Fourier transform $\operatorname{sinc}^2(100t)$

Question

$\DeclareMathOperator{\sinc}{sinc}$I try to do the Fourier transformation of $\sinc^2(100t)$ but am getting lost in between. I am using the following definitions from my study book $$\mathscr{F}[x(t)y(t)] = \frac{1}{2\pi}(X * Y)(\omega)$$ as well as $$\mathscr{F}[\sinc(t)] = \frac{\sin(\pi t)}{\pi t}$$

With that I come to the following approach

$$ \begin{align} \mathscr{F}[\sinc^2(100t)] &= \mathscr{F}[\sinc(100t)\sinc(100t)]\\&=\frac{1}{2\pi}\left[\frac{1}{100}I_{[-100\pi,100\pi]}(\omega)*\frac{1}{100}I_{[-100\pi,100\pi]}(\omega)\right]\\ &=\frac{1}{2\pi}\frac{1}{100}\operatorname{tri}\left(\frac{\omega}{100}\right)\\ &=\frac{1}{200\pi}\operatorname{tri}\left(\frac{\omega}{100}\right)\end{align}$$

While wolframalpha tells me something different

Answer 1

In terms of probability theory, the tent distribution, given by convolution of a uniform distribution over $(-c/2,c/2)$ with itself, i.e. $$ f(x)=\Big(\mathbb{1}_{(-c/2,c/2)}*\mathbb{1}_{(-c/2,c/2)}\Big)(x)=\frac{1}{c}\Big(1-\frac{|x|}{c}\Big)_+ \tag{1}\label{one} $$ has characteristic function $4\Big(\frac{\sin(tc/2)}{ct}\Big)^2$.

1. Check that $\eqref{one}$ is indeed correct.
2. Recall that $\mathcal{F}(f*g)=(\mathcal{F}f)\,(\mathcal{F}g)$
3. In probability theory, the characteristic function of a measure is defined as $\hat{\mu}(t)=\int e^{ixt}\mu(dx)$. In analysis, usually the Fourier transform of a measure is defined as $\mathscr{F}\mu(t)=\int e^{-2\pi itx}\mu(dx)$. So some constant factors are needed to get exactly what you need.