The question below is really bugging me ...Given that the following function,

$$f(x) = \Theta(x + 1) - \Theta(x - 1)\,,$$

(where $\Theta(x)$ is the [Heaviside step function](http://mathworld.wolfram.com/HeavisideStepFunction.html)) has a Fourier transform:

$$\widetilde{f}(k) = 2\frac{\sin k}{k}$$

where in this case we define the Fourier transform as

$$\widetilde{f}(k) = \int\_{-\infty}^{\infty} f(x) e^{-ikx}\,dx\,,$$

evaluate the following integral:

$$\int\_{-\infty}^{\infty}\frac{\sin^2x}{x^2}\,dx\,.$$

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So far, I've considered that the Fourier transform of $c(x) = f(x) \ast g(x)$ (with '$\ast$' denoting [convolution](http://mathworld.wolfram.com/Convolution.html)) is equal to $\widetilde{c}(k) = \widetilde{f}(k) \widetilde{g}(k)$ by the [convolution theorem](http://mathworld.wolfram.com/ConvolutionTheorem.html), hence if $g(x) = f(x)$, this gives a starting point. I have so far failed to manipulate any integrals into the correct form, though!

I feel like the solution is much simpler than I am imagining it to be ...