How do I calculate the 4th power of the Dirichlet integral with Fourier transforms? I'm pretty new to Fourier Transforms and I stumbled upon this exercise, which asks me to calculate this integral
$$\int_{-\infty}^{+\infty}\left(\frac{\sin(ax)}x\right)^4\,dx$$
The exercise suggests to use fourier transforms, but I really don't get how! I recognized that the term inside the parentheses is a Dirichlet integral, so I know its transform, which is $$\frac{1}2\sqrt\frac{\pi}2 \left(\text{sgn}(a-\lambda) +\text{sgn}(a+\lambda\right))$$ but I can't just multiply everything by $e^{-ia\lambda}$ as if were nobody's business! And then inside that integral I have the 4th power the transform, how do I deal with it?
Thanks a lot for your patience
 A: First note that enforcing the substitution $x\to x/a$, reveals
$$\int_{-\infty}^\infty \frac{\sin^4(ax)}{x^4}\,dx=a^3\int_{-\infty}^\infty \frac{\sin^4(x)}{x^4}\,dx$$

To find the integral $\int_{-\infty}^\infty \frac{\sin^4(x)}{x^4}\,dx=\int_{-\infty}^\infty \frac{\sin^2(x)}{x^2}\frac{\sin^2(x)}{x^2}\,dx$, we simply need to convolve the Fourier Transform of $\frac{\sin^2(x)}{x^2}$ with itself, and evaluate this at $\omega=0$.


PRIMER:  FOURIER-TRANSFORM PAIRS: 
We have the Fourier Transform pairs  
$$\begin{align}
f(x) &\leftrightarrow F(\omega)\\\\
f^2(x) &\leftrightarrow \frac{1}{\sqrt {2\pi}}F(\omega)*F(\omega)\\\\
\frac{\sin(x)}{x}&\leftrightarrow \sqrt{\frac{\pi}{2}}\text{rect}(\omega/2)\\\\
\frac{\sin^2(x)}{x^2}&\leftrightarrow \frac{1}{\sqrt{2\pi}}\left(\sqrt{\frac{\pi}{2}}\text{rect}(\omega/2)\right)*\left(\sqrt{\frac{\pi}{2}}\text{rect}(\omega/2)\right)=\sqrt{\frac\pi8}2 \text{tri}(\omega/2)
\end{align}$$
where $\text{rect}(t)$ and $\text{tri}(t)$ are the Rectangle Function and Triangle Function, respectively.


Therefore, we have 
$$\begin{align}
\int_{-\infty}^\infty \frac{\sin^4(x)}{x^4}\,dx &=\left.\left(\left(\sqrt{\frac\pi8}2 \text{tri}(\omega/2)\right)*\left(\sqrt{\frac\pi8}2 \text{tri}(\omega/2)\right)\right)\right|_{\omega =0}\\\\
&=\int_{-2}^2 \left(\sqrt{\frac{\pi}{8}}\frac{}{}(2-|\omega'|)\right)^2\,d\omega'\\\\
&=\left(\frac{\pi}{8}\right)\,2\int_0^2 (\omega'-2)^2\,d\omega'\\\\
&=\left(\frac{\pi}{8}\right)\,2\left(\frac83\right)\\\\
&=\frac{2\pi}{3}
\end{align}$$
Hence, we can assert that

$$\bbox[5px,border:2px solid #C0A000]{\int_{\infty}^\infty \frac{\sin^4(ax)}{x^4}\,dx=\frac{2\pi a^3}{3}}$$

