What $\alpha$ makes $\int_0^{\frac{\pi}{2}} \big{(}\frac{1}{\sin(x)}\big{)}^{\alpha}$ converge 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.
 A: 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 $\alpha<1$ .
A: $g(x)=\frac{x}{\sin x}$ 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 $1$.
