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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.

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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".


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 $\displaystyle \int_0^{\infty}\frac{\sin 2x}{x}\mathrm{Ci}\,x\,dx$.
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    $\begingroup$ at least +1 for the trade proposition $\endgroup$ – mrf May 14 '13 at 12:17
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    $\begingroup$ @O.L. I knew that $\int_0^\infty\text{Ci}\ x\ \mathrm dx = 0$ and $\int_0^\infty\text{Ci}^2x\ \mathrm dx = \pi/2$, so I thought maybe there was a general formula for $\int_0^\infty\text{Ci}^nx\ \mathrm dx,\,n\in\mathbb{N}$ and tried to find it, but failed. So I decided to solve $\int_0^\infty\text{Ci}^3x\ \mathrm dx$ for starters, and try to generalize the result after that. No other secret reasons. $\endgroup$ – Marty Colos May 14 '13 at 17:50
  • $\begingroup$ @O.L. Thanks! My story is actually true. $\endgroup$ – Marty Colos May 15 '13 at 0:38
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Expanding on Start wearing purple's answer, the following is an evaluation of $$\int_{0}^{\infty} \frac{\sin (2x)}{x} \, \text{Ci}(x) \, dx .$$

First notice that by making the substitution $ \displaystyle u = \frac{t}{x}$, we get

$$ \text{Ci}(x) = - \int_{x}^{\infty} \frac{\cos (t)}{t} \, dt = - \int_{1}^{\infty} \frac{\cos (xu)}{u} \, du.$$

Therefore,

$$ \int_{0}^{\infty} \frac{\sin 2x}{x} \, \text{Ci}(x) \, dx = - \int_{0}^{\infty} \int_{1}^{\infty} \frac{\sin (2x)}{x} \frac{\cos (xu)}{u} \, du \, dx .$$

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

$$ \begin{align} \int_{0}^{\infty} \frac{\sin (2x)}{x} \, \text{Ci}(x) \, dx &= \int_{0}^{\infty} \frac{\sin (2x) \sin (x)}{x^{2}} \, dx - \int_{0}^{\infty} \int_{1}^{\infty} \frac{\sin (2x) \sin (xu)}{x^{2}u^{2}} \, du \, dx \\ &= \int_{0}^{\infty} \frac{\sin (2x) \sin (x)}{x^{2}} \, dx - \int_{1}^{\infty} \frac{1}{u^{2}}\int_{0}^{\infty} \frac{\sin (2x) \sin(ux)}{x^{2}} \, dx \, du \, . \end{align}$$

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,

$$ \begin{align} \int_{0}^{\infty} \frac{\sin 2x}{x} \, \text{Ci}(x) \, dx &= \frac{\pi}{2} \, \text{min} \{2,1 \} - \frac{\pi}{2} \int_{1}^{\infty} \frac{1}{u^{2}} \, \text{min} \{2,u \} \, du \\ &= \frac{\pi}{2} - \frac{\pi}{2} \int_{1}^{2} \frac{u}{u^{2}} \, du - \frac{\pi}{2} \int_{2}^{\infty} \frac{2}{u^{2}} \, du \\ &= \frac{\pi}{2} - \frac{\pi}{2} \, \ln 2 - \frac{\pi}{2} \\ &= - \frac{\pi}{2} \, \ln 2 . \end{align}$$

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  • $\begingroup$ I like that you made the formula into a link, but you have made it a bit hard to right-click for opening in a new tab. ;P $\endgroup$ – J. M. is a poor mathematician Jul 8 '16 at 1:05
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    $\begingroup$ @J.M. It appears that you have to right-click to the left or right of the formula. $\endgroup$ – Random Variable Jul 8 '16 at 1:32

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