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.

  • $\begingroup$ Wolfram indicates a closed form using the Meijer G-function... However your function seems not integrable, you must consider the principal value. $\endgroup$
    – Célestin
    Jan 3, 2018 at 4:11
  • $\begingroup$ This may or may not help, but have you considered using the basic trig identity $\sin(A)\sin(B) = \frac{1}{2}\left[\cos(A - B) + \cos(A + B) \right]$ to yield $\sin(1/x)\sin(x^2) = (1/2)(\cos(1/x - x^2) + \cos(1/x + x^2))$ As before, may not be of help $\endgroup$
    – user150203
    Nov 20, 2018 at 1:46

2 Answers 2


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.

  • 2
    $\begingroup$ The simple expression for $I_{2,2}$ follows from the Glasser's master formula. While $I_{1,1}$ is a consequence of integral representation of bessel function. $\endgroup$
    – pisco
    Jan 3, 2018 at 5:04
  • $\begingroup$ @pisco125. Thanks for the interesting comment ! $\endgroup$ Jan 3, 2018 at 5:06
  • $\begingroup$ Thanks for the thoughts! $\endgroup$
    – Tom Himler
    Jan 3, 2018 at 15:46

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

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

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

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

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


  • $\begingroup$ you cannot integrate $\int_0^\infty\frac{\sin x}{x^{n+1}}\,dx$ which diverges for $n\ge 1$ $\endgroup$
    – Paul Enta
    Nov 6, 2019 at 15:07

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