TITLE: Double Integral $\int\limits_0^1\!\!\int\limits_0^1\frac{(xy)^s}{\sqrt{-\log(xy)}}\,dx\,dy$
QUESTION [12 upvotes]: Is it possible to get a closed form of the following integral?
$$I=\int_0^1\!\!\!\int_0^1\frac{(xy)^s}{\sqrt{-\log(xy)}}\,dx\,dy\quad\quad\quad(s>0).$$
My attempt:  I’ve tried a change of variables from cartesian coordinates to polar coordinates:
$$\begin{align}x&=r \cos (\theta)\\y&=r \sin (\theta).\end{align}
$$ I’ve computed the jacobian:$$
J=\left|\frac{D(x,y)}{D(r,\theta)}\right|=|\cos (\theta) r \cos (\theta)-(-r \sin (\theta)) \sin (\theta)|=r.$$ From here I'm stuck.

REPLY [7 votes]: $\sqrt{-\log(\cdot)}$ strikes me as the inverse function of a Gaussian function. 
So it may be worth to set $x=e^{-u^2}$ and $y=e^{-v^2}$ to get:
$$ I =\iint_{(0,1)^2}\frac{(xy)^s}{\sqrt{-\log(xy)}}\,dx\,dy = \iint_{(0,+\infty)^2}\frac{4uv\,e^{-(s+1)(u^2+v^2)}}{\sqrt{u^2+v^2}}\,du\,dv $$
and that looks good. If we switch to polar coordinates we have:
$$ I = \left(\int_{0}^{\pi/2}4\sin\theta\cos\theta\,d\theta\right)\cdot\left(\int_{0}^{+\infty}\rho^2 e^{-(s+1)\rho^2}d\rho\right)$$
hence:

$$ I = \frac{\sqrt{\pi}}{2}\cdot (s+1)^{-3/2}.$$