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.

Responding to the observation/comment by Random Variable from Thu Aug 8, what is missing above to turn this into a rigorous proof is an evaluation of the quantity
$$R(a,b) = \sum_{m=a}^b
\operatorname{Res}\left(\prod_{k=a}^b \frac{1}{k^2+x^2};x=im\right).$$
We put
$$g_1(x) = \prod_{k=a}^b \frac{1}{k+ix} 
\quad \text{and} \quad
g_2(x) = \prod_{k=a}^b \frac{1}{k-ix}.$$ 
The partial fraction decompositions of $g_1(x)$ and $g_2(x)$ are
$$ g_1(x) = \frac{1}{(b-a)!} 
\sum_{k=a}^b (-1)^{k+1-a} {b-a \choose k-a} \frac{i}{x-ki}$$
and
$$ g_2(x) = \frac{1}{(b-a)!} 
\sum_{k=a}^b (-1)^{k-a} {b-a \choose k-a} \frac{i}{x+ki}.$$
This implies that
$$ R(a,b) = \sum_{m=a}^b \frac{i}{(b-a)!} (-1)^{m+1-a} 
{b-a \choose m-a}
\left(\frac{1}{(b-a)!} 
\sum_{k=a}^b (-1)^{k-a} {b-a \choose k-a} \frac{1}{m+k} \right)$$
which is
$$ \frac{i}{((b-a)!)^2}
\sum_{m=a}^b \sum_{k=a}^b  (-1)^{m+1+k-2a} {b-a \choose m-a}
{b-a \choose k-a} \frac{1}{m+k}.$$
Now observe that
$$ \sum_{m=a}^b \sum_{k=a}^b  
(-1)^{m+1+k-2a} {b-a \choose m-a} {b-a \choose k-a} x^{m+k-1}
\\ = - x^{2a-1}
\sum_{m=a}^b {b-a \choose m-a} (-1)^{m-a} x^{m-a} 
\sum_{k=a}^b {b-a \choose k-a} (-1)^{k-a} x^{k-a} \\
= - x^{2a-1} (1-x)^{2(b-a)}.$$
Integrating we find that
$$ R(a,b) = - i \frac{1}{((b-a)!)^2}
\operatorname{B}(2a, 2(b-a)+1).$$
Returning to $I(a,b)$ from the other post, we get that
$$I(a,b) = \frac{1}{2} \frac{\Gamma(b+1)^2}{\Gamma(a)^2}
\times 2\pi i \times
\left(- i \frac{1}{((b-a)!)^2}
\operatorname{B}(2a, 2(b-a)+1) \right) \\=
\pi \frac{\Gamma(b+1)^2}{\Gamma(a)^2}
\frac{1}{((b-a)!)^2} \operatorname{B}(2a, 2(b-a)+1).$$
Switching to gamma functions, this becomes
$$ I(a,b) = \pi \frac{\Gamma(b+1)^2}{\Gamma(a)^2}
\frac{1}{\Gamma(b-a+1)^2} \frac{\Gamma(2a)\Gamma(2(b-a)+1)}{\Gamma(2b+1)}.$$
To conclude we apply the duplication formula several times, getting
$$ \pi \frac{\Gamma(b+1)^2}{\Gamma(a)^2 \Gamma(b-a+1)^2}
\frac{\frac{2^{2a-1}}{\sqrt{\pi}} \Gamma(a) \Gamma(a+1/2)
\frac{2^{2b-2a}}{\sqrt{\pi}} \Gamma(b-a+1/2) \Gamma(b-a+1)}
{\frac{2^{2b}}{\sqrt{\pi}} \Gamma(b+1/2) \Gamma(b+1)},$$
which is
$$  \sqrt{\pi} 2^{2a-1+2b-2a-2b} 
\frac{\Gamma(b+1)}{\Gamma(a) \Gamma(b-a+1)}\Gamma(a+1/2)
\frac{ \Gamma(b-a+1/2)}{ \Gamma(b+1/2)},$$
which is indeed
$$ \frac{\sqrt{\pi}}{2}
\frac{\Gamma(b+1)\Gamma(a+1/2)\Gamma(b-a+1/2)}
{\Gamma(a) \Gamma(b-a+1) \Gamma(b+1/2)},$$
as claimed.
