calculus - A closed-form expression for the integral $int_0^inftytext{Ci}^{3}(x) , mathrm dx$

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?

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$\text{Ci}(x)$ 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.

Answer

The answer is
$$\int_0^{\infty}\text{Ci}^3x\,dx=-\frac{3\pi\ln 2}{2}.$$
I would like to trade the method of evaluation for convincing story about what made this integral interesting for you. The story should be longer than "a friend of mine told it could be calculated in a closed form".

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Update: Not that I was really convinced by the comment below... but for those who would eventually like to figure it out:

1. Using that $\int\mathrm{Ci}\,x\,dx=x\,\mathrm{Ci}\,x-\sin x$, integrate once by parts. This yields two integrals:
   
   
   
   • $\displaystyle \int_0^{\infty}\frac{\sin 2x}{x}\mathrm{Ci}\,x\,dx=-\frac{\pi}{2}\ln 2$ (computable by Mathematica),
   
   
   • $\displaystyle \int_0^{\infty}\cos x \,\mathrm{Ci}^2x\,dx$
   
   


2. Integrating the 2nd expression once again by parts (with $u=\mathrm{Ci}^2x$, $v=\sin x$), one again reduces the problem to computing $\displaystyle \int_0^{\infty}\frac{\sin 2x}{x}\mathrm{Ci}\,x\,dx$.