Evaluate $\int_{0}^{\infty}x\left ( \operatorname{Ci}(x)^2 +\operatorname{si}(x)^2\right )\operatorname{Ci}(x)\text{d}x$

Question

Evaluate $$ I_1=\int_{0}^{\infty}x\left ( \operatorname{Ci}(x)^2 +\operatorname{si}(x)^2\right )\operatorname{Ci}(x)\text{d}x. $$ Where $\operatorname{Ci}(x)=-\int_{x}^{\infty}\frac{\cos(t)}{t}\text{d}t,\operatorname{si}(x)=-\int_{x}^{\infty}\frac{\sin(t)}{t}\text{d}t$ are cosine integral function and (modified) sine integral function respectively.

By integrating numerically, one gives $I_1=0$. My evaluation is a bit complex, so I would like to see a canonical one. Thanks for replying.

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A related version: $$ I_2=\int_{0}^{\infty}x\left ( \operatorname{Ci}(x)^2 +\operatorname{si}(x)^2\right )\operatorname{si}(x)\text{d}x=\frac\pi2\left(1-2\ln(2)\right). $$

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A sophisticated version: $$ I_3=\int_{0}^{\infty}\left ( \operatorname{Ci}(x)^2+\operatorname{si}(x)^2\right )^2\left ( \operatorname{Ci}(x)^2+\operatorname{si}(x)^2+3\pi\operatorname{si}(x)\right ) \text{d}x =6\pi^3\operatorname{Li}_2\left ( \frac{1}{4} \right ) +12\pi^3\ln(2)^2. $$ Where $\operatorname{Li}_2(z)=\sum_{k=1}^{\infty}\frac{z^k}{k^2}$ is the dilogarithm.

Answer 1

We can make use of the following result:$$\boxed{\operatorname{Ci}^2(x)+\operatorname{si}^2(x)=\int_0^\infty \frac{e^{-xy}\ln(1+y^2)}{y}dy}$$

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$$I_1=\int_0^\infty x\left(\operatorname{Ci}^2(x)+\operatorname{si}^2(x)\right)\operatorname{Ci}(x)dx$$ $$=\int_0^\infty \int_0^\infty \frac{xe^{-xy}\ln(1+y^2)\operatorname{Ci}(x)}{y}dydx$$ $$=\int_0^\infty \frac{\ln(1+y^2)}{y}\left(\int_0^\infty x e^{-xy}\operatorname{Ci}(x)dx\right)dy$$ $$=\int_0^\infty \frac{\ln(1+y^2)}{y}\left(-\frac{d}{dy}\int_0^\infty e^{-xy}\operatorname{Ci}(x)dx\right)dy$$ $$=\frac12\int_0^\infty \frac{\ln(1+y^2)}{y}\frac{d}{dy}\left(\frac{\ln(1+y^2)}{y}\right)dy=\frac14\left(\frac{\ln(1+y^2)}{y}\right)^2\bigg|_0^\infty =0$$ Above the Laplace transform of the cosine integral was used.

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Similarly, for $I_2$ we have: $$I_2=\int_0^\infty \frac{\ln(1+y^2)}{y}\left(\int_0^\infty x e^{-xy}\operatorname{si}(x)dx\right)dy$$ $$=\int_0^\infty \frac{\ln(1+y^2)}{y}\left(-\frac{d}{dy}\int_0^\infty e^{-xy}\operatorname{si}(x)dx\right)dy$$ $$=\int_0^\infty \frac{\ln(1+y^2)}{y}\frac{d}{dy}\left(\frac{\arctan y}{y}\right)dy$$ $$=\int_0^\infty \left(\frac{\ln(1+y^2)}{y^2(1+y^2)}-\color{blue}{\frac{\arctan y\ln(1+y^2)}{y^3}}\right)dy$$ $$\overset{\color{blue}{IBP}}=\int_0^\infty \left(\frac12\frac{\ln(1+y^2)}{y^2}-\frac12\frac{\ln(1+y^2)}{1+y^2}-\color{red}{\frac{\arctan y}{y(1+y^2)}}\right)dy$$ $$\overset{\color{red}{IBP}}=\int_0^\infty \left(\frac12\underbrace{\frac{\ln(1+y^2)}{y^2}}_{\pi}-\underbrace{\frac{\ln(1+y^2)}{1+y^2}}_{\pi\ln 2}+\underbrace{\frac{\ln y}{1+y^2}}_{0}\right)dy=\frac{\pi}{2}-\pi\ln 2$$ This time the Laplace transform of the sine integral was used.

