that's a question from some exam in Calculus Can someone help?

does $\int _0^{\infty }\frac{\sin\pi \:x}{\left|\ln \left(x\right)\right|^{\frac{3}{2}}}$ converge?

I proved that it converges between 1 and infinity using comparison test with the integral of $\frac{1}{x^{\frac{3}{2}}}$ Between 1/2 and 1 i used Dirichlet exam to prove it converges. Is that true?

Any thoughts aout how can I prove between 0 and 1/2?

  • 4
    $\begingroup$ I don't see what's special about $(0,1/2)$. Since $\lvert \ln x\rvert\to\infty$ as $x\to0^+$, the integrand has limit $0$ as $x\to0$. I'm more concerned about your argument for $(2,\infty)$, since $\ln^{-3/2}x\gg x^{-3/2}$. $\endgroup$ – user228113 Feb 8 '17 at 7:53
  • $\begingroup$ You are absolutely right. Thank you very much! $\endgroup$ – Dan Feb 8 '17 at 7:56
  • $\begingroup$ You primarily need to deal with the singularity at $x = 1$ which it appears you have not. $\endgroup$ – RRL Feb 8 '17 at 8:39


For $0 < x < 1$ use $|\ln x| > 1-x$ and $\sin x /x < 1$ to estimate

$$\frac{\sin \pi x}{|\ln x|^{3/2}} = \frac{\pi(1-x)}{|\ln x|^{3/2}}\frac{\sin \pi x}{\pi(1-x)} = \frac{\pi(1-x)}{|\ln x|^{3/2}}\frac{\sin \pi (1-x)}{\pi(1-x)}\leqslant \frac{\pi}{\sqrt{1-x}} $$


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.