<s>[INST] 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.
 [/INST] Expanding on Start wearing purple's answer, the following is an evaluation of $$\int_{0}^{\infty} \frac{\sin (2x)}{x} \, \operatorname{Ci}(x) \, dx .$$
First notice that by making the substitution $ u = \frac{t}{x}$, we get
$$ \operatorname{Ci}(x) = - \int_{x}^{\infty} \frac{\cos (t)}{t} \, dt = - \int_{1}^{\infty} \frac{\cos (xu)}{u} \, du.$$
Therefore,
$$ \int_{0}^{\infty} \frac{\sin (2x)}{x} \, \operatorname{Ci}(x) \, dx = - \int_{0}^{\infty} \int_{1}^{\infty} \frac{\sin (2x)}{x} \frac{\cos (xu)}{u} \, du \, dx .$$
Since the iterated integral does not converge absolutely, changing the order of integration is not justified by Fubini's theorem.
But by integrating by parts, we get
$$ \begin{align} \int_{0}^{\infty} \frac{\sin (2x)}{x} \, \operatorname{Ci}(x) \, dx &= \int_{0}^{\infty} \frac{\sin (2x) \sin (x)}{x^{2}} \, dx - \int_{0}^{\infty} \int_{1}^{\infty} \frac{\sin (2x) \sin (xu)}{x^{2}u^{2}} \, du \, dx \\ &= \int_{0}^{\infty} \frac{\sin (2x) \sin (x)}{x^{2}}  \, dx - \int_{1}^{\infty} \frac{1}{u^{2}}\int_{0}^{\infty} \frac{\sin (2x) \sin(ux)}{x^{2}} \, dx \, du \, .  \end{align}$$
In general, for $a,b \ge 0$, we have $$ \int_{0}^{\infty} \frac{\sin (ax) \sin (bx)}{x^{2}} \ dx = \frac{\pi}{2} \min \{a,b \} .$$
Therefore,
$$ \begin{align} \int_{0}^{\infty} \frac{\sin 2x}{x} \, \operatorname{Ci}(x) \, dx &= \frac{\pi}{2} \, \text{min} \{2,1 \} - \frac{\pi}{2} \int_{1}^{\infty} \frac{1}{u^{2}} \,  \text{min} \{2,u \} \, du  \\ &= \frac{\pi}{2} - \frac{\pi}{2} \int_{1}^{2} \frac{u}{u^{2}} \, du - \frac{\pi}{2} \int_{2}^{\infty} \frac{2}{u^{2}} \, du \\ &= \frac{\pi}{2} - \frac{\pi}{2} \,  \ln (2) - \frac{\pi}{2} \\ &= - \frac{\pi}{2} \, \ln (2) . \end{align}$$

UPDATE:
Integrating by parts wasn't necessary since $$- \int_{0}^{\infty} \int_{1}^{\infty} \frac{\sin (2x)}{x} \frac{\cos (xu)}{u} \, \mathrm du \, \mathrm dx = - \int_{1}^{\infty} \int_{0}^{\infty} \frac{\sin (2x)}{x} \frac{\cos (xu)}{u} \, \mathrm dx \, \mathrm  du$$ is justified by Plancherel's theorem for the Fourier transform in the form $$\int_{\mathbb{R}^{2}} \hat{f}(x) g(x) \, \mathrm dx = \int_{\mathbb{R}^{2}} f(\omega) \hat{g}(\omega) \, \mathrm d \omega. $$ (Some textbooks refer to this as the multiplication formula.)
Therefore, we can also say that $$ \begin{align}\int_{0}^{\infty} \frac{\sin (2x)}{x} \, \operatorname{Ci}(x) \, \mathrm dx &= - \int_{0}^{\infty} \int_{1}^{\infty} \frac{\sin (2x)}{x} \frac{\cos (xu)}{u} \, \mathrm du \, \mathrm dx   \\ &= - \int_{1}^{\infty} \int_{0}^{\infty} \frac{\sin (2x)}{x} \frac{\cos (xu)}{u} \, \mathrm dx \, \mathrm du \\ &= - \frac{1}{2} \int_{1}^{\infty} \frac{1}{u} \int_{0}^{\infty} \frac{\sin\left((2-u)x \right)+ \sin \left((2+u)x \right)}{x} \, \mathrm dx \, \mathrm du \\ &= -\frac{\pi}{4} \int_{1}^{\infty} \frac{1}{u} \left(\operatorname{sgn}(2-u)  +1\right) \, \mathrm du \\ &= - \frac{\pi}{2} \int_{1}^{2} \frac{\mathrm du }{u} \\  &=  - \frac{\pi}{2} \, \ln (2). \end{align}$$

In general, if $0 \le a \le 1$, then $$\int_{0}^{\infty} \frac{\sin (ax)}{x} \,  \operatorname{Ci}(x) \, \mathrm dx =  0.$$
And if $a >1$, then $$\int_{0}^{\infty} \frac{\sin (ax)}{x} \, \operatorname{Ci}(x) \, \mathrm dx = - \frac{\pi}{2} \int_{1}^{a} \frac{\mathrm du}{u} = - \frac{\pi}{2} \, \ln (a).  $$
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