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In attempting to evaluate $ \int_{0}^{\infty} [\text{Ei}(-x)]^{4} \, dx$ (which can be evaluated in terms of polylogarithm values), I determined that $$ \begin{align} \int_{0}^{\infty} [\text{Ei}(-x)]^{4} \, dx &= 8 \int_{2}^{\infty} \frac{\log(2+x) \log(x-1)}{x(1+x)} \, dx \\ &=8 \Bigg(\int_{0}^{1/2} \frac{\log(1+2x) \log(1-x)}{1+x} \, dx - \int_{0}^{1/2} \frac{\log(1+2x) \log(x)}{1+x} \, dx \\ &- \int_{0}^{1/2} \frac{\log (x) \log(1-x)}{1+x} \, dx + \int_{0}^{1/2} \frac{\log^{2}(x)}{1+x} \, dx \Bigg). \end{align}$$

I will show this at the end of my post.

I can evaluate the 3rd and 4th integrals, but I'm having difficulties with the first two integrals.

$$ \begin{align} \int_{0}^{1/2} \frac{\log(1+2x)\log(x)}{1+x} \, dx &= \int_{0}^{1/2} \log(x) \sum_{n=1}^{\infty} (-1)^{n+1} x^{n} \sum_{k=1}^{n} \frac{2^{k}}{k} \, dx \\ &= \sum_{n=1}^{\infty} (-1)^{n+1} \sum_{k=1}^{n}\frac{2^{k}}{k} \int_{0}^{1/2} x^{n} \log(x) \, dx \\ &=-\sum_{n=1}^{\infty} \left( -\frac{1}{2} \right)^{n+1} \left[\frac{\log 2}{n+1} + \frac{1}{(n+1)^{2}} \right] \sum_{k=1}^{n}\frac{2^{k}}{k} \end{align}$$

This approach doesn't seem particularly useful.

Approaching the other integral in the same manner leads to even more of a mess:

$$ \begin{align} \int_{0}^{1/2} \frac{\log(1+2x)\log(1-x)}{1+x} \, dx &= \sum_{n=1}^{\infty} (-1)^{n+1} \sum_{k=1}^{n}\frac{2^{k}}{k} \int_{0}^{1/2} x^{n} \log(1-x) \, dx \\& = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1} \left[\log (2) -\left(\frac{1}{2}\right)^{n+1} \log(2) - \sum_{m=1}^{n-1} \frac{(\frac{1}{2})^{m}}{m}\right]\sum_{k=1}^{n}\frac{2^{k}}{k} \end{align}$$


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

$(1)$ Integrate by parts.

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

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


EDIT:

Using M.N.C.E.'s suggestion in the comments, both integrals can be expressed in terms of integrals that can be evaluated using integration by parts.

I posted an answer below.

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  • 1
    $\begingroup$ Just a suggestion: Perhaps you could break the integral up with $$2\ln(1+2x)\ln(1-x)=\ln^2(1+2x)+\ln^2(1-x)-\ln^2\left(\frac{1+2x}{1-x}\right)$$ then substitute $u= \dfrac{1+2x}{1-x}$ in the third integral. $\endgroup$ – M.N.C.E. Jun 4 '15 at 23:59
  • $\begingroup$ @M.N.C.E. That would seem to result in three integrals that all can be evaluated by integrating by parts (perhaps with a bit of manipulation first). $\endgroup$ – Random Variable Jun 5 '15 at 1:06
  • $\begingroup$ Both integrands have closed form antiderivatives in terms of elementary functions and polylogs (they can be found by WolframAlpha, for example). Their correctness can be easily proved by direct differentiation and elementary transformations (polylog terms cancel in the process). The only remaining work is to calculate limit for $x\to0$ in one of the integrals and apply some polylog identities to simplify results. Not a particularly enlightening solution though. $\endgroup$ – Vladimir Reshetnikov Jun 11 '15 at 22:54
  • $\begingroup$ It almost seems as though the people here are suggesting this integral is somehow trivial because it can be evaluated. $\endgroup$ – Zach466920 Jun 12 '15 at 14:37
  • $\begingroup$ @VladimirReshetnikov Are you interested in seeing how to find an antiderivative for the first one? $\endgroup$ – Random Variable Jun 12 '15 at 16:00
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At Vladimir Reshetnikov's request I'm going to show how to find an antiderivative for $$ \frac{\log(1+2x) \log(1-x)}{1+x} $$ using 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.


$$ \begin{align} &\int \frac{\log(1+2x) \log(1-x)}{1+x} \, dx \\ &= \frac{1}{2} \int \frac{\log^{2}(1+2x)}{1+x} \, dx + \frac{1}{2} \int \frac{\log^{2}(1-x)}{1+x} \, dx - \frac{1}{2} \int \frac{\log^{2} \left(\frac{1+2x}{1-x} \right)}{1+x} \, dx \\ &= \frac{1}{2} \int \frac{\log^{2}(1+2x)}{1+x} \, dx + \frac{1}{2} \int \frac{\log^{2}(1-x)}{1+x} \, dx - \frac{3}{2} \int \frac{\log^{2}(t)}{(2+t)(1+2t)} \ dt \\ &= \frac{1}{2} \int \frac{\log^{2}(1+2x)}{1+x} \, dx + \frac{1}{2} \int \frac{\log^{2}(1-x)}{1+x} \, dx + \frac{1}{2} \int \frac{\log^{2}t}{2+t} \, dt - \int \frac{\log^{2}(t)}{1+ 2t} \ dt \end{align}$$

All 4 indefinite integrals can be evaluated by integrating by parts twice.


