Integral of $\int_{-\infty}^{+\infty}\left |{\frac{\sin{x}}{x(1+x^2)}}\right|^2\,dx $

Question

So the first part of the questions asks us to find the Fourier Transform of $$ f(x) = \left\{ \begin{array}{ll} e^{y} & \quad {-\infty}<x < 0 \\ e^{-y} & \quad 0<x < {\infty} \\ \end{array} \right. $$ which is $$ \frac{2}{1+k^2}$$

then it says work out the Fourier Transform of $$ g(x) = \left\{ \begin{array}{ll} 1 & \quad |x| \leq 2\pi\ \\ 0 & \quad \text{otherwise} \end{array} \right. $$

which is

$$ \frac{2\sin(2k\pi)}{k}$$

The final part says calculate the integral of

$$\int_{-\infty}^{+\infty} \left|{\frac{\sin{x}}{x(1+x^2)}}\right|^2\,dx $$

Now, I think I probably first have to use the convolution theorem to calculate $h(x)=f(x)\cdot g(x)$ then use Parseval's Theorem to get the integral but for some reason it doesn't seem to give a convergent answer? Can someone please help explain how I get it to converge?

Answer 1

Just in order to fix notations, let: $$ \mathcal{F}(f)(\omega) = \widehat{f}(\omega) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}f(x)\, e^{-i\omega x}\,dx.\tag{1} $$ Assuming $f_1(x)=e^{-|x|}$ and $f_2(x) = \mathbb{1}_{[-1,1]}(x)$ we have: $$ \widehat{f_1}(\omega)=\sqrt{\frac{2}{\pi}}\frac{1}{\omega^2+1},\qquad \widehat{f_2}(\omega)=\sqrt{\frac{2}{\pi}}\frac{\sin\omega}{\omega}\tag{2}$$ hence: $$ I=\int_{\mathbb{R}}\left(\frac{\sin\omega}{\omega(\omega^2+1)}\right)^2\,d\omega = \frac{\pi}{2}\|\widehat{f_1}\cdot\widehat{f_2}\|_2^2=\frac{\pi}{2}\| f_1 * f_2 \|_2^2\tag{3}$$ and we just need to compute the convolution of $f_1$ and $f_2$: $$ (f_1 * f_2)(x) = \left\{\begin{array}{rcl}e^{-|x|}\left(e+\frac{1}{e}\right)&\text{if}& |x|\geq 1,\\ 2-\frac{1}{e}\left(e^x+e^{-x}\right)&\text{if}&|x|\leq 1,\end{array}\right. \tag{4}$$ to be able to state: $$ I = \pi\left(\int_{1}^{+\infty}e^{-2x}\left(e+\frac{1}{e}\right)^2\,dx + \int_{0}^{1}\left(2-\frac{1}{e}\left(e^x+e^{-x}\right)\right)^2\,dx\right)\tag{5}$$ from which:

$$ I = \pi\left(\frac{1}{4}+\frac{5}{4e^2}\right)\tag{6} $$

follows. Notice that the convergence of the given integral is trivial since $\left|\frac{\sin x}{x}\right|\leq 1$ and $\frac{1}{x^2+1}$ belongs to $L^1\cap L^2(\mathbb{R})$.