# evaluate the Fourier transform $\int^\infty _{-\infty} \frac{sint}{t}e^{iwt}dt$

I want to evaluate $$\int^\infty _{-\infty} \frac{sin^2t}{t^2}dt$$ using Parseval's identity. For that I first have to evaluate the Fourier transform $$\int^\infty _{-\infty} \frac{sint}{t}e^{iwt}dt$$ I am stuck here. How to evaluate this integral ?

• Every Fourier Analyst must know the Fourier transform of $\operatorname{sinc}$ is the box and vice versa. Commented Feb 24, 2017 at 16:39

You can use contour integration. First since the integrand is analytic, you can deform the contour so that it goes in a small semicircle below the origin to avoid it. Then you can split the sine up, giving $$\frac{1}{2i}\left(\int e^{i(w+1)t}\frac{dt}{t}-\int e^{i(w-1)t}\frac{dt}{t} \right).$$ Now on this contour, the integral $$\int e^{ikt} \frac{dt}{t}$$ is going to depend on whether $k$ is positive or negative.

If $k$ is positive, we must close the contour above (since that's the direction $e^{ikt}$ decays) and since we avoided the origin from below, there is the pole at the origin inside the contour. It has residue $1$ so the result is $2\pi i$

If $k$ is negative we must close the contour below. There are no poles in the contour so the result is zero.

So on the deformed contour we have $$\int e^{ikt} \frac{dt}{t} = 2\pi i\theta(k)$$ where $\theta$ is the step function.

So we can write the answer $$\frac{1}{2i}\left(2\pi i \theta(w+1)- 2\pi i\theta(w-1)\right) = \pi (\theta(w+1)-\theta(w-1)).$$

This is a rectangle function, with support on $[-1,1]$. If you knew this was the answer (and that transform duality holds) then it's much easier to compute the other direction (i.e. that the inverse fourier transform of the rectangle is $\sin(x)/x$).

What you most likely want is to use Plancherel's theorem which states that if $f\in L^2(\Bbb{R})$ then $\|\hat{f}\|_2^2=2\pi\| f\|_2^2$.

As noted in the comments, it is fairly well known that $\operatorname{sinc} t$ is the Fourier transform of $\frac{1}{2}_{[-1,1]}$. Hence \begin{align} \int^\infty _{-\infty} \frac{\sin^2t}{t^2}\,dt &= 2\pi \int^\infty _{-\infty} \frac{1}{4}_{[-1,1]}\,dt\\ &=\pi \end{align}