Question

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?

Answer 1

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