Is it possible to find a closed form for this integral? $$\mathcal{S}=\int_0^\infty\frac{\sin x\cdot\operatorname{Ci}x-\cos x\cdot\operatorname{Si}x}{\sqrt{16\,x^2+1}}dx,$$ where $\operatorname{Ci}x$ is the cosine integral and $\operatorname{Si}x$ is the sine integral: $$\operatorname{Ci}x=-\int_x^\infty\frac{\cos t}t dt,\ \operatorname{Si}x=\int_0^x\frac{\sin t}t dt.$$ Numerical integration gives $$\mathcal{S}\approx0.133456902778362645676629...$$

  • $\begingroup$ The form is tantalizing, but still I want to know if there is a particular reason why you are expecting it to have a closed form. Or can you suggest a context where it arises if it have such one? $\endgroup$ Nov 19, 2013 at 8:30
  • $\begingroup$ If I may ask, in what kind of problem did you find such an integral ? $\endgroup$ Nov 19, 2013 at 8:35
  • $\begingroup$ I do not have a strong reason to believe it has a closed form, but the integrand looks kinda nice. This integral arises as part of longer calculations related to physics (I wouldn't post them here for several reasons). $\endgroup$ Nov 19, 2013 at 17:46
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    $\begingroup$ Oh, surely we can always believe that an integral arising naturally in physics either has a close form or defines a nice function... $\endgroup$ Nov 19, 2013 at 22:05
  • $\begingroup$ According to eqworld.ipmnet.ru/en/auxiliary/inttrans/FourSin2.pdf, $\sin x\cdot\operatorname{Ci}x-\cos x\cdot\operatorname{Si}x$ may convert into some improper integrals. $\endgroup$ Nov 21, 2013 at 16:56

2 Answers 2


Not sure if you consider this a closed form, but the integral $\mathcal{S}$ can be expressed in terms of the generalized Meijer $G$-function (see formula $(3)$ here for the definition) and the modified Bessel function of the $2^{nd}$ kind $K_\nu(x)$: $$\mathcal{S}=\frac{G_{2,4}^{4,2}\left(\frac18,\frac12\middle|\begin{array}{c}\frac12,\frac12\\0,0,\frac12,\frac12\\\end{array}\right)}{16\,\pi}-\frac\pi8K_0\left(\frac14\right).$$


Simplifying Cleo's answer as in here, we get $$ \mathcal{S}=-\frac{\pi}{8}\left.\frac{d}{d\nu}L_{\nu}\left(\frac14\right)\right|_{\nu=0} $$

And a more general result: $$ \mathcal{S}(a)=\int^{\infty}_{0}\frac{\sin x\operatorname{Ci}x-\cos x\operatorname{Si}x}{\sqrt{a^{-2}x^2+1}}dx=-\frac{a\pi}{2}\left.\frac{d}{d\nu}L_{\nu}\left(a\right)\right|_{\nu=0}. $$


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