integration - Contour Integral of $log(z)/(1+z^a)$ where $agt1$

I am asked to prove that:
$$ \int_{0}^{+\infty}\frac{\log z}{1+z^{\alpha}}\,dz = -\frac{\pi^2}{\alpha^2}\cdot\frac{\cos\frac{\pi}{\alpha}}{\sin^2\frac{\pi}{\alpha}},$$
provided that $\alpha > 1$, with a complex analytic method, i.e. contour integration.

However, I was not able to find a good candidate as a meromorphic function to integrate, neither a proper contour. Would you mind giving me a hand?

Answer

To do the contour integration, use a circular wedge of radius $R$ and angle $2 \pi/\alpha$ in the complex plane. This wedge encloses the pole at $z=e^{i \pi/\alpha}$. The integral about the arc vanishes as $R \to \infty$. (We technically should have a small cutout of radius $\epsilon$ about the origin, but we may ignore that piece as there is no contribution.)

The integrals that remain are over the real axis and over the line in the complex plane that is at an angle $\alpha$ with respect tot he real axis. Thus, by the residue theorem,

$$\int_0^{\infty} dx \frac{\log{x}}{1+x^{\alpha}} - e^{i 2 \pi/\alpha} \int_0^{\infty} dx \frac{\log{x}+i 2 \pi/\alpha}{1+x^{\alpha}} = i 2 \pi \operatorname*{Res}_{z=e^{i \pi/\alpha}} \frac{\log{z}}{1+z^{\alpha}} $$

which becomes

$$\left ( 1-e^{i 2 \pi/\alpha} \right ) \int_0^{\infty} dx \frac{\log{x}}{1+x^{\alpha}} - i \frac{2 \pi}{\alpha} e^{i 2 \pi/\alpha} \int_0^{\infty} \frac{dx}{1+x^{\alpha}} = \frac{2 \pi^2}{\alpha^2} e^{i \pi/\alpha}$$

To evaluate the second integral on the LHS, use the same contour and pole:

$$\left ( 1-e^{i 2 \pi/\alpha} \right )\int_0^{\infty} \frac{dx}{1+x^{\alpha}} = i 2 \pi \operatorname*{Res}_{z=e^{i \pi/\alpha}} \frac{1}{1+z^{\alpha}} = -\frac{i 2 \pi}{\alpha} e^{i \pi/\alpha}$$

so that

$$\int_0^{\infty} \frac{dx}{1+x^{\alpha}} = \frac{\pi}{\alpha \sin{(\pi/\alpha)}}$$

Thus, we now have that

$$-i 2 \sin{(\pi/\alpha)}\int_0^{\infty} dx \frac{\log{x}}{1+x^{\alpha}} - i \frac{2 \pi^2}{\alpha^2} \frac{e^{i \pi/\alpha}}{\sin{(\pi/\alpha)}} = \frac{2 \pi^2}{\alpha^2} $$

Take real and imaginary parts of the second term, and the real part cancels the RHS. Thus, dividing by the factor in front of the integral, we are left with

$$\int_0^{\infty} dx \frac{\log{x}}{1+x^{\alpha}} = -\frac{\pi^2}{\alpha^2} \frac{\cos{(\pi/\alpha)}}{\sin^2{(\pi/\alpha)}} $$

as was to be shown.