Calculate the Fourier transform of the function
$$ f(x) = \begin{cases} \sin{ax} , & |x| \le \frac{\pi}{2} \\ 0, & |x| > \frac{\pi}{2} \end{cases} $$
Hence using Parseval's formula
$\int _{-\infty}^{\infty} |f(x)|^2$ $dx$ $=\frac{1}{2\pi} \int _{-\infty}^{\infty} |\hat{f}(\xi)|^2$ $d\xi$,
Show that $\int _{-\infty}^{\infty} \frac{x^2 \cos^2(\frac{\pi x}{2})}{(x^2-1)}$ $dx$ $=\frac{\pi ^2}{4}$
$F[f(x)]=\int^{\frac{\pi}{2a}}_{-\frac{\pi}{2a}} e^{-i\xi x}.\sin{ax}$ $dx$
$=\frac{1}{2i}\int^{\frac{\pi}{2a}}_{-\frac{\pi}{2a}} e^{-i\xi x}[e^{iax}-e^{-iax}]$ $dx$
$=\frac{1}{2i}[\frac{e^{x(ai-i\xi)}}{ai-i\xi} + \frac{e^{x(-ai-i\xi)}}{ai+i\xi}]^{\frac{\pi}{2a}}_{-\frac{\pi}{2a}}$
Simplifying further,
$\frac{[ai e^{(\frac{\pi i}{2}-\frac{\pi i \xi}{2a})}+i \xi e^{(\frac{\pi i}{2}-\frac{\pi i \xi}{2a})}+ai e^{(-\frac{\pi i}{2}-\frac{\pi i \xi}{2a})}- i \xi e^{(-\frac{\pi i}{2}-\frac{\pi i \xi}{2a})}- ai e^{(-\frac{\pi i}{2}+\frac{\pi i \xi}{2a})}- i \xi e^{(-\frac{\pi i}{2}+\frac{\pi i \xi}{2a})}-ai e^{(\frac{\pi i}{2}+\frac{\pi i \xi}{2a})}+ i \xi e^{(\frac{\pi i}{2}+\frac{\pi i \xi}{2a})}]}{2i (\xi^2-a^2)}$
I do not know how to proceed longer.
I am used to evaluate F. transform of $\sin ax$ using dirac delta:
$\frac{F[\delta(x+a)-\delta(x-a)]}{2i}=\sin {a \xi}$
I then use symmetry
$ \sin \xi \leftrightarrow 2 \pi .\frac{[\delta(-\xi+a)-\delta(-\xi-a)]}{2i} $
$ \sin \xi \leftrightarrow i \pi \frac{[\delta(\xi+a)-\delta(\xi-a)]}{2i}$ (I ommitted some obvious steps since dirac delta is positive)
Please help. Following that, I should be able to carry forward for the Parseval's formula.