# Closed form for $\int_0^\infty\frac{\sin x\,\cdot\,\operatorname{Ci}x-\cos x\,\cdot\,\operatorname{Si}x}{\sqrt{16\,x^2+1}}dx$

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 $$\mathcal{S}\approx0.133456902778362645676629...$$

• 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? Nov 19, 2013 at 8:30
• If I may ask, in what kind of problem did you find such an integral ? Nov 19, 2013 at 8:35
• 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). Nov 19, 2013 at 17:46
• Oh, surely we can always believe that an integral arising naturally in physics either has a close form or defines a nice function... Nov 19, 2013 at 22:05
• 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. Nov 21, 2013 at 16:56

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