Calculate $\int_{-\infty}^{\infty} \left(\frac{\sin(ax)}{x} \right)^3\,\mathrm dx$ using properties of the Fourier transform. $\mathcal{F}$ will denote the fourier transform of $f$.
I have tried to occupy the following, putting $f(x)=\left(1-\frac{|x|}{a}\right)1_{[-a,a]}$ and $g(x)=1_{[-a,a]}$, $\forall x \in \mathbb{R}$ occupying the identity of Parseval $$\int \mathcal{F}_2f \mathcal{F}_2g=\int fg$$ But here I am stuck, any help or suggestion to solve the problem I will be very grateful.
Note that if $f(x) = \frac{1}{2}[\operatorname{sgn}(a-x) + \operatorname{sgn}(a+x)]$, then $$\mathcal{F}(f*f)(x) = [\mathcal{F}(f)]^2(x) = \frac{2}{\pi}\left[\frac{\sin(ax)}{x}\right]^2$$ Thus \begin{align*} \int_{-\infty}^\infty \left(\frac{\sin ax}{x}\right)^3\, dx &= \left(\frac{\pi}{2}\right)^{3/2}\int_{-\infty}^\infty \mathcal{F}(f*f)(x)\mathcal{F}(f)(x)\, dx\\ &= \left(\frac{\pi}{2}\right)^{3/2}\int_{-\infty}^\infty (f*f)(x)f(x)\, dx\end{align*} Compute explicitly the convolution product $f*f$ to simplify the latter integral.