Why does$ \int_{0}^{\pi} \frac{dx}{\sqrt{\sin(x)}}$ converge while $\int_{0}^{\pi} \frac{dx}{\sin(x)}$ diverges?

Question

Essentially for me, the question boils to down finding a convergent majorant for $$\left|\frac{1}{\sqrt{\sin(x)}}\right|$$ in the interval $(0, \pi)$ thereby proving convergence of $$ \int_{0}^{\pi} \frac{dx}{\sqrt{\sin(x)}}. $$ I can't think of a good majorant though.

Answer 1

Pretty much for the same reason as $$\int_0^1\frac{dt}{\sqrt t}$$ converges and $$\int_0^1\frac{dt}{t}$$ diverges. Near zero $1/\sqrt{\sin t}$ is within a constant multiple of $1/\sqrt t$ and $1/\sin t$ is within a constant multiple of $1/t$. Since $\sin(\pi-t)=\sin t$, the behaviour of these integrals near $\pi$ is the same as near $0$.

Answer 2

I would do this with $1-\frac{8}{\pi^2}\frac{(x-\frac{\pi}{2})^2}{2}$.

It has the same values in 0, $\frac{\pi}{2}$ and in $\pi$ as $\sin(x)$, while it is a polynom.

I acquired it by simply scaling the first 2 non-zero Taylor-components of $\cos(x-\frac{pi}{2})$ and scaling it to the required parameters.

Being a 2nd-grade polynom, it is not yet hopeless to integrate its recipe.