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Can we get a closed form for the following contour integral?. Let us assume that n is a non-negative integer,

$\frac{1}{2\pi i}\int^{c+i\infty}_{c-i\infty}\frac{\Gamma(n-s)\Gamma(s)\Gamma(k-s)}{\Gamma(1+n-s)}{}_1F_1(1+n-s,n+1,\frac{\alpha}{2b})\, b^s\,\mathrm{d}s$

and also how to choose the value of c in order solve the contour integral.

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    $\begingroup$ Can you tell us where this came from? This has lots of structure, close to things I know, but just a little off. Surely this contour integral did not appear out of thin air, it must have been obtained in the process of 'solving' something else - tell us that. It might be easier to see the structure in that problem than in the above expression. $\endgroup$
    – Jacques Carette
    Nov 29, 2011 at 0:52
  • $\begingroup$ Thanks for your answer. I got this as part of applying Mellin transform convolution to solve integrals of the form $\int^\infty_0 \exp\left(-\frac{\alpha}{x}\right)\exp(-b x)(x-1)^{\frac{n-1}{2}} I_{n-1}\left(\sqrt{\alpha(x-1)}\right) \mathrm{d}x $ where $\alpha$ and $b$ are non-negative, and non-zero real numbers. $\endgroup$
    – Remy
    Nov 30, 2011 at 6:32

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Remmy,

Here's some additional information:

I've recently encountered a very similar integral in my own work, with a ratio of gamma functions and a hypergeometric function and a Bessel function. The orders of both the hypergeometric function and the Bessel function depended on $s$. I was still able to produce a series solution consisting of terms containing the 1F1 and Bessel functions evaluated at the pole locations, since neither function had any poles or branch points in the $s$ plane.

I believe the same is true in your case, since it is the numerator parameter that depends upon $s$, and the 1F1 has no poles for this parameter. (Look at, for example, the functions.wolfram.com site for info on 1F1.)

Sorry for the overly introductory material in my earlier post, I was not sure of your familiarity with this method! You obviously know it well..

Hope this helps some,

Tom

Did this integral come from using Mellin transforms + Parseval to perform an integration? A standard thing to do now is to look at the large $t$ asymptotics of the integrand (where $s = \sigma + i t$) to determine if the contour can be moved across the poles of the gamma functions in the numerator, and in which direction. The constant $c$ would lie in the joint strip of analyticity of the Mellin transforms.

You then close the contour in the appropriate half plane and use the residue theorem to give either an asymptotic expansion or power series in the variable $b$. Sometimes these can be identified with known functions. This process all depends on the details of your parameters. Your situation is more complex because you have the 1F1(). Look up "Mellin-Barnes integrals" and you should find helpful information.

Good luck,

Tom

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  • $\begingroup$ @Tom: yes, this is how the integral came from, and am still trying to get the suitable contour, and yes the problem lies on the 1F1(.), which was initially a generalized Laguerre polynomial with the parameter s appearing in the order. If the parameter is not showing in the order of the Laguerre polynomial, then this can easily be solved into a finite sum that contains binomial terms of (s), and then the integral will be solved as a sum of MeijerG functions. Unfortunately this is not applicable in this case. $\endgroup$
    – Remy
    Nov 30, 2011 at 6:39
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    $\begingroup$ Remmy, I added some comments above my original answer. You can still get a series expansion for the integral by just picking up the poles of the gamma function(s) and evaluating the residues, whcih will include the 1F1 evaluated at the pole locations as well. I've gotten a useful series for a similar integral I needed in some recent work in this way. Depending on your parameters, the 1F1 may simplify to a polynomial at the pole locations. The 1F1 does not have any poles in the $s$ plane. $\endgroup$
    – Tom Dickens
    Dec 1, 2011 at 17:14
  • $\begingroup$ @Tom: Thanks Tom, I appreciate your help. This is very helpful. $\endgroup$
    – Remy
    Dec 1, 2011 at 18:33

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