Find $\int_0^\infty \frac{cos^2{k\pi / 2}}{(1-k^2)^2}$ using Parseval's theorem

Question

I'm having a little trouble finding the correct answer with this one. It is known from a previous problem that $$f(x) = \begin{cases} {{\cos{x}}} & {-\pi / 2 < x < \pi / 2} \\ {0} & {|x| > \pi / 2} \end{cases}$$ So finding this should be as simple as $$\frac{1}{2\pi}\int_{-\pi / 2}^{\pi / 2}|{\cos{x}}|^2dx\\$$ The correct answer is $\frac{\pi^2}{8}$ but I am getting the answer $\frac{1}{8}$ and am not sure why. If anyone can show me where I'm potentially going wrong that would be great!

Answer 1

You've got your normalization wrong. The Fourier transform of the function $$ f(x) = \begin{cases} \cos x & |x| < \pi/2\\0 & |x|>\pi/2\end{cases} $$ is $$ F(k) = 2\frac{\cos(k\pi/2)}{1-k^2} $$ Parseval's Theorem then says $$ \int_{-\infty}^{\infty}|f(x)|^2 dx = \int_{-\pi/2}^{\pi/2}(\cos x)^2dx = \frac{1}{2\pi}\int_{-\infty}^\infty |F(k)|^2dk = \frac{2}{\pi}\int_{-\infty}^\infty \frac{[\cos(\pi k/2)]^2}{(1-k^2)^2}dk $$ From this, we see that your integral is $$ \int_{0}^\infty \frac{[\cos(\pi k/2)]^2}{(1-k^2)^2}dk = \frac{\pi}{4}\int_{-\pi/2}^{\pi/2}(\cos x)^2dx = \frac{\pi^2}{8} $$