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?

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you mean without telling us how a depends on s? –  Carlo Beenakker Mar 22 '13 at 15:08
sorry...just fixed it –  Remy Mar 22 '13 at 15:11
What does $ii$ mean? And is $j^2=-1$? –  Daniel Loughran Mar 22 '13 at 17:00
sorry ..fixed it again ... it is hard to type using a tablet ... ii is an integer, I updated the equation –  Remy Mar 22 '13 at 19:11
Why not use $\Gamma[3+i-s]=(2+i-s)\Gamma[2+i-s]$ to cancel a Gamma function in the numerator and denominator? –  Stopple Mar 22 '13 at 22:45