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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!!!

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    $\begingroup$ What happens at $\pi$? $\endgroup$ – Daniel Fischer Feb 12 '15 at 20:55
  • $\begingroup$ Totally not considered. A critic point in which the integral should diverge: the integrand is similar to $\frac{cost}{sin^2\pi}$ so it goes to $+\infty$ I guess $\endgroup$ – Dipok Feb 12 '15 at 20:59
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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.

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$\sum_{k=1}^{n}\frac{e^{-k}}{\sin^2k} \leq 1+\int^{+\infty}_1\frac{e^{-x}}{\sin^2x}dx$. 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.

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  • $\begingroup$ This approach seems quite difficult to rigorously prove, actually, because it depends on the rather subtle character of the sequence $\sin(k)$ (there are "many" integers $k$ such that $\sin(k)$ is "small", which is really a statement about the irrationality measure of $\pi$). The other approaches seem much simpler. $\endgroup$ – Ian Feb 12 '15 at 21:48
  • $\begingroup$ @Ian, i agree,but it's not hard to see that for every $k\pi$ we have divergance of the sequence from $0$ and thus it cannot convergeto $0$ at least. It's harder to be thought and easier to be proved. This is my opinion of course:) $\endgroup$ – Haha Feb 12 '15 at 21:53
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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.

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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.

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  • $\begingroup$ So the integrand is similar to $\frac{cost}{sin^2\pi}$ so it goes to $+\infty$ right? $\endgroup$ – Dipok Feb 12 '15 at 21:05
  • $\begingroup$ @Dipok You don't even wind up considering the limit at $\infty$, the integrals over any intervals containing $\pi$ are $+\infty$. $\endgroup$ – Ian Feb 12 '15 at 21:06
  • $\begingroup$ Ok so my mistake is that I have forgotten to consider the critic point $\pi$ $\endgroup$ – Dipok Feb 12 '15 at 21:12
  • $\begingroup$ @Dipok Not just pi, also all the multiples of pi :) $\endgroup$ – imranfat Feb 12 '15 at 21:19
  • $\begingroup$ Yes of course :) $\endgroup$ – Dipok Feb 13 '15 at 8:42

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