A closed-form expression for the integral $\int_0^\infty\text{Ci}^{3}(x) \, \mathrm dx$ 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.

Using the formula I mentioned here,
first, we notice that
$$
f(x)=\operatorname{Ci}(x),
\hat{f} (\omega)=\left\{\begin{matrix}
-\frac{\pi}{\left | \omega \right | }  &, \left | \omega \right |> 1.  \\
 -\frac{\pi}{2} &,\left | \omega \right |=1. \\
 0&,\text{otherwise}.
\end{matrix}\right.
$$
Therefore,
$$
\begin{aligned}
&\int_{0}^{\infty} \operatorname{Ci}(x)^3\text{d}x\\
=&\frac{1}{2} \int_{-\infty}^{\infty} \operatorname{Ci}(x)^3\text{d}x\\
=&\frac{1}{8\pi^2} \int_{\mathbb{R}^2}\hat{f}(x)\hat{f}(y)\hat{f}(x+y)\text{d}x\text{d}y\\
=&\frac{1}{8\pi^2}\sum_{i=1}^{6} \int_{U_i}\hat{f}(x)\hat{f}(y)\hat{f}(x+y)\text{d}x\text{d}y\\
=&\frac{1}{8\pi^2}\cdot6\pi^3\cdot(-2\ln2)\\
=&-\frac{3\pi}{2}\ln2.
\end{aligned}
$$
Where
$$
\begin{aligned}
&U_1:=\left \{ (x,y)\in\mathbb{R}^2\mid x\ge1,y\ge1\right \},\\
&U_2:=\left \{ (x,y)\in\mathbb{R}^2\mid x\le-1,y\le-1\right \},\\
&U_3:=\left \{ (x,y)\in\mathbb{R}^2\mid x\ge2,1-x\le y\le-1\right \},\\
&U_4:=\left \{ (x,y)\in\mathbb{R}^2\mid x\ge1,y\le-x-1\right \},\\
&U_5:=\left \{ (x,y)\in\mathbb{R}^2\mid x\le-1,y\ge1-x\right \},\\
&U_6:=\left \{ (x,y)\in\mathbb{R}^2\mid x\le-2,1\le y\le-x-1\right \}.\\
\end{aligned}
$$
And every integral equals $-2\pi^3\ln2$.


Generalizations:

*

*$\int_{0}^{\infty}\operatorname{Ci}(x)^4\text{d}x
=3\pi\operatorname{Li}_2\left ( \frac{2}{3}  \right ) 
+\frac{3\pi}{2}\ln^23.$

*$\int_{0}^{1} \frac{\operatorname{Ci}(x)}{\sqrt{1-x^2} } 
\text{d}x
=-\frac{\pi}{16}{}_2F_3 
\left ( 1,1;2,2,2;-\frac{1}{4}  \right ) +\frac{\pi\gamma}{2}-\frac{\pi}{2}\ln2.$

*$\int_{0}^{\infty}e^{-x} \frac{\operatorname{Si}(x)}{x} 
\text{d}x=C.$
$C$ denotes Catalan's constant.

