TITLE: Closed form for $\int_0^\infty\frac{\sin x\,\cdot\,\operatorname{Ci}x-\cos x\,\cdot\,\operatorname{Si}x}{\sqrt{16\,x^2+1}}dx$
QUESTION [27 upvotes]: 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
$$\mathcal{S}\approx0.133456902778362645676629...$$

REPLY [11 votes]: 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:
$$
\mathcal{S}(a)=\int^{\infty}_{0}\frac{\sin x\operatorname{Ci}x-\cos x\operatorname{Si}x}{\sqrt{a^{-2}x^2+1}}dx=-\frac{a\pi}{2}\left.\frac{d}{d\nu}L_{\nu}\left(a\right)\right|_{\nu=0}.
$$