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

Solution 1:

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 $K_\nu(x)$: $$\mathcal{S}=\frac{G_{2,4}^{4,2}\left(\frac18,\frac12\middle|\begin{array}{c}\frac12,\frac12\\0,0,\frac12,\frac12\\\end{array}\right)}{16\,\pi}-\frac\pi8K_0\left(\frac14\right).$$

Solution 2:

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