# Using the inversion formula of a Fourier Transform to calculate a limit

Define $f:\mathbb{R}\rightarrow\mathbb{R}$ by

$$f(x)=\begin{cases} 1-|x|/2 \quad \text{if} \ |x|\leq 2, \\ 0 \qquad \quad\quad\ \text{otherwise}. \end{cases}$$ Calculate the fourier transform of $f$ and hence, using inversion formula, show that
$$\int_{-\infty}^\infty \frac{\sin^2 (t)}{t^2}dt =2\pi.$$ I computed the fourier transform to be
$$\hat{f}(t) = \frac{1-\cos(2t)}{t^2} = \frac{2\sin^2(t)}{t^2}.$$
I'm not sure how to use it now. I thought that the inversion would give us our old function $f$ back... let alone a constant value.

The definition of the inversion formula is
$$f(x) = \frac{1}{2\pi}\int_{\mathbb{R}}\hat{f}(t)e^{ixt}dt.$$

• Just take $x=0$. – uniquesolution Oct 17 '17 at 12:00

The inversion formula give you : $$\forall x\in \mathbb{R},\quad f(x) =\frac{1}{2\pi}\int_{-\infty}^{+\infty}\hat{f}(t)e^{ixt}\mathrm{d}t =\frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{\sin^2(t)}{t^2}e^{ixt}\mathrm{d}t$$ Hence, $$f(0) =1= \frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{\sin^2(t)}{t^2}\mathrm{d}t$$ So $$\int_{-\infty}^{+\infty}\frac{\sin^2(t)}{t^2}\mathrm{d}t = \pi .$$
It appears that there is not $2$ if your Fourier transform is correct.
• Thank you. Could you please double check my Fourier Transform? I used wolframalpha.com/input/?i=integral+from+-2+to+0+of+(1%2Bx%2F2)*e%5E(-ixa)dx+%2B+integral+from+0+to+2+of+(1-x%2F2)*e%5E(-ixa)dx to compute my integral and it seems to get what I got... The definition the fourier transform given is $$\hat{f}(t) = \int_{\mathbb{R}} f(x)e^{-ixt}dx$$ – Twenty-six colours Oct 18 '17 at 0:48
• I've checked and your Fourier Transform is correct. The answer of $\int_\mathbb{R}\sin^2(t)/t^2 dt$ is $\pi$ (I've checked it with maple). – Zanzi Oct 18 '17 at 9:32