TITLE: A closed-form expression for the integral $\int_0^\infty\text{Ci}^{3}(x) \, \mathrm dx$
QUESTION [31 upvotes]: Is there a closed-form expression for this integral:
$$\int_0^\infty\text{Ci}^{3}(x) \, \mathrm dx,$$
where $\text{Ci}(x)=-\int_x^\infty\frac{\cos z}{z}\mathrm dz$ is the cosine integral?

$\text{Ci}(x)$ and $\text{Ci}^{2}(x)$ have primitives/antiderivatives that can be expressed in terms of the trigonometric integral functions.
So it's not too difficult to show that  $$\int_0^\infty\text{Ci}(x) \, \mathrm dx =0$$ and $$\int_0^\infty\text{Ci}^{2}(x) \, \mathrm dx = \frac{\pi}{2}.$$
But $\text{Ci}^{3}(x)$ doesn't appear to have a primitive that can be expressed in terms of known functions.

REPLY [2 votes]: EDIT: I added justification for changing the order of integration.

The following is an evaluation of $$I(4) = \int_{0}^{\infty}\operatorname{Ci}^{4}(x) \, \mathrm dx. $$
Integrating by parts twice, we get
$ \begin{align} I(4) &= 12 \int_{0}^{\infty} \frac{\sin (x) \cos (x)\operatorname{Ci}^{2}(x)}{x} \, \mathrm dx \\ &= 6 \int_{0}^{\infty}\frac{\sin(2x)\operatorname{Ci}^{2}(x) }{x} \, \mathrm dx \\ &= -6 \int_{0}^{\infty} \frac{\sin (2x)\operatorname{Ci}(x)}{x}\int_{1}^{\infty} \frac{\cos(xv) }{v} \, \mathrm dv \mathrm \, \mathrm dx \\& \overset{(1)}{=} - 6 \int_{1}^{\infty}\frac{1}{v} \int_{0}^{\infty} \frac{\sin(2x) \cos(vx)\operatorname{Ci}(x)}{x} \, \mathrm dx \, \mathrm dv \\ &=   6 \int_{1}^{\infty} \frac{1}{v} \left( \int_{0}^{\infty} \frac{\sin(2x) \cos(vx) }{x} \int_{1}^{\infty} \frac{\cos(xu)}{u}\, \mathrm du \, \mathrm dx \, \right)\mathrm dv \\& \overset{(2)}{=} 6 \int_{1}^{\infty} \frac{1}{v} \int_{1}^{\infty} \frac{1}{u} \int_{0}^{\infty} \frac{\sin (2x) \cos(ux) \cos(vx)}{x} \, \mathrm dx \, \mathrm du \, \mathrm dv \\ &= \scriptsize \frac{3}{2} \int_{1}^{\infty}\frac{1}{v} \int_{1}^{\infty} \frac{1}{u} \int_{0}^{\infty} \frac{\sin\left((2+u+v)x \right)+ \sin\left((2-u+v)x \right)+\sin\left((2+u-v)x\right)+\sin\left((2-u-v)x \right)}{x} \, \mathrm dx \, \mathrm du \, \mathrm dv \\ &\overset{(3)}{=} \frac{3 \pi}{4} \int_{1}^{\infty} \frac{1}{v} \int_{1}^{\infty} \frac{1+\operatorname{sgn}(2+u-v)+\operatorname{sgn}(2-u+v)+\operatorname{sgn}(2-u-v)}{u} \, \mathrm du \, \mathrm dv \\ &\overset{(4)}{=} \frac{3 \pi}{2} \left( \int_{1}^{3} \frac{1}{v}\int_{1}^{v+2} \frac{1}{u} \, \mathrm du \, \mathrm dv + \int_{3}^{\infty} \frac{1}{v} \int_{v-2}^{v+2} \frac{1}{u} \, \mathrm du \mathrm \, dv \right) \\ &= \frac{3 \pi}{2} \left(\int_{1}^{3} \frac{\ln(2+v)}{v} \, \mathrm dv + \int_{3}^{\infty} \ln \left(\frac{v+2}{v-2} \right)   \frac{\, \mathrm dv}{v} \right) \\ &\overset{(5)}{=} \frac{3 \pi}{2} \left(\int_{1}^{3} \frac{\ln(2+v)}{v} \, \mathrm dv+ \int_{0}^{1/3} \frac{\ln(1+2w)}{w} \, \mathrm dw - \int_{0}^{1/3} \frac{\ln(1-2w)}{w} \, \mathrm dw \right) \\ &= \small \frac{3 \pi}{2} \left(\int_{1}^{3} \frac{\ln(2)}{v} \, \mathrm dv + \int_{1}^{3}\frac{\ln \left(1 + \frac{v}{2} \right)}{v} \, \mathrm dv + \int_{0}^{1/3} \frac{\ln(1+2w)}{w} \, \mathrm dw - \int_{0}^{1/3} \frac{\ln(1-2w)}{w} \, \mathrm dw  \right) \\ &\overset{(6)}{=} \frac{3 \pi}{2} \left(\ln(2) \ln(3) - \operatorname{Li}_{2} \left(- \frac{3}{2}   \right) + \operatorname{Li}_{2} \left(- \frac{1}{2} \right)-\operatorname{Li}_{2} \left(- \frac{2}{3} \right) + \operatorname{Li}_{2} \left(\frac{2}{3}\right) \right) \\ & \overset{(7)}{=}\frac{3 \pi}{2} \left(\ln(2) \ln(3)  + \zeta(2)+ \frac{1}{2} \ln^{2} \left(\frac{2}{3} \right)  +\operatorname{Li}_{2} \left(-\frac{1}{2} \right) + \operatorname{Li}_{2}\left(\frac{2}{3}\right) \right)\\ & \overset{(8)}{=} \frac{3 \pi}{2} \left(\ln(2) \ln(3)  + \zeta(2) -\operatorname{Li}_{2} \left(\frac{1}{3} \right) + \operatorname{Li}_{2}\left(\frac{2}{3}\right) \right) \\ &\overset{(9)}{=} \frac{3 \pi}{2} \left(\ln(2) \ln(3) +  \ln \left(\frac{1}{3} \right) \ln \left(\frac{2}{3} \right)+2  \operatorname{Li}_{2}\left(\frac{2}{3}\right)   \right) \\ &= \frac{3 \pi}{2} \left(\ln^{2}(3) + 2 \operatorname{Li}_{2} \left(\frac{2}{3} \right) \right). \end{align}$

