Improper Integral : $\int_{0}^{\infty } \frac{dx}{x\sin x}$?

Question

$$\int\limits_{0}^{\infty}\frac{dx}{x\sin x}$$

How can I explain that this integral diverges?

Answer 1

On $(0,\pi/2)$ we have $$ 0<\sin x\leq x\quad\mbox{so}\quad\frac{1}{x\sin x}\geq \frac{1}{x^2}. $$

Now $$ \int_0^{\pi/2}\frac{1}{x^2}dx $$ diverges.

So your integral diverges at $0$.

Answer 2

The easiest way to see that this integral diverges is by looking at $x=0$: there, $\sin{x} \sim x$, so the integral near $x=0$ goes as $1/x^2$, which is a non-integrable singularity there, i.e. $\lim_{\epsilon \rightarrow o}\int_\epsilon^1 dx/x^2 = \lim_{\epsilon \rightarrow o} (-1/\epsilon)$, which diverges.

Answer 3

Let $0<b<\pi$ and consider $\int_0^b\frac{dx}{x\sin x}$. Nothe that $$\lim_{x\to 0^+}(x-0)^1\times\frac{1}{x\sin x}$$ tends to infinity, so by comparison test, it diverges.