An extraordinary sin integral

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.


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

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

where a, b>0. Taking the imaginary part of the last expression we get that

    \[\int_{-\infty}^{\infty} \sin \left ( x^2 + \frac{1}{x^2} \right ) \, {\rm d}x = \sqrt{\frac{\pi}{2}} \left ( \sin 2 + \cos 2 \right )\]

and this is the final answer. See, no e!. Of course we can also extract the real part and calculate the corresponding integral involving \cos.

Read more


Leave a Reply