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Proof for the mentioned result

Since $2\cos x= e^{ix}+e^{-ix}$, we can write: $$\operatorname{Ci}(x)=-\int_x^\infty \frac{\cos t}{t}dt=-\frac12\int_x^\infty \frac{e^{it}}{t}dt-\frac12\int_x^\infty \frac{e^{-it}}{t}dt$$ $$=-\frac12\int_x^\infty \int_0^\infty e^{it}e^{-ty}dydt-\frac12\int_x^\infty \int_0^\infty e^{-it}e^{-ty}dydt$$ $$=-\frac12 \int_0^\infty\int_x^\infty e^{-(y-i)t}dtdy-\frac12\int_0^\infty\int_x^\infty e^{-(y+i)t}dtdy$$ $$=-\frac12 e^{ix} \int_0^\infty \frac{e^{-xy}}{y-i}dy-\frac12 e^{-ix} \int_0^\infty \frac{e^{-xy}}{y+i}dy$$ By the same approach, using $2i\sin x= e^{ix}-e^{-ix}$, we arrive at: $$\operatorname{si}(x)=-\frac1{2i} e^{ix} \int_0^\infty \frac{e^{-xy}}{y-i}dy+\frac1{2i} e^{-ix} \int_0^\infty \frac{e^{-xy}}{y+i}dy$$

Finally, writting $2(a^2+b^2)$ as $(a+b)^2+(a-b)^2$ yields: $$\small \operatorname{Ci}^2(x)+\operatorname{si}^2(x)=\int_0^\infty \frac{e^{-xy}}{y-i}dy\int_0^\infty \frac{e^{-xz}}{z+i}dz\overset{y+z\to y}=\int_0^\infty \int_z^\infty \frac{e^{-xy}}{(y-z-i)(z+i)}dydz$$ $$\small =\int_0^\infty e^{-xy} \int_0^y \frac{1}{(y-z-i)(z+i)}dzdy=\int_0^\infty \frac{e^{-xy}\ln(1+y^2)}{y}dy$$

Answer 2

We can actually use $I_{1}$ and $I_{2}$ to find closed-form expressions for $$\int_{0}^{\infty} x \operatorname{Ci}(x)^{3} \, \mathrm dx$$ and $$\int_{0}^{\infty} x \operatorname{si}(x)^{3} \, \mathrm dx.$$

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Let $E_{1}(z)$ be the exponential integral function defined in the right half-plane by the integral $$E_{1}(z) = \int_{1}^{\infty} \frac{e^{-zt}}{t} \, \mathrm dt, \quad \left( \Re(z) \ge 0, z \ne 0 \right). $$

For positive values of $x$, $E_{1}(x) = - \operatorname{E}_{i}(-x)$.

Similar to the my answer here, if we integrate the function $zE_{1}(z)^{3}$ around an infinitely-large closed quarter circle in the first quadrant of the complex plane, we get $$ \begin{align} \oint zE_{1}(z)^{3} dz &=\int_{0}^{\infty} x E_{1}(x)^{3} \, \mathrm dx - \int_{0}^{\infty} (ix) E_{1}(ix)^{3} \, i \, \mathrm dx \\ &= \small\int_{0}^{\infty} x E_{1}(x)^{3} \, \mathrm dx + \int_{0}^{\infty}x\left(-\operatorname{Ci}(x)^{3}+3i \operatorname{Ci}(x)^{2}\operatorname{si}(x) +3 \operatorname{Ci}(x) \operatorname{si}(x)^{2} -i \operatorname{si}(x)^{3}\right) \, \mathrm dx \\ &=0. \end{align}$$

Equating the real parts on both side of the equation, we have $$\int_{0}^{\infty} x E_{1}(x)^{3} \, \mathrm dx+ \int_{0}^{\infty} x\left(-\operatorname{Ci}(x)^{3}+3\operatorname{Ci}(x) \operatorname{si}(x)^{2} \right) \, \mathrm dx =0.$$

And combining the above equation with $I_{1}$, we get $$\int_{0}^{\infty}x \operatorname{Ci}(x)^{3} \, \mathrm dx = \frac{1}{4} \int_{0}^{\infty}x E_{1}(x)^{3} \, \mathrm dx. $$

(If you ask WolframAlpha to approximate the value of $\int_{0}^{a} x \left(\operatorname{Ci}(x)^{3}+ \frac{1}{4} \operatorname{E}_{i}(-x)^{3} \right) \, \mathrm dx $ for increasing positive values of $a$, it returns results that appear to be approaching zero. But if you ask it to approximate $\int_{0}^{\infty} x \left(\operatorname{Ci}(x)^{3}+ \frac{1}{4} \operatorname{E}_{i}(-x)^{3} \right) \, \mathrm dx $, it strangely returns an approximation that is nowhere near zero. This could simply be the result of WolframAlpha needing more computation time.)

To evaluate $$\int_{0}^{\infty}xE_{1}(x)^{3}\, \mathrm dx, $$ we can do what I did here to evaluate $\int_{0}^{\infty}E_{1}(x)^{3}\, \mathrm dx $.

