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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.


How can we prove this result?

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    $\begingroup$ The famous Jewish Rabbi Hillel once said that the greatest commandment is loving God above all, and one's neighbor as oneself, and that the rest of the whole Torah is but a footnote to this. In that same spirit, we might also say that almost all calculus and integration related posts on MSE are but a footnote to Cleo's answers. $\endgroup$
    – Lucian
    May 28, 2015 at 5:54
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    $\begingroup$ $$\int^\infty_0\operatorname{Ei}^4(-x)dx=12\int^2_1\frac{2\ln(1+t)\ln(2+t)-2Li_2(-t)+Li_2(-t(2+t))}{t}dt$$ $\endgroup$
    – M.N.C.E.
    May 30, 2015 at 7:25
  • $\begingroup$ @M.N.C.E. The latter integral can be evaluated with Mathematica. How did you get it? $\endgroup$ May 30, 2015 at 19:38
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    $\begingroup$ $$ \begin{align}\int_{0}^{\infty} [\text{Ei}(-x)]^{4} \, dx = 8 \Bigg(&\int_{0}^{1/2} \frac{\log(1-x) \log(1+2x)}{1+x} \, dx - \int_{0}^{1/2} \frac{\log(x) \log(1+2x)}{1+x} \, dx \\ &- \int_{0}^{1/2} \frac{\log (1-x) \log(x)}{1+x} \, dx + \int_{0}^{1/2} \frac{\log^{2}(x)}{1+x} \, dx \Bigg)\end{align}$$ I haven't completed the evaluation. $\endgroup$ Jun 3, 2015 at 19:37

2 Answers 2

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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.


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}


$(1)$ 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 $\frac{1}{u}$.


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.

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  • $\begingroup$ i would be really interested in the missing steps! nice work btw. (+1) $\endgroup$
    – tired
    Jun 5, 2015 at 18:19
  • $\begingroup$ @tired Are you referring to evaluating those 6 integrals at the end? Except for the one I mentioned, it's just integration by parts. Nothing fancy. $\endgroup$ Jun 5, 2015 at 18:32
  • $\begingroup$ ok, i believe that, put posting at least the corresponding results would make the answer much more complete i think! :) $\endgroup$
    – tired
    Jun 5, 2015 at 18:35
  • $\begingroup$ @tired I edited my post and showed how to do the only one that requires a bit of a trick at the start. The others are very similar. Wolfram Alpha can provide them if you want them quickly. $\endgroup$ Jun 5, 2015 at 19:24
  • $\begingroup$ thanks ! the reduction to @Cleos form looks really nasty.. ^^ $\endgroup$
    – tired
    Jun 6, 2015 at 10:02
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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}

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