3
$\begingroup$

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.

Thanks

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $p>1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.