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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 '15 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 '15 at 7:25
  • $\begingroup$ @M.N.C.E. The latter integral can be evaluated with Mathematica. How did you get it? $\endgroup$ – Vladimir Reshetnikov May 30 '15 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. $\endgroup$ – Random Variable Jun 3 '15 at 19:37
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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 real positive values.


As I stated here,

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

$(1)$ Integrate by parts.

$(2)$ Make the change of variables $u= y+z$, $v=yz$.

$(3)$ Make the substitution $ t^{2}=u^2-4v$.

$(4)$ Replace $u$ with $\frac{1}{u}$.

Then using the identity I mentioned at the beginning of this post,

$$ \begin{align} &\int_{0}^{\infty} [\text{Ei}(-x)]^{4} \, dx \\ &= -4\int_{0}^{1/2} \frac{\log^{2} \left(\frac{1-u}{1+2u} \right)}{1+u} \, du + 4 \int_{0}^{1/2} \frac{\log^{2} \left(\frac{u}{1+2u} \right)}{1+u} \, du + 4 \int_{0}^{1/2} \frac{\log^{2} \left(\frac{u}{1-u} \right)}{1+u} \, du \\& = -12 \int_{1/4}^{1} \frac{\log^{2} (u)}{(1+2u)(2+u)} \, du +4 \int_{0}^{1/4} \frac{\log^{2} (u)}{(1-2u)(1-u)} \, du + 4 \int_{0}^{1} \frac{\log^{2}(u)}{(1+2u)(1+u)} \, du \\& \approx 14.0394 \end{align}$$

After performing partial fraction decomposition, you can evaluate all the integrals by integrating by parts twice. In some cases you need to pick a particular "$v$" for it to work. Wolfram Alpha can provide the antiderivatives if needed.

It will require additional work to get it in the form provided by Cleo.

EDIT:

Actually, the only tricky one is $$\int \frac{\log^{2}(x)}{2+x} \, dx. $$

Let $u=\log^{2}(x)$ and $dv = \frac{dx}{2+x}$. For $v$ choose the antiderivative $\log \left(\frac{2+x}{2} \right)$.

Then $$ \int \frac{\log^{2}(x)}{2+x} \, dx = \log^{2}(x) \log \left(\frac{2+x}{2} \right) - 2 \int \frac{\log (x) \log \left( \frac{2+x}{x} \right)}{x} \, dx.$$

Now let $u = \log(x)$ and $dv= \frac{ \log \left(1 + \frac{x}{2}\right)}{x} \, dx$. Then by definition $v = - \text{Li}_{2} \left( - \frac{x}{2}\right)$.

So

$$ \begin{align} \int \frac{\log^{2}(x)}{2+x} \, dx &= \log^{2}(x) \log \left(\frac{2+x}{2} \right) +2 \text{Li}_{2} \left(- \frac{x}{2} \right) \log(x) - 2\int \frac{\text{Li}_{2} \left(- \frac{x}{2} \right)}{x} \, dx \\ &= \log^{2}(x) \log \left(\frac{2+x}{2} \right) +2 \text{Li}_{2} \left(- \frac{x}{2} \right) \log(x) -2 \text{Li}_{3} \left(- \frac{x}{2} \right)+C. \end{align} $$

The other 5 are very similar and don't require any tricks at the start.

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  • $\begingroup$ i would be really interested in the missing steps! nice work btw. (+1) $\endgroup$ – tired Jun 5 '15 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. $\endgroup$ – Random Variable Jun 5 '15 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 '15 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. $\endgroup$ – Random Variable Jun 5 '15 at 19:24
  • $\begingroup$ thanks ! the reduction to @Cleos form looks really nasty.. ^^ $\endgroup$ – tired Jun 6 '15 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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