Nice integral $\int_{0}^{\infty}\frac{x\log(x)}{e^{x^2}+1}dx=?$

Question

Prove that :

$$\int_{0}^{\infty}\frac{x\log(x)}{e^{x^2}+1}dx=\frac{-1}{8}\log^2(2)$$

This result is surprisingly elegant and sober .

I have tried integration by part but I don't see a good way .

I have tried power series with :

$$ \frac{1}{\mathrm{e}^{x^2}+1}=\sum_{n=1}^{+\infty}{\left(-1\right)^{n-1}\mathrm{e}^{-nx^2}} $$

And get a infinite sum instead of an integral but I become quickly stuck .

furthermore I don't see an obvious substitution to this integral .

Finally we have: $$\frac14\int_0^\infty\frac{\log x}{e^x+1}\,dx=\int_0^\infty\frac{x\log x}{e^{x^2}+1}\,dx$$ I prefer hints but a detailed answer is also good .

Any helps is greatly appreciated

Thanks a lot .

Answer 1

Your idea of a series expansion $$\frac{1}{\mathrm{e}^{x^2}+1}=\sum_{n=1}^{+\infty}{\left(-1\right)^{n-1}\mathrm{e}^{-nx^2}}$$ is very good since $$\int x \log (x)\,e^{-n x^2} \,dx=\frac{\text{Ei}\left(-n x^2\right)-2 e^{-n x^2} \log (x)}{4 n}$$ $$\int_0^\infty x \log (x)\,e^{-n x^2} \,dx=-\frac{\log (n)+\gamma }{4 n}$$ and performing the summation $$-\frac{1}{8} \log (2) (\log (2)-2 \gamma )-\frac{1}{4} \gamma \log (2)=-\frac{1}{8} \log ^2(2)$$

Answer 2

The last relation given in the question is a substitution $u=x^2$. Then any of the answers here show that $\int_0^\infty\frac{\ln x}{e^x+1}=-\frac12\log^22$, and the original integral follows.