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

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?

Solution 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 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.

Solution 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}