I hope I used the correct tag (definite integral).
I ran across an integral that is rather tough and I am wondering if anyone could give me a shove in the right direction.
$\displaystyle\int_{0}^{1}\frac{\tanh^{-1}(x)\ln(x)}{x(1-x^{2})}\text{ d}x=\frac{-7}{16}\zeta(3)-\frac{{\pi}^{2}}{8}\ln(2)$
This solution is almost exactly like the solution to $\displaystyle \int_{0}^{\frac{\pi}{2}}x\ln(\sin(x))\text{ d}x$, which is $\displaystyle \frac{7}{16}\zeta(3)-\frac{{\pi}^{2}}{8}\ln(2)$
I solved the latter integral by using the identity $\displaystyle -\ln(\sin(x))-\ln(2)=\sum_{k=1}^{\infty}\frac{\cos(2kx)}{k}$, then integrating:
$\displaystyle -\int_{0}^{\frac{\pi}{2}}x\ln(\sin(x))\text{ d}x-\frac{{\pi}^{2}}{8}\ln(2)=\int_{0}^{\frac{\pi}{2}}x\cos(2x)\text{ d}x+\frac{\int_{0}^{\frac{\pi}{2}}\cos(4x)\text{ d}x}{2}+\frac{\int_{0}^{\frac{\pi}{2}}\cos(6x)}{3}\cdot\cdot\cdot\cdot$
But, $\displaystyle \int_{0}^{\frac{\pi}{2}}x\cdot \cos(2kx)\text{ d}x=-\left(\frac{1+(-1)^{k+1}}{(2k)^{2}}\right)$
Thus: $\displaystyle \int_{0}^{\frac{\pi}{2}}x\ln(\sin(x))\text{ d}x=\frac{1}{4}\sum_{k=1}^{\infty}\frac{1+(-1)^{k+1}}{k^{3}}-\frac{{\pi}^{2}}{8}\ln(2)$
$\displaystyle =\frac{1}{2}\displaystyle\sum_{k=0}^{\infty}\frac{1}{(2k+1)^{3}}-\frac{{\pi}^{2}}{8}\ln(2)$
and so on. Which results in the solution I mentioned in the beginning.
Sorry for all that, but I wanted to show you what I was using in order to some how relate it to the integral I am wanting to solve. I have been trying and trying to relate the aforementioned $\displaystyle \tanh$ integral with this one. The solutions are so nearly the same, I figured there has to be a way to relate them and solve the integral. Does anyone have some ideas?. I have tried the identity $\displaystyle \tanh^{-1}(x)=\frac{1}{2}\left[\ln(1+x)-\ln(1-x)\right]$, then breaking it up:
Resulting in $\displaystyle \frac{1}{2}\int_{0}^{1}\frac{\ln(x)\ln(1+x)}{x(x^{2}-1)}\text{ d}x-\frac{1}{2}\int_{0}^{1}\frac{\ln(x)\ln(1-x)}{x(x^{2}-1)} \text{ d}x$, then I used the various series representations for $\displaystyle \ln(1+x)$, $\displaystyle \frac{1}{1-x^{2}}$, etc. I tried double integrals, but I always get stuck.
I even broke it up per partial fraction expansion, but several of the resulting integrals were still nasty.
Does anyone have some clever ideas?.