TITLE: $ \int_{-\infty}^{+\infty}\sin(x^2)\cos2\alpha x \, \mathrm{d}x $
QUESTION [6 upvotes]: 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!

REPLY [7 votes]: 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)