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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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4 Answers 4

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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
    Commented May 14, 2013 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. $\endgroup$ Commented May 14, 2013 at 17:50
  • $\begingroup$ @O.L. Thanks! My story is actually true. $\endgroup$ Commented May 15, 2013 at 0:38
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    $\begingroup$ "Computable by Mathematica" is very unsatisfactory when one seeks to evaluate integrals for oneself $\endgroup$
    – FShrike
    Commented Jan 8, 2023 at 15:58
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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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  • $\begingroup$ 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 $\endgroup$ Commented Jul 8, 2016 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. $\endgroup$ Commented Jul 8, 2016 at 1:32
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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.
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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.

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