Let $a \notin \mathbb{Z}$ and $a \neq \frac{1}{2}$. Prove that
$$\sum_{n=0}^{\infty} \frac{2}{\Gamma \left ( a + n \right ) \Gamma \left ( a – n \right )} = \frac{2^{2a-2}}{\Gamma \left ( 2a – 1 \right )} + \frac{1}{\Gamma^2 (a)}$$
Attempt
Using the fact that
\begin{align*}
\frac{1}{\Gamma\left ( a+x \right ) \Gamma \left ( \beta – x \right )} &= \frac{1}{\left ( a+x-1 \right )! \left ( \beta-x-1 \right )!} \\ &=\frac{1}{\Gamma \left ( a + \beta – 1 \right )} \frac{\left ( a + \beta-2 \right )!}{\left ( a + x -1 \right )! \left ( \beta – x -1 \right )!} \\ &=\frac{1}{\Gamma \left ( a + \beta – 1 \right )} \binom{a + \beta – 2}{a + x -1}
\end{align*}
the question really boils down to the sum
$$\mathcal{S} = \sum_{n=0}^{\infty} \binom{2a-2}{a+n-1}$$
To this end,
\begin{align*}
\sum_{n=0}^{\infty} \binom{2a-1}{a+n-1} &=\frac{1}{2\pi i} \sum_{n=0}^{\infty} \oint \limits_{|z|=1} \frac{\left ( 1+z \right )^{2a-1}}{z^{a+n}}\, \mathrm{d}z \\
&= \frac{1}{2\pi i} \oint \limits_{|z|=1} \frac{\left ( 1 + z \right )^{2a-1}}{z^a} \sum_{n=0}^{\infty} \frac{1}{z^n} \, \mathrm{d}z \\
&= \frac{1}{2\pi i} \oint \limits_{|z|=1} \frac{\left ( 1+z \right )^{2a-1}}{z^{a-1} \left ( z-1 \right )} \, \mathrm{d}z
\end{align*}
using the handy identity $\displaystyle \binom{n}{k} = \frac{1}{2\pi i } \oint \limits_{\gamma} \frac{\left ( 1+z \right )^n}{z^{k+1}} \, \mathrm{d}z$. I think I'm on the right track, but I'm having a difficult time evaluating the last contour integral. Any help?
Best Answer
If $a\notin\mathbb{Z}$ then the sum converges if and only if $\color{red}{a>1/2}$, which is obtained from $$\lim_{x\to+\infty}\frac{x^\lambda\Gamma(x)}{\Gamma(\lambda+x)}=1\implies\lim_{n\to\infty}\frac{(-1)^n\color{red}{n^{2a-1}}}{\Gamma(a+n)\Gamma(a-n)}=\frac{\sin a\pi}{\pi}$$ using the reflection formula for $\Gamma(a-n)$.
In this case, a possible solution is to apply the Poisson summation formula $$\sum_{n\in\mathbb{Z}}f(n)=\sum_{n\in\mathbb{Z}}\hat{f}(n),\qquad\hat{f}(y):=\int_{-\infty}^\infty f(x)e^{-2i\pi xy}\,dx$$ to $f(x)=\cos^{2(a-1)}\pi x$ for $|x|<1/2$ (and $f(x)=0$ elsewhere); then we get $$\hat{f}(y)=\frac{2^{2-2a}\Gamma(2a-1)}{\Gamma(a+y)\Gamma(a-y)}$$ as a "reciprocal beta" integral (see DLMF or e.g. this question), and the result follows.