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I was trying to find the values of $r>0$ for which the integral

$\int^{\pi/2}_{0}\csc^{r}x\,\mathrm{d}x$

exists. Unfortunately, I'm short on ideas on how this could be done. If this may serve as any guidance, to be crude, I tried punching different values for $r$ in Mathematica, and it looks like we are looking for $r\in(0,1)$. In any case, I would be thankful for advice on how we can determine and prove that.

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    Informally, $\csc x$, that is, $\frac{1}{\sin x}$, behaves like $\frac{1}{x}$ near $0$. The formal argument is not much different.2012-10-10

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Hint: For $x\ne 0$ in our interval, $\frac{1}{2}\lt \frac{\sin x}{x}\lt 1,$ so $\dfrac{1}{x}\lt \csc x \lt \dfrac{2}{x}$.

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    The divergence for $x\ge 1$ follows from the left-hand inequality for $\sec x$. The convergence for $r\lt 1$ follows from the right-hand inequality.2012-10-10
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Note that the problem with the integrand is at $x=0$. Using the fact that $\sin(x) \approx x$ as $x \rightarrow0 $, then $\frac{1}{\sin(x)^r} \approx \frac{1}{x^r}$. Now, you can derive some conditions on $r$.