Test the convergence of the integral Test the convergence of $$\int_0^\pi{\sqrt x\over \sin x}dx$$
I have to do it using comparison test. There is another test mentioned called the $\mu-test$ but its definition in the book doesn't seem right, cannot find it on the internet too. Is someone familiar with it ? It is a corollary of comparison test.

We have
$$I = \underbrace{\int_0^{\pi} \dfrac{\sqrt{x}}{\sin(x)}dx = \int_0^{\pi} \dfrac{\sqrt{\pi-x}}{\sin(x)}dx}_{x \mapsto \pi-x}$$
Also, $\sin(x) \leq x$. Hence,
$$I \geq \int_0^{\pi} \dfrac{\sqrt{\pi-x}}{x}dx \geq \int_0^{\pi/2} \dfrac{\sqrt{\pi-x}}{x}dx \geq \int_0^{\pi/2} \dfrac{\sqrt{\pi/2}}xdx = \infty$$
