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Can we get a closed form for the following contour integral?. Let us assume that n is a non-negative integer,

$\frac{1}{2\pi i}\int^{c+i\infty}_{c-i\infty}\frac{\Gamma(n-s)\Gamma(s)\Gamma(k-s)}{\Gamma(1+n-s)}{}_1F_1(1+n-s,n+1,\frac{\alpha}{2b})\, b^s\,\mathrm{d}s$

and also how to choose the value of c in order solve the contour integral.

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1 Answer 1

Clearly the confluent hypergeometric is independent on $s$, thus: $$ \begin{eqnarray} && \frac{1}{2\pi i}\int^{c+i\infty}_{c-i\infty}\frac{\Gamma(s+n)\Gamma(1-s+n)\Gamma(s+1)}{(-s+n)\Gamma(s)\Gamma(1+s-n)}{}_1F_1(n-1,n+1,\frac{c}{2b})\,\mathrm{d}s = \\ &&{}_1F_1(n-1,n+1,\frac{c}{2b}) \cdot \frac{1}{2\pi i}\int^{c+i\infty}_{c-i\infty}\frac{\Gamma(s+n)\Gamma(1-s+n)\Gamma(s+1)}{(-s+n)\Gamma(s)\Gamma(1+s-n)}\,\mathrm{d}s \end{eqnarray} $$

The integral multiplying the hypergeometric function is known as Barnes integral. Its value will depend on the value of $c$ relative to the patter of poles of integrand. Let $$ G_n(s) = \frac{\Gamma(s+n)\Gamma(1-s+n)\Gamma(s+1)}{(-s+n)\Gamma(s)\Gamma(1+s-n)} = \frac{\Gamma(s+n)\Gamma(-s+n)}{\Gamma(1+s-n)} \cdot \frac{\Gamma(s+1)}{\Gamma(s)} = \frac{\Gamma(s+n)\Gamma(-s+n)}{\Gamma(1+s-n)} \cdot s $$ I will further assume $n$ to be a non-negative integer, and $c$ to be a real number. Let's list zeros and poles of $G(s)$ explicitly, with $k \in \mathbb{Z}^+$: $$ p_k^{(1)} = -n - k \qquad p_k^{(2)} = n + k \qquad z^{(1)} = 0 \qquad z_k^{(2)} = n-1-k $$ Consider $n=0$, first. Then $G_0(s) = \Gamma(-s)$. Then if $c > 0$, the Barnes integral can be evaluated via residue theorem (you have to see that complete the integration contour into a closed loop with arc extending from $c+i\infty$ to $c-i \infty$ to the left is permitted): $$ \forall c > 0, \frac{1}{2 \pi i} \int_{c-i \infty}^{c + i \infty} \Gamma(-s) \mathrm{d} s = -\sum_{k=0}^\infty \operatorname{Res}_{s=k} \Gamma(-s) = \sum_{k=0}^\infty \frac{(-1)^k}{k!} = \frac{1}{\mathrm{e}} $$ If $n>0$, notice that for every pole $p_k^{(1)}$ there exists a zero $z^{(2)}_m$ that cancels it. So, again, assuming $c > -n$ we have: $$ \frac{1}{2 \pi i} \int_{c-i \infty}^{c + i \infty} G_n(s) \mathrm{d} s = -\sum_{k=0}^\infty \operatorname{Res}_{s=n+k} G_n(s) = \sum_{k=0}^\infty (-1)^{k} (k+n) \frac{(k+2n-1)!}{k!^2} $$ The sum is easily evaluated in terms of hypergeometric function: $$ \sum_{k=0}^\infty (-1)^{k} (k+n) \frac{(k+2n-1)!}{k!^2} = (2 n)! \left(\, _1F_1(2 n+1;1;-1)-\frac{1}{2} n \, _1F_1(2 n;1;-1)\right) $$

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Thanks Sasha for your reply, I have just updated the integral. Now the function is dependent on s. –  Remy Nov 28 '11 at 16:18
    
I think now it looks more like an inverse mellin transform... –  Remy Nov 28 '11 at 16:24
    
Thanks for the explanation of the poles and zeros, I always struggle with residues... –  Remy Nov 28 '11 at 16:34

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