Is $\int_0^{\infty} \frac{1}{e^{x^2}|\sin{x}|} dx $ convergent? I uaed comparison test $$\frac{1}{e^{x^2}|\sin{x}|} < \frac{1}{|\sin{x}|}$$
But $\int\frac{1}{|\sin{x}|} dx$ is divergent.
I don't know how to show whether the integral is convergent or not.
 A: $e^{x^{2}} \sin x  \geq x/2$ for $x$ near $0$, say $0< x <t$,  and $\int_0^{t} \frac  2 { x}dx =\infty$ so the integral is divergent.  
I have used the facts that $e^{x^{2}} \to 1$ and $\frac  {\sin x } x \to 1$ as $x \to 0$. 
A: Since $\lim_{x \rightarrow 0} \frac{e^{x^2} |\sin x|}{2x} = \frac{1}{2}$,
so there exists an interval $(0,~t)$, $\frac{e^{x^2} |\sin x|}{2x} < 1$,
which means $\frac{1}{e^{x^2} |\sin x|} > \frac{1}{2x}$.
Hence, 
$\int_0^\infty \frac{1}{e^{x^2} |\sin x|} > 
\int_0^t \frac{1}{e^{x^2} |\sin x|} > \int_0^t \frac{1}{2x} = + \infty$
A: I'd like to add to the other answers that concentrating on what happens near $x=0$ may give the incorect impression that only $x=0$ is a problem point of the integral. The denominator has a singularity at every point that $\sin(x)=0$, that means $x=k\pi, k \in \mathbb Z$.
Because $|\sin(x)|$ is $\pi$-periodic, the behavior of $|\sin(x)| \sim |x|$ that holds near $x=0$ also holds near $x=k\pi, k \in \mathbb Z$ in the general form of $|\sin(x)| \sim |x-k\pi|$. So near enough those points, we have $|\sin(x)| \le 2|x-k\pi|$
Near each such points, $e^{x^2}$ is bounded above by say $e^{x^2} \le 2e^{(k\pi)^2}$ simply due to continuity of $e^{x^2}$.
That means that near enough $x=k\pi, k \in \mathbb Z$, then we have for the integrand
$$\frac1{e^{x^2}|\sin(x)|} \ge \frac1{2e^{(k\pi)^2}2|x-k\pi|} = \frac1{4e^{(k\pi)^2}}\frac1{|x-k\pi|}.$$
The first factor is a constant, the second factor is known to not be integrable near $x=k\pi$, as the antiderivative is $\ln|(x-k\pi)|$.
So not only is $\int_{0}^{\infty}\frac1{e^{x^2}|\sin(x)|}$ not convergent, any $\int_{a}^{b}\frac1{e^{x^2}|\sin(x)|}$ where there does not exist an integer $k$ with $k\pi < a,b < (k+1)\pi$ is divergent, while if such $k$ exists, the intgral exists.