For the first one let $u = \log^{2}(1+2x)$ and $dv= \frac{dx}{1+x}$. Then for $v$ choose $\log(2+2x)$.

$$ \begin{align} \int \frac{\log^{2}(1+2x)}{1+x} \, dx &= \log^{2}(1+2x) \log(2+2x) -4 \int \frac{\log(1+2x) \log(2+2x)}{1+2x} \\ &= \log^{2}(1+2x) \log(2+2x) - 2 \int \frac{ \log(1+w) \log(w) }{w} \ dw \\ &= \log^{2}(1+2x) \log(2+2x) -2 \left(-\log(w) \text{Li}_{2} (-w) + \int \frac{\text{Li}_{2}(-w)}{w} \ dt\right) \\ &=\log^{2}(1+2x) \log(2+2x) + 2 \log(1+2x) \text{Li}_{2} (-1-2x) - 2 \text{Li}_{3}(-1-2x) + C \end{align}$$


For the second one let $ u = \log^{2}(1-x)$ and $dv = \frac{dx}{1+x}$. This time choose $\log \left(\frac{1+x}{2} \right)$ for $v$.

$$ \begin{align} \int \frac{\log^{2}(1-x)}{1+x} \, dt &= \log^{2}(1-x) \log \left(\frac{x+1}{2} \right) + 2 \int \frac{\log(1-x) \log \left(\frac{1+x}{2} \right)}{1-x} \ dx \\ &= \log^{2}(1-x) \log \left(\frac{1+x}{2} \right) - 2 \int \frac{\log(w) \log \left(1- \frac{w}{2} \right)}{w} \, dw \\ &= \log^{2}(1-x) \log \left(\frac{1+x}{2} \right) - 2 \left(- \log (w) \text{Li}_{2} \left( \frac{w}{2} \right)+ \int \frac{\text{Li}_{2}(\frac{w}{2})}{w} \right) \ dw \\ &= \log^{2}(1-x) \log \left(\frac{1+x}{2} \right) + 2 \log(1-x) \text{Li}_{2} \left(\frac{1-x}{2} \right) - 2 \text{Li}_{3} \left(\frac{1-x}{2} \right) + C \end{align}$$


The third one can be found at the end of this post.


The fourth one is more straightforward.

$$ \begin{align} \int \frac{\log^{2}(t)}{1+2t} \, dt &= \frac{1}{2} \log^{2}(t) \log(1+2t) - \int \frac{\log(t) \log(1+2t)}{t} \, dt \\ &= \frac{1}{2} \log^{2}(t) \log(1+2t) - \left(- \log(t) \text{Li}_{2}(-2t) + \int \frac{\text{Li}_{2}(-2t)}{t} \, dt \right) \\ &=\frac{1}{2} \log^{2}(t) \log(1+2t) + \log(t) \text{Li}_{2}(-2t) - \text{Li}_{3}(-2t) + C . \end{align}$$

And don't forget that for the 3rd and 4th integrals, $t = \frac{1+2x}{1-x}$.


All 4 antiderivatives above agree with Wolfram Alpha.

If we combine everything, the antiderivative for $\frac{\log(1+2x) \log(1-x)}{1+x}$ appears to be a bit different from the one provided by Wolfram Alpha. This might be because of the identity I used at the beginning.

Plugging in the limits I get

$$ \begin{align} &\int_{0}^{1/2} \frac{\log(1+2x) \log(1-x)}{1+x} \, dx \\ &= -3 \text{Li}_{3}(-2) + 3 \text{Li}_{2}(-2) \log(2) + \frac{\zeta(3)}{8} - \text{Li}_{3} \left(\frac{1}{4} \right) - \text{Li}_{2} \left(\frac{1}{4} \right) \log(2) - \frac{5}{6} \log^{3}(2) \\ &- \frac{\pi^{2}}{12} \log(2) + \text{Li}_{3} \left(- \frac{1}{2} \right) + \text{Li}_{3} (-8) + 2 \text{Li}_{2}(-8) \log(2) - \log(3) \log^{2}(2) \\ &\approx -0.05738655697. \end{align}$$

I'll leave it to others to figure out how to express the result in the nicest form possible.


A slightly nicer form of the result is

$$ \begin{align} &\int_{0}^{1/2} \frac{\log(1+2x) \log(1-x)}{1+x} \, dx \\ &= \text{Li}_{3} \left( - \frac{1}{8}\right) - \frac{3}{2} \text{Li}_{3} \left(\frac{1}{4} \right) + \frac{15}{8} \zeta(3) + \frac{3}{2} \log^{3}(2) - \log(3) \log^{2}(2)- \frac{\pi^{2}}{6} \log(2) \\ &+ 2 \text{Li}_{2} \left(-\frac{1}{8} \right) \log(2) - \frac{5}{2} \text{Li}_{2} \left( \frac{1}{4}\right) \log(2). \end{align}$$

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