How to prove that $\sum_{k=0}^\infty \binom{x}{x-k}\cdot\binom{x}{k-x} = 1$? How to prove this:
$$\sum_{k=0}^\infty \binom{x}{x-k}\cdot\binom{x}{k-x} = 1$$
For all $x\in\mathbb R_{\ge0}$ and with $\binom{x}{r}=\frac{\Gamma(x+1)}{\Gamma(r+1)\cdot\Gamma(x-r+1)}$
It is obviously true for all $x\in \mathbb N_0$, because then, all the values for $k≠x$ become $0$ and the one for $k=x$ becomes $1$. But after a few random calculations with Wolframalpha, I think it should hold for all positive real $x$.
 A: For $x>0$, ${2x\choose k} = (-1)^k{k-2x-1\choose k}\sim \frac{(-1)^k}{k^{2x+1}\Gamma(-2x)}$, $k\to\infty$. Then the power series for $(1+z)^{2x}$ converges uniformly for $|z|\le 1$ by Weierstrass test; in particular, it can be integrated termwise. Therefore,
$$
\sum_{k=0}^\infty \binom{x}{x-k}\cdot\binom{x}{k-x} = \sum_{k=0}^\infty \frac{\Gamma(x+1)^2}{\color{blue}{\Gamma(x-k+1)\Gamma(k-x+1)}\color{green}{\Gamma(2x-k+1)\Gamma(k+1)}} \\
= \frac{\Gamma(x+1)^2}{\color{green}{\Gamma(2x+1)}\color{blue}{\pi} }\sum_{k=0}^\infty\frac{\color{blue}{\sin \pi(k-x)}}{\color{blue}{k-x}}\color{green}{{2x \choose k}} = \frac{\color{red}{\Gamma(x+1)}^2}{\color{red}{\Gamma(2x+1)}\pi }\sum_{k=0}^\infty \mathrm{Re}\int_0^\pi e^{iy(k-x)}{2x \choose k}dy  \\
= \frac{\Gamma(x+1)2^{-2x}}{\Gamma(x+\frac12)\sqrt{\pi} }\mathrm{Re}\int_0^\pi e^{-i x y} \sum_{k=0}^\infty e^{iky}{2x \choose k}dy =  \frac{\Gamma(x+1)2^{-2x}}{\Gamma(x+\frac12)\sqrt{\pi} }\mathrm{Re}\int_0^\pi e^{-i x y} (1+e^{iy})^{2x}dy \\
= \frac{\Gamma(x+1)2^{-2x}}{\Gamma(x+\frac12)\sqrt{\pi} }\mathrm{Re}\int_0^\pi \big(e^{-iy/2}(1+e^{iy})\big)^{2x}dy = \frac{\Gamma(x+1)2^{-2x}}{\Gamma(x+\frac12)\sqrt{\pi} }\int_0^\pi  \left(2\cos \frac{y}2\right)^{2x}dy\\
= \frac{2\Gamma(x+1)}{\Gamma(x+\frac12)\sqrt{\pi} }\int_0^{\pi/2}  \left(\cos z\right)^{2x}dz  = \frac{2\Gamma(x+1)}{\Gamma(x+\frac12)\sqrt{\pi} }\frac{\sqrt{\pi}\Gamma(x+\frac12)}{2\Gamma(x+1)} = 1.
$$
