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So the question is for what $\alpha$ our integrand is integrable. I started with the fact that $$\sin(x)\leq x$$ in this interval $$\therefore \ \frac{1}{\sin(x)}\geq \frac{1}{x}$$ So the function diverges for all ${\alpha\geq 1}$ I suspect that it might be convergent for $\alpha \in (0,1)$ but I cannot bound it from above with a function for which it is true.

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    $\begingroup$ Why don't you bound it with $\frac{2x}{\pi}$ in the other direction? $\endgroup$ – Diger Jun 29 '18 at 12:58
  • $\begingroup$ @Diger Will do my friend. thanks $\endgroup$ – Kran Jun 29 '18 at 13:12
  • $\begingroup$ @Kacper Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – user Aug 6 '18 at 21:43
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We have that for $x\to 0^+$

$$\frac{1}{\sin x}\sim \frac{1}{x}$$

therefore by limit comparison test with $\int_0^{\frac{\pi}{2}} \big{(}\frac{1}{x}\big{)}^{\alpha}dx$ the given integral diverges for $\alpha\ge1$ and converges for $\alpha<1$ .

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  • $\begingroup$ yeah that was my motivation, but I cannot that "the function behaves like [...]" I need rigorous proof. $\endgroup$ – Kran Jun 29 '18 at 13:11
  • $\begingroup$ @Kacper The rigorous proof can be done by the limit comparison test, it is a theorem which allow us to conclude the behaviour of the given integral by the behaviour of the other. $\endgroup$ – user Jun 29 '18 at 13:13
  • $\begingroup$ @Kacper It is a powerful and useful criteria which avoid the direct comparison test based on inequalities which can be more difficult to be applied in general. $\endgroup$ – user Jun 29 '18 at 13:14
  • $\begingroup$ Refer also to math.toronto.edu/mat137/files/137_LCT2.pdf $\endgroup$ – user Jun 29 '18 at 13:17
  • $\begingroup$ Yeaj sorry I did not get what you wanted to do, I understood it even before I read your comments. I take bak what I said ;) $\endgroup$ – Kran Jun 29 '18 at 13:20
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$g(x)=\frac{x}{\sin x}$ is increasing and bounded between two positive constants for $x\in(0,\pi/2)$. It follows that $$ K^{\alpha}_1\int_{0}^{\pi/2}\frac{dx}{x^\alpha}\leq\int_{0}^{\pi/2}\sin(x)^{-\alpha}\,dx \leq K^{\alpha}_2\int_{0}^{\pi/2}\frac{dx}{x^\alpha}$$ and the middle integral is convergent iff $\alpha<1$ as expected. You may compute it its closed form in terms of the $\Gamma$ function through Euler's Beta function, leading to $$ \int_{0}^{\pi/2}\sin(x)^{-\alpha}\,dx = \frac{\sqrt{\pi}}{2}\cdot\frac{\Gamma\left(\frac{1}{2}-\frac{\alpha}{2}\right)}{\Gamma\left(1-\frac{\alpha}{2}\right)}\qquad \text{for Re}(\alpha)<1.$$ If you are just interested in discussing real values of $\alpha$, you may exploit the fact that due to the non-negativity and continuity of the sine function over $(0,\pi/2)$, $M(\alpha)=\int_{0}^{\pi/2}\sin(x)^{-\alpha}\,dx = \int_{0}^{1}\frac{dz}{z^\alpha \sqrt{1-z^2}}$ is non-negative and log-convex where it is defined, as a consequence of the Cauchy-Schwarz inequality and the fact that $\text{continuous}+\text{midpoint-convex}$ implies $\text{convex}$. $\lim_{\alpha\to 1^-}M(\alpha)=+\infty$ is fairly straightforward, hence the domain of $M$ cannot extend beyond $1$.

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