Concept of Residue Cancellation

I am trying to understand how to apply the residue theorem to solve

$\frac{1}{2\pi j}\int^{\gamma+j\infty}_{\gamma-j\infty}\Gamma(n-s)\Gamma(s)\Gamma(1-s) {}_1F_1(s;b;c) \left(\frac{b}{c}\right)^s\,\mathrm{d}s$

where $n\in\mathbb{N}$, $\,n\ge0$, $\,(b,c)\in\mathbb{R}$ and $\,(b,c)>0$. I understand that I need to sum the residues in points at which the above expression is not analytic. I also understood that the function ${}_1F_1$ is analytic over the s-plane, and hence I need to look for points for the Gamma functions.

Is the residue from $\Gamma(1-s)$ cancelling the residue from $\Gamma(s)$ and vise versa?

how to get the right residues for this combination?

Is any ${}_pF_q$ analtyic over the s plane?

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Your mixing of semicolons and commas in your hypergeometric functions is confusing. The definition I'm accustomed to is $${}_1 F_1(a;b;z)=\sum_{k=0}^\infty \frac{(a)_k}{(b)_k}\frac{z^k}{k!}$$ Note the delimiters. What definition are you using? –  Ｊ. Ｍ. Dec 5 '11 at 9:12
@J.M. : Thanks for pointing this out. I fixed the hypergeometric function. The $\gamma$ within the integration limits is a real. –  Remy Dec 5 '11 at 9:34
I don't think you have. Why is there still a comma? –  Ｊ. Ｍ. Dec 5 '11 at 9:36
@J.M.: Just fixed... –  Remy Dec 5 '11 at 9:40