Prove $\int_0^{\infty} \int_0^{\infty} \frac{\sqrt{xy} ~dxdy}{(x+y)(1+x y)^s}=\frac{\pi}{2(s-1)}$ for $s>1$ By evaluation with WolframAlpha for different values of $s$ it is apparent that:

$$I(s)=\int_0^{\infty} \int_0^{\infty} \frac{\sqrt{xy} ~dxdy}{(x+y)(1+x y)^s}=\frac{\pi}{2(s-1)},~~~~~s>1$$

I'm not really familiar with this kind of integrals. Can we introduce new coordinates like:
$$u=xy,~~~~~v=x+y$$
But then the expressions for $x(u,v)$ and $y(u,v)$ become conditional on $x>y$ or $x<y$, so I'm not sure how to transform the integral correctly.
We can also try something like this to build a recurrence:
$$I(s)=\int_0^{\infty} \int_0^{\infty} \frac{xy ~dxdy}{\sqrt{xy}(x+y)(1+x y)^s}= \\ =\int_0^{\infty} \frac{dxdy}{\sqrt{xy}(x+y)(1+x y)^{s-1}}-\int_0^{\infty} \frac{dxdy}{\sqrt{xy}(x+y)(1+x y)^{s}}$$
But this doesn't seem to help either.
 A: If you let $u=xy$ and $v=y/x$ you get
$$
\int_0^{+\infty}\frac{1}{(1+u)^s}\,du\int_0^{+\infty}\frac{1}{2\sqrt{v}(1+v)}\,dv
$$
which is easy to integrate.
A: By symmetry the integral over $y\geq x$ equals the integral over $y\leq x$, hence:
$$ I(s) = 2\int_{0}^{+\infty}\int_{0}^{x}\frac{\sqrt{xy}}{(x+y)(1+xy)^s}\,dy\,dx= 4\int_{0}^{+\infty}\int_{0}^{1}\frac{x z^2}{(1+z^2)(1+x^2 z^2)^s}\,dz\,dx $$
through the substitution $y=x z^2$. The substitution $x=\frac{t}{z}$ and Fubini's theorem then lead to:
$$ I(s) = 4\int_{0}^{+\infty}\int_{0}^{1}\frac{t}{(1+z^2)(1+t^2)^s}\,dz\,dt =\pi\int_{0}^{+\infty}\frac{t\,dt}{(1+t^2)^s}=\color{red}{\frac{\pi}{2(s-1)}}$$
as wanted.
A: An alternative approach is to transform coordinates to polar coordinates.  Then, we have
$$\begin{align}
I(s)&=\int_0^{\pi/2} \frac{\sqrt{\sin(\phi)\cos(\phi)}}{\sin(\phi)+\cos(\phi)}\left(\int_0^\infty \frac{r}{\left(1+r^2\sin(\phi)\cos(\phi)\right)^s}\,dr\right)\,d\phi\\\\
&=\frac{1}{2(s-1)}\int_0^{\pi/2} \frac{1}{\sqrt{\sin(\phi)\cos(\phi)}(\sin(\phi)+\cos(\phi))}\,d\phi\\\\
&=\frac{1}{2(s-1)}\int_0^{\pi/2} \frac{1}{\sqrt{\cos(2(\phi-\pi/4))}\,\cos(\phi-\pi/4)} \,d\phi\\\\
&=\frac{1}{2(s-1)}\,2\,\int_0^{\pi/4} \frac{1}{\sqrt{\cos(2\phi)}\,\cos(\phi)} \,d\phi\\\\
&=\frac{1}{(s-1)}\,\left.\left(\arctan\left(\frac{\sin(\phi)}{\sqrt{\cos(2\phi)}}\right)\right)\right|_{0}^{\pi/4}\\\\
&=\frac{\pi}{2(s-1)}
\end{align}$$
as expected!
