Integral $\int\limits_0^\infty \prod\limits_{k=0}^\infty\frac{1+\frac{x^2}{(b+1+k)^2}}{1+\frac{x^2}{(a+k)^2}} \ dx$ Does anybody know how to prove this identity?
$$\int_0^\infty \prod_{k=0}^\infty\frac{1+\frac{x^2}{(b+1+k)^2}}{1+\frac{x^2}{(a+k)^2}} \ dx=\frac{\sqrt{\pi}}{2}\frac{\Gamma \left(a+\frac{1}{2}\right)\Gamma(b+1)\Gamma \left(b-a+\frac{1}{2}\right)}{\Gamma(a)\Gamma \left(b+\frac{1}{2}\right)\Gamma(b-a+1)}$$
I found this on Wikipedia.

To conclude this we actually do the simplification in terms of the $\Gamma$ function. We have
$$\prod_{k=0}^{b-a-1}\frac{1}{2a+2k+1} = 
\frac{1}{2^{b-a}} \frac{\Gamma(a + 1/2)}{\Gamma(b+1/2)}.$$
Furthermore $$\prod_{k=0}^{b-a} (a+k) = \frac{\Gamma(b+1)}{\Gamma(a)}.$$
Finally, $$ \frac{1}{2^{b-a}} {2(b-a) - 1 \choose b-a} =
\frac{1}{2^{b-a}} \frac{\Gamma(2b-2a)}{\Gamma(b-a+1)\Gamma(b-a)}.$$
Putting these all together, we have
$$I(a, b) = \frac{\pi}{4^{b-a}} \frac{\Gamma(a + 1/2)}{\Gamma(b+1/2)} 
\frac{\Gamma(b+1)}{\Gamma(a)} \frac{\Gamma(2b-2a)}{\Gamma(b-a+1)\Gamma(b-a)}$$
Now by the duplication formula,
$$ \Gamma(2b-2a) = \frac{2^{2(b-a)-1}}{\sqrt\pi} \Gamma(b-a) \Gamma(b-a+1/2)$$
so that finally
$$I(a, b) = \frac{\pi}{4^{b-a}} \frac{\Gamma(a + 1/2)}{\Gamma(b+1/2)} 
\frac{\Gamma(b+1)}{\Gamma(a)} \frac{\Gamma(b-a+1/2)}{\Gamma(b-a+1)}
\frac{2^{2(b-a)-1}}{\sqrt\pi} \\ 
=  \frac{\sqrt\pi}{2} \frac{\Gamma(a + 1/2)}{\Gamma(b+1/2)} 
\frac{\Gamma(b+1)}{\Gamma(a)} \frac{\Gamma(b-a+1/2)}{\Gamma(b-a+1)}.$$
We have verfied the formula for $a$ and $b$ positive integers with $b>a.$ It should now follow by a continuity argument that we can extend it to real $a,b$ with $b>a>1.$
