Double Integral $\int\limits_0^1\!\!\int\limits_0^1\frac{(xy)^s}{\sqrt{-\log(xy)}}\,dx\,dy$ 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.
 A: Our aim is to show that

$$
\int_0^1\int_0^1\frac{(xy)^s}{\sqrt{-\log xy}}\,dx\,dy
=2\int_0^1\int_y^1\frac{(xy)^s}{\sqrt{-\log xy}}\,dx\,dy
=\frac{\sqrt{\pi}}{2(1+s)^{3/2}}.
$$

We will need the error function,
$$
\def\erf{\,\text{erf}\,}
\erf(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2}\,dt.
$$
Let us start to calculate a primitive with respect to $x$. With
$$
u=\sqrt{1+s}\sqrt{-\log(xy)}
$$
we find that
$$
\int\frac{(xy)^s}{\sqrt{-\log xy}}\,dx=-\frac{\sqrt{\pi}}{y\sqrt{1+s}}\erf\bigl(\sqrt{1+s}\sqrt{-\log(xy)}\bigr).
$$
Inserting limits, we find that
$$
\int_y^1\frac{(xy)^s}{\sqrt{-\log xy}}\,dx
=-\frac{\sqrt{\pi}}{y\sqrt{1+s}}\erf\bigl(\sqrt{1+s}\sqrt{-\log(y)}\bigr)
+\frac{\sqrt{\pi}}{y\sqrt{1+s}}\erf\bigl(\sqrt{1+s}\sqrt{-2\log(y)}\bigr)
$$
Next, we want to integrate with respect to $y$. The integrals are very similar. We find that ($u=\sqrt{-a\log y}$ and integrating by parts twice)
$$
\begin{split}
\int \frac{1}{y}\erf\bigl(\sqrt{1+s}\sqrt{-a\log y}\bigr)\,dy
&=-\frac{2}{a}\int u\erf(\sqrt{1+s}u)\,du\\
&=-\frac{1}{a}u^2\erf(\sqrt{1+s}u)+\frac{\sqrt{1+s}}{a\sqrt{\pi}}\int u^2 e^{-(1+s)u^2}\,du\\
&=-\frac{1}{a}u^2\erf(\sqrt{1+s}u)
-\frac{1}{2a\sqrt{\pi}\sqrt{1+s}}ue^{-(1+s)u^2}\\
&\qquad+\frac{1}{4a(1+s)}\erf(\sqrt{1+s}u).
\end{split}
$$
I leave it to you to insert $a=1$ and $a=2$, and to work with the limits.
A: $\sqrt{-\log(\cdot)}$ 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:

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

A: \begin{align}
u & = xy \\[10pt]
v & = \frac y x \\[10pt]
du\,dv = \frac{\partial(u,v)}{\partial(x,y)} \, dx \, dy & = \frac {2y} x \, dx\,dy = 2v\,dx\,dy \\[10pt]
\frac{du\,dv}{2v} & = dx\,dy 
\end{align}
\begin{align}
& \int_0^1 \int_u^{1/u} \frac{u^s}{\sqrt{-\log u}} \, \frac{dv \, du}{2v} \\[10pt]
= {} & \int_0^1 \frac{u^s}{\sqrt{-\log u}} \cdot \frac 1 2 \left( \log \frac 1 u - \log u   \right) \, du \\[10pt]
= {} & \int _0^1 u^s \sqrt{-\log u} \, du \tag 1
\end{align}
\begin{align}
w & = \sqrt{-\log u} \\[10pt]
e^{-w^2} & = u \\[10pt]
-2w e^{-w^2}\,dw & = du
\end{align}
So $(1)$ becomes $\displaystyle \int_\infty^0 e^{-sw^2} w ( -2w e^{-w^2} \, dw) = 2\int_0^\infty w^2 e^{-(s+1)w^2} \, dw$.
We can reduce this to the expected value of a function of a random variable with a standard normal distribution:
\begin{align}
(s+1) w^2 & = \frac{z^2} 2 \\[10pt]
dw & = \frac{dz}{\sqrt2\sqrt{s+1}}
\end{align}
Our integral becomes
$$
2\int_0^\infty \frac{z^2}{2(s+1)} e^{-z^2/2} \frac{dz}{\sqrt2\sqrt{s+1}} = \frac 1 {(s + 1)^{3/2}\sqrt 2} \int_0^\infty z^2 e^{-z^2/2}\,dz. \tag 1
$$
The last integral is
$$
\int_0^\infty z^2 \left( \frac{e^{-z^2/2}}{\sqrt{2\pi}} \right) \, dz \cdot \sqrt{2\pi}.
$$
This integral is half the expected value of a $\chi^2_1$ random variable; thus it is equal to $1/2$, so we get $(1/2)\sqrt{2\pi},$ so the whole expression $(1)$ is
$$
\frac 1 {(s + 1)^{3/2}\sqrt 2} \cdot \frac 1 2 \cdot \sqrt{2\pi} = \frac{\sqrt \pi}{2(s + 1)^{3/2}}.
$$
A: Here is an approach.
We have, for $s>0$, $\sigma>0$, 

