I am trying to figure out is following integral converge or diverge $$\int_0^{\infty}\frac{|\sin x|}{e^{x^2 \sin^2 x}}dx$$
I have tried to find inequalities on $$\int_{k\pi}^{(k+1)\pi}\frac{|\sin x|}{e^{x^2 \sin^2 x}}dx$$ but with no success.
Any hints?
You were on the right track. In order to prove that the original integral is convergent, it is enough to show that the sequence $$ a_{k}=\int_{k\pi-\pi/2}^{k\pi+\pi/2}\left|\sin x\right|\exp\left(-x^2\sin^2 x\right)\,dx=\int_{-\pi/2}^{\pi/2}\left|\sin x\right|\exp\left(-(x+k\pi)^2\sin^2 x\right)\,dx $$ is summable. In a $\frac{\pi}{2}$-neighbourhood of the origin we have $x^2\geq \sin^2 x\geq \frac{4}{\pi^2}x^2$ and $e^{-k^2x^2}\leq \frac{1}{1+k^2x^2}$, hence it is enough to show that $$ b_k = \int_{-\pi/2}^{\pi/2}\frac{\left|x\right|}{1+\frac{4}{\pi^2}(x+k\pi)^2 x^2}\,dx$$ is summable. This is simple since $b_k$ turns out to be $O\left(\frac{\log k}{k^2}\right)$ as $k\to +\infty$. Indeed: $$ b_k \leq \int_{0}^{\pi/2}\frac{x}{1+4k^2 x^2}\,dx+\int_{0}^{\pi/2}\frac{x}{1+4\left(k-\frac{1}{2}\right)^2 x^2}\,dx $$ and $\int_{0}^{\pi/2}\frac{x}{1+4a^2 x^2}\,dx = \frac{\log(1+\pi^2 a^2)}{8a^2}\leq \frac{\sqrt{\pi}}{4a\sqrt{a}}$ for any $a>0$.