Evaluate $\int_{0}^\infty \sin\left(\frac{1}{x}\right)\sin\left(x^2\right)\,dx$

Question

I don't really have much a reason for giving this question other than curiosity. I am looking to evaluate

$$\int_{0}^\infty \sin\left(\frac{1}{x}\right)\sin\left(x^2\right)\,dx$$

Using the complex definition of $\sin(x)$ doesn't seem to work, nor can I give a reasonable differentiating under the integral sign. Also, it doesn't seem Wolfram Alpha has a closed answer, but I would be curious to see if there is one.

Answer 1

This is too long for a comment.

Just out of curiosity, I used a CAS to see what could be the result of $$I_{p,q}=\int_{0}^\infty \sin(\frac{1}{x^p})\sin(x^q)dx$$ and observed (for limited ranges of $p$ and $q$) that, if $p\neq q$, the result is either a Meijer G-function (as Phoenix already commented) or a nasty linear combination of hypergoemetric functions.

To my surprise, the very first cases where $p=q$ leads to quite simple results $$I_{1,1}=\frac{\pi }{2}J_1(2)$$ $$I_{2,2}=\frac{1}{4} \sqrt{\frac{\pi }{2}} \left(\frac{1}{e^2}+\sin (2)-\cos (2)\right)$$ $$I_{3,3}=\frac{1}{6} \left(\pi \left(J_{\frac{1}{3}}(2)-J_{-\frac{1}{3}}(2)\right)+\sqrt{3} K_{\frac{1}{3}}(2)\right)$$ $$I_{4,4}=-\frac{\sqrt{2+\sqrt{2}}}{16} \left(\sqrt{2} \pi \left(J_{-\frac{1}{4}}(2)-J_{\frac{1}{4}}(2)\right)-2 K_{\frac{1}{4}}(2)\right)$$ where appear Bessel functions.

Answer 2

$\int_0^\infty\sin\dfrac{1}{x}\sin x^2~dx$

$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin x^2}{(2n+1)!x^{2n+1}}~dx$

$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin x}{(2n+1)!(\sqrt x)^{2n+1}}~d(\sqrt x)$

$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin x}{2(2n+1)!x^{n+1}}~dx$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^{n+1}(n+1)!\sin\dfrac{n\pi}{2}}{2(2n+1)!}$ (according to http://eqworld.ipmnet.ru/en/auxiliary/inttrans/FourSin2.pdf)

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^{n+1}(n+1)\sin\dfrac{n\pi}{2}}{2^{2n+1}\left(\dfrac{3}{2}\right)_n}$ (according to http://mathworld.wolfram.com/PochhammerSymbol.html)

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^{n+1}(2n+2)}{2^{4n+3}\left(\dfrac{3}{2}\right)_{2n+1}}$