Convergence of improper integral $\to\pi$

Question

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

$\sin$ is converging, but the rest has infinite limit in this segment, while $ \to\pi$, so Abel doesn't work. Dirichlet is not working either. I tried to bound this integral, but unsuccessfully. Not sure what else to use. Any throughts?

Answer 1

Denote $u= \pi -x$. You have $$f(x)= \frac{\sin^2x}{\pi^2 - x ^2} = f(\pi - u) = \frac{\sin^2 u}{\pi^2-(\pi-u)^2}= \frac{\sin^2 u}{2 \pi u-u^2}.$$

Therefore $$\lim\limits_{u \to 0} \frac{\sin^2 u }{2 \pi u - u^2} = 0$$ as $\sin u \sim u$ around $0$.

Conclusion, the function under the integral sign is continuous at $\pi$ and the integral converges.

Answer 2

Split the integral in half and apply $x\mapsto\pi-x$ to the $x>\pi/2$ integral. You'll see each integral ends up with an integrand whose numerator and denominator are respectively $O(x^2)$ and $O(x^{0\,\text{or}\,1})$ for small $x$, and the integrals converge.