0

Let $\alpha > 1$, how to compute this integral:

$$ I(\alpha)=\int_{0}^{\infty} \frac{\log z}{1+z^{\alpha}}dz $$

My try:

When $\alpha = 2$, I find that $$ I(\alpha) = \int_{0}^{1} \frac{\log z}{1+z^{2}}dz + \int_{1}^{\infty} \frac{\log z}{1+z^{2}}dz = \int_{0}^{1} \frac{\log z}{1+z^{2}}dz + \int_{1}^{0} \frac{\log z}{1+z^{2}}dz = 0 $$ when I replace $z$ with $\frac{1}{t}$ of the integration. But I'm stuck when $\alpha \neq 2.$

Any help will be greatly appreciated.

Gary
  • 31,845
ppphy
  • 81

1 Answers1

3

First compute

$$ \int_0^{\infty } \frac{x^{s-1}}{x^a+1} \, dx=\frac{\pi \csc \left(\frac{\pi s}{a}\right)}{a} $$

with Beta function. Then

$$ \begin{aligned} \int_0^{\infty } \frac{\ln x}{x^a+1} \, dx&=\lim_{s\to 1}\frac{\partial }{\partial s}\int_0^{\infty } \frac{x^{s-1}}{x^a+1} \, dx \\&=-\frac{\pi ^2 }{a^2}\cot \left(\frac{\pi }{a}\right) \csc \left(\frac{\pi }{a}\right) \end{aligned} $$

Wahrscheinlichkeitsrechnung
  • 371