How to compute $\int_{-\infty}^{+\infty}(\frac{\sin x}{x})^2(\frac{\sin ax}{ax})^2dx, a>0$ please?
I want to use Fourier transform, Fourier transform define as $$\mathcal{F}[f(t)] = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}f(t)e^{ist}dt.$$
first, I write $$(\frac{\sin x}{x})^2(\frac{\sin ax}{ax})^2 = -\frac{e^{-2iax}}{8a^2x^4}-\frac{e^{2iax}}{8a^2x^4}-\frac{e^{-2ix}}{8a^2x^4}-\frac{e^{2ix}}{8a^2x^4}+\frac{e^{2iax-2ix}}{16a^2x^4}+\frac{e^{2iax+2ix}}{16a^2x^4}+\frac{e^{-2iax-2ix}}{16a^2x^4}+\frac{e^{2ix-2iax}}{16a^2x^4}+\frac{1}{4a^2x^4} (*)$$
then, I know $\mathcal{F}[\frac{1}{x}] = i\sqrt{\frac{\pi}{2}}\mathrm{sign}(x)$, and $\mathcal{F}[f^{(n)}(x)] = (-is)^n\mathcal{F}[f(x)]$, so I get $\mathcal{F}[\frac{1}{x^4}] = (-is)^3\mathcal{F}[-\frac{1}{6x}] = \frac{s^3}{6}\sqrt{\frac{\pi}{2}}\mathrm{sign}(s)$. Now I use $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\frac{1}{x^4}e^{isx}dx = \mathcal{F}[\frac{1}{x^4}] = \frac{s^3}{6}\sqrt{\frac{\pi}{2}}\mathrm{sign}(s),$$
so $$\int_{-\infty}^{+\infty}\frac{1}{x^4}e^{isx}dx = \frac{\pi s^3}{6}\mathrm{sign}(s).$$
Let $s = -2a$ and so on, I compute all the parts of (*). The result is same as Mathematica gives.
I use $\int_{-\infty}^{+\infty}\frac{1}{x^4}e^{isx}dx = \frac{\pi s^3}{6}\mathrm{sign}(s)$, but I know the integral can not converge, so how to fix it please?