Double Integral on $[0,1]\times[0,1]$ I want to calculate the double integrl is given by
$$\int_{0}^{1}\int_{0}^{1}\frac{x^3y^3\ln(xy)}{1-xy}dxdy$$
I set $u=xy$ and $v=1-xy$,then calculate Jacobian but my changing variable was not useful.
How can we choos $u$ and $v$ ? Is there a other way?
Thanks
 A: Take $xy=u
 $. We have $$I=\int_{0}^{1}\int_{0}^{1}\frac{x^{3}y^{3}\log\left(xy\right)}{1-xy}dxdy=\int_{0}^{1}\frac{1}{y}\int_{0}^{y}\frac{u^{3}\log\left(u\right)}{1-u}dudy
 $$ and $$\int_{0}^{y}\frac{u^{3}\log\left(u\right)}{1-u}du=\sum_{k\geq0}\int_{0}^{y}u^{k+3}\log\left(u\right)du
 $$ $$=\sum_{k\geq0}\frac{y^{k+4}\log\left(y\right)}{k+4}-\sum_{k\geq0}\frac{y^{k+4}}{\left(k+4\right)^{2}}
 $$ hence $$I=\sum_{k\geq0}\frac{1}{k+4}\int_{0}^{1}y^{k+3}\log\left(y\right)dy-\sum_{k\geq0}\frac{1}{\left(k+4\right)^{2}}\int_{0}^{1}y^{k+3}dy
 $$ $$=-2\sum_{k\geq0}\frac{1}{\left(k+4\right)^{3}}=\color{red}{-2\left(\zeta\left(3\right)-\frac{251}{216}\right).}$$
A: $\displaystyle J=\int_0^1 \dfrac{x^3(\ln x)^2}{1-x}dx=\Big[R(x)\ln x\Big]_0^1-\int_0^1 \dfrac{R(x)}{x}dx=-\int_0^1\int_0^1 \dfrac{x^3t^3\ln(tx)}{1-xy}dtdx$
For $x\in[0,1]$, $\displaystyle R(x)=\int_0^x\dfrac{t^3\ln(t)}{1-t}dt=\int_0^1 \dfrac{x^4t^3\ln(tx)}{1-tx}dt$ and $\lim_{x\rightarrow 0}R(x)\ln x=0$
$J=\displaystyle \sum_{n=0}^{\infty}\left(\int_0^1 x^{n+3}(\ln x)^2dx\right)$
$J=\displaystyle 2\sum_{n=0}^{\infty} \dfrac{1}{(n+4)^3}$
$J=\displaystyle 2\left(\zeta(3)-1-\dfrac{1}{2^3}-\dfrac{1}{3^3}\right)$
$J=\displaystyle 2\left(\zeta(3)-\dfrac{251}{216}\right)$
Therefore,
$\displaystyle \int_0^1\int_0^1 \dfrac{x^3t^3\ln(tx)}{1-xy}dtdx=-2\left(\zeta(3)-\dfrac{251}{216}\right)$
