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

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?

• 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 $\ln^{-3/2}x\gg x^{-3/2}$. – user228113 Feb 8 '17 at 7:53
• You are absolutely right. Thank you very much! – Dan Feb 8 '17 at 7:56
• You primarily need to deal with the singularity at $x = 1$ which it appears you have not. – RRL Feb 8 '17 at 8:39

For $0 < x < 1$ use $|\ln x| > 1-x$ and $\sin x /x < 1$ to estimate
$$\frac{\sin \pi x}{|\ln x|^{3/2}} = \frac{\pi(1-x)}{|\ln x|^{3/2}}\frac{\sin \pi x}{\pi(1-x)} = \frac{\pi(1-x)}{|\ln x|^{3/2}}\frac{\sin \pi (1-x)}{\pi(1-x)}\leqslant \frac{\pi}{\sqrt{1-x}}$$