Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Given the following contour integral

$$\frac{1}{2\pi j}\int^{c+j\infty}_{c-j\infty} \frac{\Gamma(-1+a+s)\Gamma(b+s)}{\Gamma(3+a-s)}\cos(-1+a+s)\, {}_2F_1\Big(-1-a+s,-1+a+s;\frac{1}{2};z\Big) y^s\: \mathrm{d}s ,$$

where $a,b,z,y \in \mathbb{R}$, as noticed there are two poles given by

$$P^{(1)}_k = 1-a-k \quad P^{(2)}_k=-b-k \quad\text{ where }\quad k=0,1,2,\dotsc,\infty.$$

The questions are:

  1. What is the suitable contour to sum the residues and solve the integral?
  2. Does the zero caused by $\Gamma(3+a-s)$ cancel any of the poles?
share|improve this question
add comment

1 Answer 1

up vote 2 down vote accepted

1) I imagine this integral came from a Mellin transform approach to compute another integral. You would come up with an appropriate 'c' lying in the overlap region of definition of your two original Mellin transforms (e. g. one might be defined for $c > 1/2$ and the other for $0 < c < 1$, in which case you would choose $1/2 < c < 1$.

Then you need to estimate the behavior of your integrand for large $\Im(s) $, to determine which direction you could move the integration contour and pick up poles to develop a series for your integral.

2) Yes, the poles of the $\Gamma$ function in the denominator can cancel poles in the numerator, you just need to determine those $s$ where that will occur. This all depends on the relative values of $a$ and $b$. Also, the $\cos$ function could cancel poles.

I hope this is not too vague to be of help!

Regards, Tom

share|improve this answer
    
yes, @Tom, you are definitely right. This has come as a result from applying the mellin transform method for integral evaluation. I see....the cos function will cancel poles. I get what you mean. As I understood from what you said, I need to remove the removable poles (those cancelled by the zeros) and then based on what is left I have to choose either the right or the left part of the plane for the contour. –  Remy Mar 1 '12 at 17:40
    
You will need to remove the poles, and you will also need to determine which direction(s) you can push the contour. This involves using asymptotic estimates for the functions in your integrand (e. g. Stirling's formula) to derive the large $t$ behavior of the integrand, where $s =\sigma + i t$. See the book Asymptotics and Mellin-Barnes Integrals. For example, if you find that the behavior is $e^{-\pi |t|}$, you can move the contour either direction since the integrand will decay to zero for any value of $\sigma$, but if you find $|t|^{\sigma-3}$ you would not be able to move past $\sigma=3$. –  Tom Dickens Mar 1 '12 at 18:45
    
Thanks Tom, this is really helpful. –  Remy Mar 2 '12 at 11:11
    
@Remy, I've noticed your series of questions about integrals and series, and it seems like they've been related to a common goal. How is your progress? Reading between the lines, I would guess that you are working on a problem I would find interesting! –  Tom Dickens Feb 9 at 4:21
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.