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
$$f(x) = \frac{1}{2\pi}\int_{\mathbb{R}}\hat{f}(t)e^{ixt}dt.$$
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.