Do integrals $\int_0^\pi \frac{x}{\sin{x}}dx$ and $\int_0^\pi \frac{x}{\sqrt{\sin{x}}}dx$ converge? Do integrals $$\int_0^\pi \frac{x}{\sin{x}}dx$$ and $$\int_0^\pi \frac{x}{\sqrt{\sin{x}}}dx$$
converge?
I don't know how to solve it because we have only done comparison test, but only with $\int_1^\infty \frac{1}{x^p}dx$. I don't know if you can apply it here. First thing i would do here is split them onto two integrals.
$$\int_0^\pi \frac{x}{\sin{x}}dx=\int_0^{\frac{\pi}{2}} \frac{x}{\sin{x}}dx+\int_{\frac{\pi}{2}}^\pi \frac{x}{\sin{x}}dx$$
and 
$$\int_0^\pi \frac{x}{\sqrt{\sin{x}}}dx=\int_0^{\frac{\pi}{2}} \frac{x}{\sqrt{\sin{x}}}dx+\int_{\frac{\pi}{2}}^\pi \frac{x}{\sqrt{\sin{x}}}dx$$
 A: It is enough to consider the behaviour of the integrand functions at the endpoints of the integration range. $x=0$ is a removable singularity for both $\frac{x}{\sin(x)}$ and $\frac{x}{\sqrt{\sin x}}$, but while $x=\pi$ is an integrable singularity for $\frac{x}{\sqrt{\sin x}}$, it is a simple pole of $\frac{x}{\sin x}$, hence a non-integrable singularity. It follows that the second integral is convergent but the first one is not. We may also notice that
$$ I_2 = \int_{0}^{\pi}\frac{x\,dx}{\sqrt{\sin x}}= \int_{0}^{\pi/2}\frac{x}{\sqrt{\sin x}}\,dx + \int_{\pi/2}^{\pi}\frac{x}{\sqrt{\sin x}}\,dx = \pi\int_{0}^{\pi/2}\frac{dx}{\sqrt{\sin x}}, $$
and since $\sin(x)>\frac{2x}{\pi}$ over $\left(0,\frac{\pi}{2}\right)$ holds by concavity, we have
$$ 0< I_2 < \pi^2. $$ 

The exact value of $I_2$ can be computed through the substitution $x=\arcsin t$ and Euler's beta function:
$$\boxed{\; I_2 = \int_{0}^{\pi}\frac{x\,dx}{\sqrt{\sin x}}=\int_{0}^{1}\frac{\pi\,dt}{\sqrt{t(1-t^2)}} = \frac{\pi}{2}\,B\left(\frac{1}{4},\frac{1}{2}\right)=\color{red}{\sqrt{\frac{\pi}{8}}\,\Gamma\left(\frac{1}{4}\right)^2}.\;}$$
