$ \int_{-\infty}^{+\infty}\sin(x^2)\cos2\alpha x \, \mathrm{d}x $

Question

How to compute $$ \int_{-\infty}^{+\infty}\sin(x^2)\cos2\alpha x\,\mathrm{d}x $$

The integrating by parts does not work. Write $$ I(\alpha)=\int_{-\infty}^{+\infty}\sin(x^2)\cos2\alpha x\,\mathrm{d}x, $$ then $$ I'(\alpha)=-\int_{-\infty}^{+\infty}2x\sin(x^2)\sin2\alpha x\,\mathrm{d}x. $$

I do not kwow how to go on. Appreciate any help!

Answer 1

First, let's notice that $$ \int_{-\infty}^{\infty}{\sin}\left( x^2 \right) \cos 2ax\,dx=\frac{1}{2}\int_{-\infty}^{\infty}{\sin}\left( x^2+2ax \right) +\sin \left( x^2-2ax \right) \,dx $$ For the first integral, substitute $u=x+a$, we have $$ \int_{-\infty}^{\infty}{\sin}\left( x^2+2ax \right) \,dx=\int_{-\infty}^{\infty}{\sin}\left( u^2-a^2 \right) \,du = \int_{-\infty}^{\infty}{\sin}\left( u^2 \right) \cos a^2-\cos \left( u^2 \right) \sin a^2\,du $$ And a similar process works for the second integral. Finally, $$\begin{aligned} \int_{-\infty}^{\infty}{\sin}\left( x^2 \right) \cos 2ax&=\cos a^2\int_{-\infty}^{\infty}{\sin}\left( u^2 \right) \,du-\sin a^2\int_{-\infty}^{\infty}{\cos}\left( u^2 \right) \,du\\&\overset{(1)}=\sqrt{\frac{\pi}{2}}\left( \cos a^2-\sin a^2 \right)\end{aligned} $$ _{Proof of (1)}