TITLE: Show that $\int_{0}^{\infty }\frac {\ln x}{x^4+1}\ dx =-\frac{\pi^2 \sqrt{2}}{16}$
QUESTION [17 upvotes]: I could prove it using the residues but I'm interested to have it in a different way (for example using Gamma/Beta or any other functions) to show that
$$
\int_{0}^{\infty}\frac{\ln\left(x\right)}{x^{4} + 1}\,{\rm d}x
=-\frac{\,\pi^{2}\,\sqrt{\,2\,}\,}{16}.
$$
Thanks in advance.

REPLY [3 votes]: We can use the following way to solve. It is very simple. In fact
\begin{eqnarray}
I&=&\int_0^\infty \frac{\ln x}{1+x^4}dx=\int_0^1 \frac{\ln x}{1+x^4}dx+\int_1^\infty \frac{\ln x}{1+x^4}dx\\
&=&\int_0^1 \frac{\ln x}{1+x^4}dx-\int_0^1 \frac{x^2\ln x}{1+x^4}dx\tag{1}\\
&=&\int_0^1\sum_{n=0}^\infty(-1)^n(x^{4n}-x^{4n+2})\ln xdx\tag{2}\\
&=&\sum_{n=0}^\infty(-1)^{n+1}\left(\frac{1}{(4n+1)^2}-\frac{1}{(4n+3)^2}\right)\\
&=&\sum_{n=-\infty}^\infty(-1)^{n+1}\frac{1}{(4n+1)^2}\\
&=&-\frac{\sqrt{2}\pi}{16}\tag{3}.
\end{eqnarray}
Here, for (1), (2), and (3), we used the substitute $x\to\frac{1}{x}$, $\int_0^1x^n\ln xdx=-\frac1{(n+1)^2}$, and
$$ \sum_{n=-\infty}^\infty(-1)^{n}\frac{1}{(an+b)^2}=\frac{\pi^2a^2\cos\frac{b\pi}{a}}{\sin^2\frac{b\pi}{a}} $$
respectively.