For which values of $p$ does the improper integral $$\int_0^\infty \frac{\log x}{1+x^p}\ dx$$ converge? I tried integration parts and various tricks, but it does not seems to work.


  • $\begingroup$ Comparison. The integrand is continuous on $(0,+\infty)$, so the questions to be answered are: how does it behave at $0$ and how at $+\infty$? $\endgroup$ – Daniel Fischer Dec 8 '14 at 19:53
  • $\begingroup$ As $x\to 0$, the denominator is almost $1$, hence you are integrating something comparable with $\log x$: $\int_0^1 \log x dx $ does not give problems. So as Daniel Fischer said, you should check for $x\to +\infty$. (use the fact that $\int_1^\infty x^p dx<+\infty$ iff $p<-1$). $\endgroup$ – Milly Dec 8 '14 at 20:07

We have $$\int_0^\infty\dfrac{x^{\large a-1}}{1+x^b}\ dx=\frac{\pi}{b}\csc\left(\frac{a\pi}{b}\right)$$ for $0<a<b$. Differentiating with respect to $a$ then setting $a=1$ and $b=p$, we get $$\int_0^\infty\dfrac{\ln x}{1+x^p}\ dx=-\frac{\pi^2}{p}\csc\left(\frac{\pi}{p}\right)\cot\left(\frac{\pi}{p}\right)$$ for $p>1$.


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