Evaluate the integral $\int_{-\infty}^{\infty}{\left| 2t\cdot\text{sinc}^2(2t)\right|^2}\,dt$ I have a question in solving the integral
$$\int_{-\infty}^{\infty}{\left| 2t\cdot\text{sinc}^2(2t)\right|^2}\,dt.$$
I know that you can use Parseval's Theorem to prove that $\int_{-\infty}^{\infty}\text{sinc}^4(kt) = \frac{2\pi}{3k}$ (I can provide the proof if you would like). While this might seem to help, I don't know that I can really use that fact with the $2t$ in the equation because you'll have to integrate by parts.
Is there another way that I could do this integral to make it easier? I feel like I might be overthinking it and this can be simplified down to something much more manageable.
Thank you so much for your help in advanced, I really appreciate it!!! 

Integration by parts is enough. Our integral equals
$$ \int_{-\infty}^{+\infty}\frac{\sin^4(2x)}{4x^2}\,dx=\frac{1}{2}\int_{-\infty}^{+\infty}\frac{\sin^4(u)}{u^2}\,du=\int_{-\infty}^{+\infty}\frac{2\cos(u)\sin^3(u)}{u}\,du $$
but:
$$ 2\cos(u)\sin^3(u)=\frac{1}{2}\sin(2u)-\frac{1}{4}\sin(4u)\,du $$
hence:
$$ \int_{-\infty}^{+\infty}\frac{\sin^4(2x)}{4x^2}\,dx = \left(\frac{1}{2}-\frac{1}{4}\right)\pi = \color{red}{\frac{\pi}{4}}.$$