Specifically, we have

$\begin{align} \int_{0}^{\infty}xE_{1}(x)^{3}\, \mathrm dx &\overset{(1)}{=} \frac{3}{2} \int_{0}^{\infty} x e^{-x}E_{1}(x)^{2} \, \mathrm dx \\ &= \frac{3}{2} \int_{0}^{\infty}x e^{-x} \int_{1}^{\infty} \int_{1}^{\infty} \frac{e^{-xy}}{y} \frac{e^{-xz}}{z} \, \mathrm dy \, \mathrm dz \, \mathrm dx \\ &= \frac{3}{2} \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{yz} \int_{0}^{\infty} x e^{-(1+y+z)x} \, \mathrm dx \, \mathrm dy \, \mathrm dz \\ &= \frac{3}{2} \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{yz} \frac{1}{(1+y+z)^{2}} \, \mathrm dy \, \mathrm dz \\ &= \frac{3}{2} \int_{1}^{\infty} \frac{1}{z} \int_{1}^{\infty}\left(\frac{1}{(1+z)^2} \left(\frac{1}{y}-\frac{1}{1+y+z} \right) -\frac{1}{1+z} \frac{1}{(1+y+z)^{2}}\right) \, \mathrm dy \, \mathrm dz \\ &= \frac{3}{2} \int_{1}^{\infty} \frac{1}{z} \left(\frac{\ln(2+z)}{(1+z)^{2}} -\frac{1}{(1+z)(z+2)} \right) \, \mathrm dz \\ &= \frac{3}{2} \left( \int_{1}^{\infty} \frac{1}{z} \frac{\ln(2+z)}{(1+z)^{2}} \, \mathrm dz - \ln(2) + \frac{\ln(3)}{2} \right) \\ &\overset{(2)}{=} \frac{3}{2} \left(\int_{0}^{1} \frac{u \ln(1+2u)}{(1+u)^{2}} \, \mathrm du - \int_{0}^{1} \frac{u \ln(u)}{(1+u)^{2}} \, \mathrm du - \ln(2) + \frac{\ln(3)}{2} \right) \\ &=\frac{3}{2} \left(\int_{0}^{1} \frac{u \ln(1+2u)}{(1+u)^{2}} \, \mathrm du + \frac{\pi^{2}}{12} -2 \ln (2) + \frac{\ln(3)}{2} \right) \\ &\overset{(3)}{=} \small \frac{3}{2} \left(\frac{\ln(3)}{2} + \ln(2) \ln(3) -2 \int_{0}^{1} \frac{\mathrm du}{(1+2u)(1+u)} -2 \int_{0}^{1} \frac{\ln(1+u)}{1+2u} \, \mathrm du + \frac{\pi^{2}}{12} -2 \ln (2) + \frac{\ln(3)}{2} \right) \\ &=\frac{3}{2} \left(\ln(2) \ln(3) -\ln(3) -2 \int_{0}^{1} \frac{\ln(1+u)}{1+2u} \, \mathrm du + \frac{\pi^{2}}{12} \right) \end{align}$

where $$ \begin{align} \int_{0}^{1} \frac{\ln(1+u)}{1+2u} \, \mathrm du &= \frac{1}{2} \int_{1}^{3} \frac{\ln (1+w) - \ln(2)}{w} \, \mathrm dw \\ &= \frac{1}{2} \left( \operatorname{Li}_{2}(-1)- \operatorname{Li}_{2}(-3) -\ln(2) \ln(3) \right) \\ & \overset{(4)}{=} \frac{\pi^{2}}{24} + \frac{\ln^{2}(3)}{4}+\frac{1}{2} \, \operatorname{Li}_{2}\left(-\frac{1}{3} \right) - \frac{\ln(2) \ln(3)}{2}. \end{align}$$

Therefore, $$ \begin{align} \int_{0}^{\infty}xE_{1}(x)^{3}\, \mathrm dx &= -\int_{0}^{\infty} x \operatorname{E}_{i}(-x)^{3} \mathrm dx \\ &= \frac{3}{2} \left(2 \ln(2) \ln(3) -\ln(3) -\frac{\ln^{2}(3)}{2} -\operatorname{Li}_{2} \left(-\frac{1}{3} \right) \right) \\ &= 0.1949195673... \end{align}$$

and $$\int_{0}^{\infty} x \operatorname{Ci}(x)^{3} \, \mathrm dx = \frac{3}{8} \left(2 \ln(2) \ln(3) -\ln(3) -\frac{\ln^{2}(3)}{2} -\operatorname{Li}_{2} \left(-\frac{1}{3} \right) \right). $$

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$(1)$ Integrate by parts.

$(2)$ Let $z= \frac{1}{u}$.

$(3)$ Integrate by parts.

$(4)$ Use the dilogarithm inversion formula.

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If we equate the imaginary parts on both sides of the equation $$\small\int_{0}^{\infty} x E_{1}(x)^{3} \, \mathrm dx + \int_{0}^{\infty}x\left(-\operatorname{Ci}(x)^{3}+3i \operatorname{Ci}(x)^{2}\operatorname{si}(x) +3 \operatorname{Ci}(x) \operatorname{si}(x)^{2} -i \operatorname{si}(x)^{3}\right) \, \mathrm dx =0, $$ we have $$\int_{0}^{\infty} \left(3x\operatorname{Ci}(x)^{2}\operatorname{si}(x) - x \operatorname{si}(x)^{3} \right) \, \mathrm dx =0. $$

And combining the above equation with $I_{2}$, we get $$\int_{0}^{\infty} x \operatorname{si}(x)^{3} \, \mathrm dx = \frac{3 \pi}{8} \left(1-2 \ln(2) \right).$$