A Gamma summation

Let $a \notin \mathbb{Z}$ and $a > \frac{1}{2}$ . Prove that

$\sum_{n=0}^{\infty} \frac{2}{\Gamma \left ( a + n \right ) \Gamma \left ( a - n \right )} = \frac{2^{2a-2}}{\Gamma \left ( 2a - 1 \right )} + \frac{1}{\Gamma^2 (a)}$

Solution

Consider the function

$f(x) = \left\{\begin{matrix} \cos^{2a-2} \pi x & , & \left | x \right |< \frac{1}{2} \\ 0 & , & \text{elsewhere} \end{matrix}\right.$

We easily evaluate that

$\hat{f}(y)= \int_{-\infty}^{\infty} f(x) e^{-2i \pi x y} \, \mathrm{d}x = \frac{2^{2-2a} \Gamma(2a-1)}{\Gamma(a+y) \Gamma(a-y)}$

Hence it follows from Poisson summation formula that

$\begin{align*} 1 &= \sum_{n=-\infty}^{\infty} f(n) \\ &= \sum_{n=-\infty}^{\infty} \hat{f}(n) \\ &= \sum_{n=-\infty}^{\infty} \frac{2^{2-2a} \Gamma(2a-1)}{\Gamma(a+n) \Gamma(a-n)} \end{align*}$

Thus,

$\begin{align*} \frac{1}{2^{2-2a} \Gamma(2a-1)} &= \sum_{n=-\infty}^{\infty} \frac{1}{\Gamma(a+n) \Gamma(a-n)} \\ &=\sum_{n=-\infty}^{-1} \frac{1}{\Gamma(a+n) \Gamma(a-n)} + \frac{1}{\Gamma^2(\alpha)} + \\ &\quad \quad \quad + \sum_{n=1}^{\infty} \frac{1}{\Gamma(a+n) \Gamma(a-n)} \\ &= 2 \sum_{n=1}^{\infty} \frac{1}{\Gamma(a+n) \Gamma(a-n)} + \frac{1}{\Gamma^2(a)} \\ &= 2 \sum_{n=0}^{\infty}\frac{1}{\Gamma(a+n) \Gamma(a-n)} - \frac{1}{\Gamma^2(a)} \end{align*}$