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

How can I explain that this integral diverges?

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3  
It diverges for an infinite number of reasons! (Every zero of $\sin x$ will be problematic.) –  mrf Feb 19 '13 at 20:27
2  
@mrf But only countably many... –  Did Feb 19 '13 at 20:37

3 Answers 3

up vote 20 down vote accepted

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 $0$.

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Who knows? Yours is the best answer. –  Pete L. Clark Feb 19 '13 at 19:56
    
At present, zero downvote. –  Did Feb 19 '13 at 20:36
    
+1 well done. You are nearly 6k. ;-) –  B. S. Feb 20 '13 at 2:29
    
@BabakSorouh Thanks Babak! –  1015 Feb 20 '13 at 2:30
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@sbr Note that $\sin x$ happens to be negative on $\mathbb{R}_+$. So the inequality $\frac{1}{x\sin x}\geq \frac{1}{x^2}$ can't hold on the whole $\mathbb{R}_+$. –  1015 Mar 7 '13 at 13:22

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.

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You must take one "$" off of your text :) –  Marra Feb 19 '13 at 19:43
    
@Gustavo: Thank you! –  Ron Gordon Feb 19 '13 at 19:50

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.

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Nice.......and +1 –  amWhy Feb 20 '13 at 0:25

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