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

Question

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 1

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 $\displaystyle \int_0^{\infty}\frac{\sin 2x}{x}\mathrm{Ci}\,x\,dx$.

Answer 2

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.

Answer 3

EDIT: I added justification for changing the order of integration.

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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}$

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$(1)$ 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.