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

Solution 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".

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

Solution 2:

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,

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

Solution 3:

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.