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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!

Michael Hardy
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user823011
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  • @Falcon Probably you should write square$^2$ before $x$ :) – VIVID Dec 24 '20 at 07:02
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    Use $\sin(x^2)\cos(2ax)=\frac 12 (\sin (x^2+2ax)+\sin(x^2-2ax))$. Then $\sin (x^2+2ax)=\sin((x+a)^2-a^2)=\sin((x+a)^2)\cos a^2-\cos((x+a)^2)\sin a^2$. Then use the known fact that $\sin(x^2)$ integral on $(-\infty, \infty)$ is computable... – Yuval Dec 24 '20 at 07:04
  • After @Yuval 's hint, you can use: https://math.stackexchange.com/questions/2062974/integral-of-sinx2-from-0-to-infinity?noredirect=1 – VIVID Dec 24 '20 at 07:06
  • Same approach as mine. I wrote the full proof in my answer. – FFjet Dec 24 '20 at 07:09

1 Answers1

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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)

VIVID
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FFjet
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