TITLE: Calculate $\int_0^\infty \frac{\ln x}{1 + x^4} \mathrm{d}x$ using residue calculus
QUESTION [18 upvotes]: I need to evaluate this integral using calculus of residues: 
$$\int_0^\infty\frac{\ln(x)}{1+x^4}\mathrm{d}x$$
I know I need to consider $\displaystyle \int_0^\infty$$\frac{\ln(z)}{1+z^4}\mathrm{d}z$.
Then the integrand has singularities at at $-e^{i\pi/4}$, $e^{i\pi/4}$, $-e^{3i\pi/4}$, and $e^{3i\pi/4}$. Then I believe I should find the residues of the integrand, however I am not sure how to proceed.
Could anyone point me in the right direction?

REPLY [3 votes]: From the link Calculating the integral $\int_0^{\infty}{\frac{\ln x}{1+x^n}}$ using complex analysis, we have for $n=4$
$$ \int_0^\infty\frac{\ln(x)}{1+x^4}\mathrm{d}x=-\frac{\pi^2\cos\frac{\pi}{4}}{4^2\sin^2\frac{\pi}{4}}=-\frac{\pi^2\sqrt2}{16}. $$