Is the improper integral $\int_0^{\pi/2} \sqrt{\cot x}\, dx$ convergent?

Question

Is the improper integral $\int_0^{\pi/2} \sqrt{\cot x} \,dx$ convergent? I am unable to use any kind of comparison test or anything.

Answer 1

A potential problem of convergence is as $x \to 0^+$, in this case, by Taylor expansion we have $$ \sqrt{\cot x} \sim \frac1{\sqrt{x}} $$ which converges near $0^+$.

Answer 2

Note that $\frac{\tan x}{x}$ is bounded near the origin, so you can estimate

$$\sqrt{\cot x} \lesssim \frac{1}{\sqrt x}$$

where the implicit constant can actually be computed as $1$ (although that's much more than what's actually necessary).