$$
\int_0^1\!\int_0^1\frac{x^{s-1}y^{\sigma-1}}{\sqrt{-\log (xy)}}\,dx\,dy=\frac{\sqrt{\pi}}{s\sqrt{\sigma}+\sigma\sqrt{s}}.\tag1
$$

Proof. One may observe that
$$
\frac1{\sqrt{a}}=\frac2{\sqrt{\pi}}\int_0^\infty e^{\large-at^2}dt, \qquad a>0. \tag2
$$ Assume $s>0$, $\sigma>0$ and $\sigma\neq s$. Then we obtain
$$
\begin{align}
\iint_{[0,1]\times[0,1]} \frac{x^{s-1}y^{\sigma-1}}{\sqrt{-\log (xy)}}\,dx\,dy&=\iint_{[0,1]\times[0,1]} x^{s-1}y^{\sigma-1}\left(\frac1{\sqrt{-\log (xy)}}\right)dx\,dy\\\\
&=\iint_{[0,1]\times[0,1]} x^{s-1}y^{\sigma-1}\left(\frac2{\sqrt{\pi}}\int_0^\infty e^{\large t^2\log(xy)}dt\right)dx\,dy\\\\
&=\frac2{\sqrt{\pi}}\int_0^\infty\iint_{[0,1]\times[0,1]} x^{s-1}y^{\sigma-1}\times x^{t^2}y^{t^2}dx\,dy\,dt\\\\
&=\frac2{\sqrt{\pi}}\int_0^\infty\left(\int_0^1x^{t^2+s-1}dx\right)\times \left(\int_0^1y^{t^2+\sigma-1}dy\right)\,dt\\\\
&=\frac2{\sqrt{\pi}}\int_0^\infty\frac1{(t^2+s)(t^2+\sigma)}\,dt\\\\
&=\frac2{\sqrt{\pi}(\sigma-s)}\left(\int_0^\infty\frac{1}{\left(t^2+s\right)}\,dt-\int_0^\infty\frac{1}{\left(t^2+\sigma\right)}\,dt\right)\\\\
&=\frac2{\sqrt{\pi}(\sigma-s)}\left(\frac{\pi}{2\sqrt{s}}-\frac{\pi}{2\sqrt{\sigma}}\right)\\\\
&=\frac{\sqrt{\pi}}{s\sqrt{\sigma}+\sigma\sqrt{s}}.
\end{align}
$$
We obtain the desired integral by setting $s:=s+1$ in $(1)$ and $\sigma :=s+1$  giving

$$
\int_0^1\!\int_0^1\frac{(xy)^s}{\sqrt{-\log (xy)}}\,dx\,dy=\frac{\sqrt{\pi}}{2(1+s)^{3/2}}.\tag3
$$

