TITLE: How can one show that $\int_{0}^{1}{x\ln{x}\ln(1-x^2)\over \sqrt{1-x^2}}\mathrm dx=4-{\pi^2\over 4}-\ln{4}?$
QUESTION [6 upvotes]: How can one show that 

$$\int_{0}^{1}{x\ln{x}\ln(1-x^2)\over \sqrt{1-x^2}}\mathrm dx=4-{\pi^2\over 4}-\ln{4}?\tag1$$

I have tried IBP but it is seem too complicate. I am not sure what to do or how to tackle this problem. Please any help? Thank!

REPLY [2 votes]: Let $x=\sin t$ and then
$$I=\int_0^1 \frac{x \ln x \ln (1 - x)}{\sqrt{1 - x^2}} \, dx =2\int_0^{\pi/2}\sin t\ln\sin t\ln\cos tdt=-2\int_0^{\pi/2}\ln\sin t\ln\cos td\cos t.$$
Letting $u=\cos t$ gives
\begin{eqnarray}
I&=&-2\int_0^{\pi/2}\ln\sin t\ln\cos td\cos t\\
&=&\int_0^1\ln(1-u^2)\ln udu\\
&=&-\int_0^1\sum_{k=1}^\infty\frac1{k}u^{2k}\ln udu\\
&=&\sum_{k=1}^\infty\frac1{k(2k+1)^2}\\
&=&\sum_{k=1}^\infty\left[\frac{1}{k(2k+1)}-\frac{2}{(2k+1)^2}\right].
\end{eqnarray}
Now using
$$　\sum_{k=1}^\infty\frac{1}{k(2k+1)}=2-2\ln2 $$
and
$$ \sum_{k=1}^\infty\frac{1}{(2k-1)^2}=\frac{\pi^2}{8} $$
it is easy to get
$$ I=4-2\ln2-\frac{\pi^2}{4}.$$