Convergence of $\int^{+\infty}_1\frac{e^{-x}}{\sin^2x}$

Question

I'm studying the convergence of $$\int^{+\infty}_1\frac{e^{-x}}{\sin^2x}dx$$ The solution is "It is divergent" But I can't figure it out. For $+\infty$ it should converge because $\sin^2x$ is $1$ and $e^{-x}$ goes to $0$. Obviously it is not a correct reasoning but I just can't understand how it diverges. Can someone explain me clearly how it diverges? thanks in advance!!!

Answer 1

Hint Without loss of generality, consider the integral $$\int_\pi^{\pi + 1}\frac{e^{-x}}{\sin^2 x}\mathrm dx$$

and show that, for $x \to \pi^+$, $$\frac{e^{-x}}{\sin^2 x} > \frac1{x - \pi}.$$ Combine it with the fact that the integral $$\int_\pi^{\pi + 1}\frac1{x - \pi}\mathrm dx$$ is divergent.

Answer 2

If you know that $\lim_{x \to \pi} \sin(x - \pi) / (x - \pi) = 1$, then you should be able to show divergence by considering the integral $\int_{\pi - 1}^{\pi} e^{-x}/(x-\pi)^2$, and use the fact that $e^{-x}$ is positive bounded away from zero on the interval, i.e. $e^{-x} > c$ for some $c > 0$ on the interval.

Answer 3

$\sum_{k=1}^{n}\frac{e^{-k}}{\sin^2k} \leq 1+\int^{+\infty}_1\frac{e^{-x}}{\sin^2x}dx$. If your integral is convergent then we can take $n\to \infty$ and thus we have that $\sum_{k=1}^{\infty}\frac{e^{-k}}{\sin^2k}<\infty$. But $\frac{e^{-k}}{\sin^2k}\not \to 0$ for $k\to \infty$ and thus the series diverge. Contradiction.

Answer 4

At $x =\pi$ this integrand exhibits the same boorish behavior exhibited by $x\mapsto 1/ x^2$ at $0$. The integral is toast right there.