begingroup$ You probably know this, but the formula $e^{m}=\frac{1}{2\pi i}\int_{C}(-m)^{-x}\Gamma(x)dx$ is usually written with $-z$ instead of $m$, and represents the statement that $e^{-z}$ is the inverse Mellin transform of the gamma function. Wikipedia calls this the Cahen-Mellin integral. The way it's written there suggests that the formula is still valid for general $z$ with $C$ being a vertical line (but that wouldn't make sense for $z=0$, so clearly the Wikipedia article is a bit sloppy with its phrasing). $\endgroup$
    – Dan Romik
    Feb 24 at 8:25
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    $\begingroup$ You can try to check if the usual proof of this formula in the case of $z$ a positive real (or $m$ a negative real in your notation) with $C$ a vertical line works in the more general case. As I recall, that proof works by shifting the contour to the left in steps of 1 unit successively, each time skipping over one of the singularities of the integrand (which are due to the poles of the gamma function). The change in the value of the integral is given by the residue of the integrand at the pole, and you get the successive terms in the Taylor series for $e^{-z}$. $\endgroup$
    – Dan Romik
    Feb 24 at 8:33
  • $\begingroup$ ... What you'll need to check is that after $n$ such shifts, the shifted integral decays to $0$ as a function of $n$. It was a long time ago that I looked at Paris and Kaminski's book, but I think they give asymptotic formulas for the behavior of $\Gamma(s)$ in all the different asympotic regimes that you might need to check something like that. $\endgroup$
    – Dan Romik
    Feb 24 at 8:37
  • $\begingroup$ @DanRomik Your suggestion to change the notation was good, I did it both here and in my paper. $\endgroup$
    – ThomasJr
    Feb 26 at 3:53

1 Answer 1

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Might be helpful to accentuate the poles by expressing the M-B contour integral in more standard notation as

$$ e^{-p \cdot x} = \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} \frac{(p x)^{-s}}{(-s)!}ds$$

for $x>0$ and $\sigma > 0$.

This is the inverse modified Mellin transform for the modified Mellin transform

$$ \int_{0}^{\infty}e^{-p x} \frac{x^{s-1}}{(s-1)!} dx = p^{-s}$$

for $Re(p) > 0$ and $Re(s) > 0$.

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  • $\begingroup$ That formula you gave it not suitable for my problem (especially having the parameter $s$ showing in a gamma). $\endgroup$
    – ThomasJr
    Feb 24 at 19:44
  • $\begingroup$ The Euler reflection formula is $(-s)!s!= \frac{\pi s}{\sin(\pi s)} = \Gamma(s+1)\Gamma(-s+1)$, so $\Gamma(s) = \frac{\pi }{\sin(\pi s)} \frac{1}{(-s)!}$. Changes of variables are no problem--change $x $ to $t$ in my formulas and $s$ to $\sigma + ix$. Then let $t=\sigma =1$ and $p=-m$ to obtain your formula. $\endgroup$ Feb 24 at 20:49
  • $\begingroup$ That is not the issue. At a first glance the formula you gave is not the solution to the problem I posted. Think of a parametrization of the contour $C$. When replaced in the MB integral it should not change the position of $m$ in the integral and thus defeat the purpose of using that integral representation of the exponential function in the first place. To make it even simpler, just look at the example I gave for when $m$ is a negative number. For the general case, $m$ should not slip into a function (it doesn't help that you changed the notation I used in the enunciation of the problem). $\endgroup$
    – ThomasJr
    Feb 24 at 21:55
  • $\begingroup$ I've just edited the original post to specify more clearly what I'm looking for. $\endgroup$
    – ThomasJr
    Feb 25 at 2:48

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