Using a Fourier transform to evaluate a sinc^2 integral I am working through some questions in preparation for an upcoming exam and I am really struggling with the following question:
Find the Fourier transform of the function: 
$f(x) = \Theta(x+1) - \Theta(x-1)$
and use it to evaluate $\int_{-\infty}^{\infty} sin^2(x)/x^2 dx$
The first step appeared to be simple; $f(x)$ is a top hat function $T(x;-1,1)$ and I worked the FT out to be $F(k) = 2sin(k)/k$.
This meant the integral could be written as $0.25\int_{-\infty}^\infty F(k)^2 dk$. This is where I became lost; this screams convolution to me but I don't understand how to apply the convolution here to get a solution. I am interested in the method that needs to be used. 
I understand that $\hat C(k) = F(k)F(k)$ but again what method must be used to make this solve the problem?
Thanks in advance! 
 A: Let the Fourier Transform pair $f(x)\leftrightarrow F(k)$ be given by
$$\begin{align}
F(k)&=\int_{-\infty}^{\infty}f(x)e^{ikx}\,dx\\\\
f(x)&=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(k)e^{-ikx}\,dk
\end{align}$$
Then, we have the transform pair 
$$\Theta(x+1)-\Theta(x-1)\leftrightarrow \frac{2\sin(k)}{k}$$
Denote by $P(x)$ the pulse function $P(x)=\Theta(x+1)-\Theta(x-1)$.   Now, let $I(x)$ be given by 
$$\begin{align}
I(x)&=\int_{-\infty}^{\infty}\frac{\sin^2(k)}{k^2}e^{-ikx}\,dk\\\\
&=\frac{\pi}{2}\,\color{blue}{\underbrace{ \frac{1}{2\pi}\int_{-\infty}^\infty \left(\frac{2\sin(k)}{k}\right)\left(\frac{2\sin(k)}{k}\right)e^{-ikx}\,dk}_{\text{Inverse Fourier Transform of a Product}}}\\\\
&=\frac{\pi}{2}\color{blue}{\underbrace{(P*P)(x)}_{\text{Convolution of the Inverse Fourier Transforms}}}\\\\
&=\frac{\pi}{2}\color{blue}{\int_{-\infty}^\infty P(x-x')P(x')\,dx'}\\\\
&=\frac{\pi}{2}\color{blue}{\int_{-1}^1 P(x-x')\,dx'}\\\\
\end{align}$$
Finally, we have
$$\int_{-\infty}^\infty\frac{\sin^2(x)}{x^2}\,dx=I(0)=\pi$$
And we are done!
