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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.

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  • $\begingroup$ Is this a Mellin transform? $\endgroup$ – cactus314 Jun 11 '16 at 15:56
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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.

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  • 1
    $\begingroup$ Your first line was most helpful! $\endgroup$ – Yuriy S Jun 11 '16 at 14:34
  • $\begingroup$ @YuriyS: symmetry tricks are often very powerful :) $\endgroup$ – Jack D'Aurizio Jun 11 '16 at 14:35
  • $\begingroup$ Jack, apology for the "incorrect" edit. I've reversed it to the original. And +1 $\endgroup$ – Mark Viola Jun 11 '16 at 15:23
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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.

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  • $\begingroup$ This is even better. Should've tried this substitution. Thank you! $\endgroup$ – Yuriy S Jun 11 '16 at 14:48
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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!

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  • $\begingroup$ Nice job @Dr.MV. Maybe later you can help me on a double integral. $\endgroup$ – gymbvghjkgkjkhgfkl Jun 12 '16 at 1:40
  • $\begingroup$ @Chinacat Thank you! And I'd be happy to help if I can. $\endgroup$ – Mark Viola Jun 12 '16 at 2:25

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