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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?

Jack D'Aurizio
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Maths Meister
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  • 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! https://math.stackexchange.com/questions/2950772/using-laplace-transforms-to-evaluate-int-0-infty-frac-sin2xx2x2 –  Dec 14 '18 at 10:27

1 Answers1

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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 $L^1\cap L^2(\mathbb{R})$.

Jack D'Aurizio
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  • Thanks a lot for the explanation. I understand everything here except for how you got the convolution integral, seems like that was the problem I was having. What in general, would you say are steps in solving a convolution integral? – Maths Meister Apr 12 '15 at 04:03
  • @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