Calculate $\int _0^\infty \frac{\sin^2 (\pi \omega)}{(\omega ^2 -1)^2}d \omega$

Question

We define $f: \mathbb{R}\to \mathbb{R} $ by: $$f(x)=\left\{\begin{matrix} \sin x, & |x|\leq \pi\\ 0, & |x|>\pi \end{matrix}\right.$$ A. Find the Fourier transform of $f$. Answer: $$F(\omega)=\frac{2 i \sin(\omega \pi)}{\sqrt{2\pi}(\omega^2-1)}$$ B. Calculate the integral $$\int _0^\infty \frac{\sin^2 (\pi \omega)}{(\omega ^2 -1)^2}d \omega$$ Any help? I am referred this "identity" but I don't know what it means and how to use it. $$||f||_2=||F(\omega)||_2$$

Answer 1

A. $$\widehat{f}(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\pi}^{\pi}e^{-i\omega x}\sin x\,dx = \frac{1}{\sqrt{2\pi}}\Im\left(\int_{-\pi}^{+\pi}e^{(1-i\omega)x}\,dx\right)=\frac{2i\sin(\pi\omega)}{\sqrt{2\pi}(\omega^2-1)}.$$ B. Since Parseval's theorem gives: $$\int_{-\infty}^{+\infty}|\widehat{f}(\xi)|^2\,d\xi = \int_{-\infty}^{+\infty}|f(x)|^2\,dx = \int_{-\pi}^{\pi}\sin^2 x\,dx = \pi,$$ but the LHS of the last identity is: $$\frac{2}{\pi}\int_{-\infty}^{+\infty}\frac{\sin^2(\pi\omega)}{(\omega^2-1)^2}\,d\omega = \frac{4}{\pi}\int_{0}^{+\infty}\frac{\sin^2(\pi\omega)}{(\omega^2-1)^2}\,d\omega,$$ it follows that:

$$\int_{0}^{+\infty}\frac{\sin^2(\pi\omega)}{(\omega^2-1)^2}\,d\omega=\frac{\pi^2}{4}.$$