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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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    What's the definition of $\text{csc}$?2012-10-10
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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

2 Answers 2

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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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    Ok, now if we take the power of $r$ from all the sides, we get: 1. In the middle - the expression we are integrating; 2. On the left - $1/x^{r}$, which is integrable on $(0,\frac{\pi}{2})$ for $r\in(0,1)$ (which can be easily shown), and 3. On the right, we get $\frac{2^{r}}{x^{r}}$, the integral of which diverges for $r\geq{1}$. (which can again be shown the same way as in 2.) Is that right?2012-10-10
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    Or rather, it should be quite the opposite :) But essentially we are proving the same.2012-10-10
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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 $r$.