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An extraordinary sin integral

By Tolaso in Uncategorized on February 10, 2018.

I was surfing the net today and I fell on this cute integral

$\mathcal{J}= \int_{-\infty}^{\infty} \sin \left( x^2 + \frac{1}{x^2} \right) \, {\rm d}x$

I have seen integrals of such kind before like for instance this $\displaystyle \int_{0}^{\infty} \sin \left( x^2 - \frac{1}{x^2} \right) \, {\rm d}x$ .In fact something more general holds

$\int_0^{\infty}\sin \left( ax^2 - \frac{b}{x^2} \right) \, {\rm d}x =\frac{1}{2} \sqrt{\frac{\pi}{2a}} e^{-2 \sqrt{ab}}$

where $a, b>0$ .

Solution

The original integral does not fall into this category which is a real shame. Yet it does have a closed form and it does not contain an $e$ in its final answer. Strange, huh? So similar but so different at the same time these two integrals. A sign changes everything.

We begin by exploring the integral

$\mathcal{J}' = \int_0^\infty \exp \left( \frac{ia}{x^2} + ib x^2 \right) \, {\rm d}x$

Manipulating the integral ( substitutions and known Gaussian results) reveals that

$\int_{0}^{\infty} \exp \left (\frac{ia}{x^2} + i b x^2 \right ) \, {\rm d}x =\frac{1}{2} \sqrt{\frac{\pi}{b}}\exp\left ( \frac{i \pi}{4} \right ) e^{2i \sqrt{ab}}$