Prove convergence of an improper integral.

Question

How do I prove convergence of a following improper integral?

$$\int_0^\pi \frac {\sin^2x}{\pi^2-x^2} $$

I've already tried Abel's and Dirichlet's tests but they don't work here.

Answer 1

Near $\pi$, the integrand is $$\frac{\sin^2(\pi-x)}{(\pi-x)(\pi+x)}=\frac{\sin^2 y}{y(2\pi-y)}$$ where $y=\pi-x$. This tends to zero as $y\to 0$, that is as $x\to\pi$.

Answer 2

This answer is very similar to the answer of @Lord, my intention is just put some more context to it. The function $$f:(0,\pi)\to\Bbb R,\quad x\mapsto \frac{\sin^2(x)}{\pi^2-x^2}$$

can be continuously extended to the domain $[0,\pi]$ because the lateral limits at the boundary points exists, so the improper integral is equivalent to a proper integral of Riemann of the function $f$ continuously extended to $[0,\pi]$.