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$$\frac{\ln(z)}{1+z^4}\mathrm{d}z$.
Then the integrand has singularities at at $-e^{i\pi/4}$, $e^{i\pi/4}$, $-e^{3i\pi/4}$, and $e^{3i\pi/4}$. Then I believe I should find the residues of the integrand, however I am not sure how to proceed.
Could anyone point me in the right direction?
There are a number of ways to attack this using the residue theorem. One way is to consider the following contour integral:
$$\oint_C dz \frac{\log{z}}{1+z^4} $$
where $C$ is a quarter circle of radius $R$ in the upper-right quadrant, modified by a small quarter-circle of radius $\epsilon$ so as to avoid the branch point at $z=0$. Thus the contour integral is equal to
$$\int_{\epsilon}^R dx \frac{\log{x}}{1+x^4} + i R \int_0^{\pi/2} d\theta \, e^{i \theta} \frac{\log{(R e^{i \theta})}}{1+R^4 e^{i 4 \theta}} \\ + i \int_R^{\epsilon} dy \, \frac{\log{y} + i \pi/2}{1+y^4} + i \epsilon \int_{\pi/2}^0 d\phi \, e^{i \phi} \frac{\log{(\epsilon e^{i \phi})}}{1+\epsilon^4 e^{i 4 \phi}}$$
As $R\to\infty$, the second integral vanishes as $\log{R}/R^3$; as $\epsilon \to 0$, the fourth integra. vanishes as $\epsilon \log{\epsilon}$. In these limits, the contour integral becomes
$$(1-i) \int_0^{\infty} dx \frac{\log{x}}{1+x^4} + \frac{\pi}{2} \int_0^{\infty} \frac{dx}{1+x^4}$$
By the residue theorem, the contour integral is $i 2 \pi$ times the residue of the pole inside $C$, namely $e^{i \pi/4}$, so that we have
$$(1-i) \int_0^{\infty} dx \frac{\log{x}}{1+x^4} + \frac{\pi}{2} \int_0^{\infty} \frac{dx}{1+x^4} = i 2 \pi \frac{i \pi/4}{4 e^{i 3 \pi/4}} = \frac{\pi^2}{8} e^{i \pi/4} $$
By equating imaginary parts, we find that
$$\int_0^{\infty} dx \frac{\log{x}}{1+x^4} = -\frac{\sqrt{2}}{16} \pi^2 $$
We can also find the other integral as well from the real part:
$$\int_0^{\infty} \frac{dx}{1+x^4} = \frac{\sqrt{2}}{4} \pi $$