Looking for a proof of Cleo's result for ${\large\int}_0^\infty\operatorname{Ei}^4(-x)\,dx$

Question

In this answer Cleo posted the following result without a proof: $$\begin{align}\int_0^\infty\operatorname{Ei}^4(-x)\,dx&=24\operatorname{Li}_3\!\left(\tfrac14\right)-48\operatorname{Li}_2\!\left(\tfrac13\right)\ln2-13\,\zeta(3)\\&-32\ln^32+48\ln^22\cdot\ln3-24\ln2\cdot\ln^23+6\pi^2\ln2,\end{align}\tag1$$ where $\operatorname{Ei}$ is the exponential integral: $$\operatorname{Ei}(x)=-\int_{-x}^\infty\frac{e^{-t}}tdt.\tag2$$ The result confirms numerically with at least 1000 decimal digits of precision.

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How can we prove this result?

Answer 1

I would like to thank M.N.C.E. for suggesting the use of the identity $$2\log(x)\log(y) = \log^{2}(x) + \log^{2}(y) - \log^{2} \left(\frac{x}{y} \right), $$ where $x$ and $y$ are positive real values.

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As I stated here, we have

\begin{align} &\int_{0}^{\infty} [\operatorname{Ei}(-x)]^{4} \, dx \\ &= -4 \int_{0}^{\infty} [\operatorname{Ei}(-x)]^{3} e^{-x} \, dx \tag{1} \\ &= 4 \int_{0}^{\infty} \int_{1}^{\infty} \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{wyz} e^{-(w+y+z+1)x} \, dw \, dy \, dz \,dx \tag{2} \\& =4 \int_{1}^{\infty} \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{wyz} \int_{0}^{\infty} e^{-(w+y+z+1)x} \, dx \, dw \, dy \, dz \\ &= 4 \int_{1}^{\infty} \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{wyz} \frac{1}{w+y+z+1} \, dw \, dy \, dz \\ &= 4 \int_{1}^{\infty} \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{yz} \frac{1}{y+z+1} \left(\frac{1}{w} - \frac{1}{w+y+z+1} \right) \, dw \, dy \, dz \\ &= 4 \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{yz} \frac{1}{y+z+1} \log(y+z+2) \, dy \, dz \\ &= 8 \int_{1}^{\infty} \int_{1}^{z} \frac{1}{yz} \frac{1}{y+z+1} \log(y+z+2) \, dy \, dz \tag{3} \\ &= 8 \int_{2}^{\infty} \int_{u-1}^{u^{2}/4} \frac{1}{v} \frac{1}{u+1} \log(u+2) \frac{dv \, du}{\sqrt{u^{2}-4v}} \tag{4} \\& =16 \int_{2}^{\infty} \frac{\log(u+2)}{u+1} \int_{0}^{u-2} \frac{1}{u^{2}-t^{2}} \, dt \, du \tag{5} \\& =16 \int_{2}^{\infty} \frac{\log(u+2)}{u+1} \frac{1}{u} \, \operatorname{artanh} \left( \frac{u-2}{2}\right) \, du \\ &= 8 \int_{2}^{\infty} \frac{\log(u+2) \log(u-1)}{u(u+1)} \, du \\ &= 8 \Bigg(\int_{0}^{1/2} \frac{\log(1-u)\log(1+2u)}{1+u} \, du - \int_{0}^{1/2} \frac{\log(u) \log(1+2u)}{1+u} \, du \\ &- \int_{0}^{1/2} \frac{\log(u) \log (1-u) }{1+u} \, du + \int_{0}^{1/2} \frac{\log^{2}(u)}{1+u} \, du \Bigg). \tag{6} \end{align}

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$(1)$ Integrate by parts.

$(2)$ $\operatorname{Ei}(-x) = - \int_{x}^{\infty} \frac{e^{-t}}{t} \, dt = - \int_{1}^{\infty} \frac{e^{-xu}}{u} \, \mathrm du $

$(3)$ The integrand is symmetric about the line $z=y$.

$(4)$ Make the change of variables $u= y+z$, $v=yz$.

$(5)$ Make the substitution $ t^{2}=u^2-4v$.

$(6)$ Replace $u$ with $\frac{1}{u}$.

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Then using the identity I mentioned at the beginning of this answer, we have

$$ \begin{align} &\int_{0}^{\infty} [\text{Ei}(-x)]^{4} \, dx \\ &= -4\int_{0}^{1/2} \frac{\log^{2} \left(\frac{1-u}{1+2u} \right)}{1+u} \, du + 4 \int_{0}^{1/2} \frac{\log^{2} \left(\frac{u}{1+2u} \right)}{1+u} \, du + 4 \int_{0}^{1/2} \frac{\log^{2} \left(\frac{u}{1-u} \right)}{1+u} \, du \\& = -12 \int_{1/4}^{1} \frac{\log^{2} (w)}{(1+2w)(2+w)} \, dw +4 \int_{0}^{1/4} \frac{\log^{2} (y)}{(1-2y)(1-y)} \, dy + 4 \int_{0}^{1} \frac{\log^{2}(z)}{(1+2z)(1+z)} \, dz \\&= -6 \int_{1/4}^{1} \frac{\log^{2} (w)}{(1+2w)(1+\frac{w}{2})} \, dw +4 \int_{0}^{1/4} \frac{\log^{2} (y)}{(1-2y)(1-y)} \, dy + 4 \int_{0}^{1} \frac{\log^{2}(z)}{(1+2z)(1+z)} \, dz. \end{align}$$

EDIT:

Partial fraction decomposition followed by two applications of integration by parts shows that $ \begin{align} \int \frac{\log^{2}(x)}{(1+ax)(1+bx)} \, dx = \frac{1}{a-b} \Big(&2\big(\operatorname{Li}_{3}(-bx) - \operatorname{Li}_{3}(-ax) \big) + 2 \log (x) \big( \operatorname{Li}_{2}(-ax) - \operatorname{Li}_{2}(-bx)\big) \\ &+\log^{2}(x) \big(\ln(1+ax) - \log(1+bx) \big) \Big) + C. \end{align}$

This primitive can be used to determine the value of all three definite integrals above.

The final result won't immediately be in the form given by Cleo. Getting it in that form will require the use of polylogarithm identities.

Answer 2

Here are some useful components to developing a solution to the proposed integral. \begin{align}\tag{1} \int_{0}^{\infty} Ei^{4}(-x) \, dx &= -4 \, \int_{0}^{\infty} e^{-x} \, Ei^{3}(-x) \, dx \\ \int_{0}^{\infty} e^{-x} \, Ei^{3}(-x) \, dx &= - \ln^{2}2 + \int_{0}^{\infty} x \, e^{-x} \, Ei^{2}(-x) \, dx - 2 \, \int_{0}^{\infty} e^{-x} \, Ei(-x) \, Ei(-2x) \, dx \tag{2.1}\\ &= - \ln^{2}2 - \frac{2}{3} \int_{0}^{\infty} x \, Ei^{3}(-x) \, dx + \int_{0}^{\infty} e^{-2x} \, Ei^{2}(-x) \, dx \\ & \hspace{15mm} + \int_{0}^{\infty} Ei(-2x) \, Ei^{2}(-x) \, dx \tag{2.2} \end{align}