# 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.

• There is an interval around zero where $\sin(x)\le x$.
– lcv
May 15, 2020 at 5:22
• How do you integrate through the poles at $\pi,2\pi,3\pi,\ldots$?
– Gary
May 15, 2020 at 7:10

$$e^{x^{2}} \sin x \geq x/2$$ for $$x$$ near $$0$$, say $$0< x , and $$\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 $$x \to 0$$.

• Why $e^{x^2}sinx \geq \frac{x}{2}$?
– Mina
May 15, 2020 at 5:40
• Let $f(x)=\frac {e^{x^{2}} \sin x } x$. Then $f(x) \to 1$ as $x \to 0$. This implies that $f(x) >\frac 1 2$ if $x$ is some interval of the type $(0,t)$. @Mina May 15, 2020 at 5:42

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, $$\int_0^\infty \frac{1}{e^{x^2} |\sin x|} > \int_0^t \frac{1}{e^{x^2} |\sin x|} > \int_0^t \frac{1}{2x} = + \infty$$

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.