$(1)$ Apply Plancherel's theorem for the Fourier transform in the form $$\int_{\mathbb{R}} f(x) \hat{g}(x) \, \mathrm dx = \int_{\mathbb{R}} \hat{f}( \omega) g(\omega) \, \mathrm d \omega , $$ where $f$ and $g$ are square-integrable functions on $\mathbb{R}$.
$(2)$ Apply Plancherel's theorem again.
$(3)$ $\int_{0}^{\infty} \frac{\sin (ax)}{x} \, \mathrm dx = \frac{\pi}{2} \operatorname{sgn}(a)$
$(4)$ See the contour plot here.  Inside the yellow strip, $$1+\operatorname{sgn}(2+u-v)+\operatorname{sgn}(2-u+v)+\operatorname{sgn}(2-u-v) =2.$$  Inside the two red triangles, the value of the above function is $0$.
$(5)$ Make the substituion $w = \frac{1}{v}$ in the second integral.
$(6)$ $- \int_{0}^{x} \frac{\ln(1-yt)}{t} \, \mathrm d t= - \int_{0}^{xy} \frac{\ln(1-u)}{u} \, \mathrm du = \operatorname{Li}_{2} (xy)$
$(7)$  Apply the dilogarithm inversion formula.
$(8)$ Apply Landen's identity.
$(9)$ Apply the dilogarithm reflection formula.