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

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?

• The convolution theorem is the right way, maybe if you included your work we can explain where id you made a mistake – Elaqqad Apr 11 '15 at 17:51
• I recently posted a similar question! math.stackexchange.com/questions/2950772/… – user150203 Dec 14 '18 at 10:27

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 $L^1\cap L^2(\mathbb{R})$.
• @MathsMeister: there is no general path in computing $$(f * g)(x)=\int_{\mathbb{R}} f(y)g(x-y)\,dy,$$ but in this case one of the two functions is pretty easy to handle since it is the characteristic function of an interval, and every polynomial times an exponential function is not difficult to integrate. – Jack D'Aurizio Apr 12 '15 at 12:43
• Are you sure the convolution in line (4) is $$(f_1 * f_2)(x) = \left\{\begin{array}{rcl}e^{-|x|}\left(e+\frac{1}{e}\right)&\text{if}& |x|\geq 1,\\ 2-\frac{1}{e}\left(e^x+e^{-x}\right)&\text{if}&|x|\leq 1,\end{array}\right. \tag{4}$$ because shouldn't it be $$(f_1 * f_2)(x) = \left\{\begin{array}{rcl}e^{-|x|}\left(e-\frac{1}{e}\right)&\text{if}& |x|\geq 1,\\ 2-\frac{1}{e}\left(e^x+e^{-x}\right)&\text{if}&|x|\leq 1,\end{array}\right. \tag{4}$$ ? – Maths Meister Apr 25 '15 at 21:26