# Coutour Integral of Gamma Functions

How do I solve the Integral $$\frac{1}{2\pi j} \oint \frac{b^{ - s} \Gamma[2 + i - s] \Gamma[s] \Gamma[-1 - i + s]}{ (2 + i - s) \Gamma[3 + i - s]} \:\mathrm{d}s$$

This integral is an inverse Mellin transform. Therefore, the contour extends from $l+j\infty$ to $l-j\infty$, where $l\in\mathbb{R}$.

$j=\sqrt{-1}$

$b \in\mathbb{R},\quad b>0$

$i\in\mathbb{R}, \quad i\ge 0$

$\Gamma(.)$ is the gamma function. Does it make any difference when $i$ becomes an integer?

• Why not use $\Gamma[3+i-s]=(2+i-s)\Gamma[2+i-s]$ to cancel a Gamma function in the numerator and denominator? Mar 22 '13 at 22:45
• I agree with you . I can substitute $$\frac{1}{2+i-s} = \frac{\Gamma(2+i-s)}{\Gamma(3+i-s)}$$, but would this result in a Meijer G-function? because two of the poles will be coinciding and have a difference of 1.
– Remy
Mar 23 '13 at 16:36