Why the norm of this convolution can be written like this? Let $f_a : \mathbb{R}\setminus \{0\}\to \mathbb{R}$ be given by
$$f_a(x) = \dfrac{\sin(ax)}{\pi x}.$$
I've read that it is possible to write the $L^2$ norm of the convolution $f_a\ast f_b$ as follows:
$$\|f_a\ast f_b\|^2=\int_{-\infty}^\infty \dfrac{\sin(ax)\sin(bx)}{\pi^2 x^2}dx$$
Now why is this true? I mean, by definition
$$f_a\ast f_b(x)=\int_{-\infty}^\infty f_a(y)f_b(x-y)dy$$
So the norm is the integral of one integral and I don't see why it reduces to that simple formula. I mean, I tried writing the convolution explicitly
$$\|f_a\ast f_b\|^2=\int_{-\infty}^\infty \left(\int_{-\infty}^\infty \dfrac{\sin(ay)}{\pi y}\dfrac{\sin(b(x-y))}{\pi(x-y)}dy\right)^2 dx$$
but this doesn't seem to lead anywhere.
Why this convolution can be simplified like that? What is the point with this?
 A: Let $f(x)\iff F(k)$ be the Fourier Transform pair defined by
$$\begin{align}
F(k)&=\int_{-\infty}^\infty f(x)e^{ikx}\,dx\tag1 \\\\f(x)&=\int_{-\infty}^\infty F(k)e^{-ikx}\,\frac{dk}{2\pi}\tag 2
\end{align}$$

Parseval's Theorem states that for the Fourier Transform pair $f(t)\iff F(k)$ as defined in $(1)$ and $(2)$ we have
$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty |f(x)|^2\,dt=\frac{1}{2\pi}\int_{-\infty}^\infty |F(k)|^2\,dk} \tag 3$$
From the convolution theorem, if $f(x)=(f_a*f_b)(x)$, then $F(k)=F_a(k)F_b(k)$.  Using $(3)$ we find that
$$\int_{-\infty}^\infty \left| (f_a*f_b)(x)\right|^2\,dx=\frac{1}{2\pi} \int_{-\infty}^\infty \left|F_a(k)F_b(k)\right|^2\,dk \tag 4$$
Inasmuch as the Fourier Transform of $f_a(x)$ is $F_a(k)=\text{rect}(k/2a)$, where $\text{rect}(t)$ is the rectangle function given by
$$\text{rect}(t)=\begin{cases}1&,|t|<1/2\\\\0&,|t|>1/2\end{cases}$$
then applying $(4)$ reveals
$$\begin{align}
\int_{-\infty}^\infty \left| (f_a*f_b)(t)\right|^2\,dt&=\frac{1}{2\pi}\int_{-\infty}^\infty \left|\text{rect}(\omega/2a)\text{rect}(\omega/2b) \right|^2\,d\omega\\\\
&=\frac{\min(a,b)}{\pi} \tag 5
\end{align}$$

Now, we let $I(a,b)=\int_{-\infty}^\infty \frac{\sin(ax)\sin(bx)}{\pi^2x^2}\,dx$.  Then, Differentiating Under The Integral yields
$$\begin{align}
\frac{\partial I(a,b)}{\partial b}&=\frac{1}{\pi}\int_{-\infty}^\infty \cos(bx)\frac{\sin(ax)}{\pi x}\,dx\\\\
&= \frac{1}{\pi}\int_{-\infty}^\infty e^{ibx}\frac{\sin(ax)}{\pi x}\,dx\\\\
&=\frac{1}{\pi}F(b)\\\\
&=\frac1\pi \text{rect}(b/2a) \tag 6
\end{align}$$
Integrating $\frac{\partial I(a,b)}{\partial b}$ as given in $(6)$, we obtain
$$\begin{align}
I(a,b)&=\int_0^b \frac1\pi \text{rect}(x/2a)\,dx\\\\
&=\frac{\min(a,b)}{\pi}\tag 7
\end{align}$$
which is identical to $(5)$.  

Hence, we have from equating $(5)$ and $(7)$

$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty \left| (f_a*f_b)(t)\right|^2\,dt=\int_{-\infty}^\infty \frac{\sin(ax)\sin(bx)}{\pi^2x^2}\,dx}$$

as was to be shown!
