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Let $\operatorname{Ei}x$ denote the exponential integral: $$\operatorname{Ei}x=-\int_{-x}^\infty\frac{e^{-t}}tdt.\tag1$$ It's not difficult to find that $$\int\operatorname{Ei}x\,dx=x\,\operatorname{Ei}x-e^x,\tag2$$ $$\int\operatorname{Ei}^2x\,dx=x\,\operatorname{Ei}^2x-2\,e^x\operatorname{Ei}x+2\,\operatorname{Ei}(2x)\tag3$$ and $$\int_{-\infty}^0\operatorname{Ei}x\,dx=-1,\tag4$$ $$\int_{-\infty}^0\operatorname{Ei}^2x\,dx=\ln4.\tag5$$


Is it possible generalize these results for higher powers of $\operatorname{Ei}x$?
In particular, are there closed forms for $$\int\operatorname{Ei}^3x\,dx\tag6$$ or $$\int_{-\infty}^0\operatorname{Ei}^3x\,dx\ ?\tag7$$

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I added an alternative approach at the end of my post.


First note that $$\int_{-\infty}^{0} \text{Ei}^{3}(x) \ dx = -\int_{0}^{\infty} \text{E}_{1}^{3}(x) \ dx. $$

Then integrating by parts,

$$ - \int_{0}^{\infty} \text{E}_{1}^{3}(x) \ dx = -x \text{E}_{1}^{3}(x) \Big|^{\infty}_{0} - 3 \int^{\infty}_{0} \text{E}_{1}^{3}(x) e^{-x} \ dx. $$

Since $E_{1}(x)$ behaves like $-\log x$ near $x=0$ and $ \displaystyle\frac{e^{-x}}{x}$ near $\infty$, the boundary terms vanish.

And

$$ \begin{align} \int_{0}^{\infty} \text{E}_{1}(x) \text{E}_{1}(x) e^{-x} \ dx &= \int_{0}^{\infty} \int_{1}^{\infty} \int_{1}^{\infty} \frac{e^{-xy}}{y} \frac{e^{-xz}}{z} e^{-x} \ dy \ dz \ dx \\ &= \int_{1}^{\infty} \int_{1}^{\infty} \int_{0}^{\infty} \frac{1}{yz} e^{-(y+z+1)x} \ dx \ dy \ dz \\ &= \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{yz} \frac{1}{y+z+1} \ dy \ dz \\ &= \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{z} \frac{1}{1+z} \left( \frac{1}{y} - \frac{1}{y+z+1} \right) \ dy \ dz \\ &= \int_{1}^{\infty} \frac{\log(z+2)}{z(1+z)} \ dz \\ &= \log 2 \int_{1}^{\infty} \left( \frac{1}{z} - \frac{1}{1+z} \right) \ dz + \lim_{b \to \infty} \Bigg( \int_{1}^{b} \frac{\log (1+ \frac{z}{2})}{z} \ dz \\ &- \int_{1}^{b} \frac{\log (1+ \frac{z}{2})}{1+z} \ dz \Bigg) \\ &= \log^{2}(2) + \lim_{b \to \infty} \left[- \text{Li}_{2} \left(-\frac{z}{2} \right) \Big|^{b}_{1} - \int_{1}^{b} \frac{\log(1+ \frac{z}{2})}{1+z} \ dz \right] \end{align}$$

where

$$ \begin{align} \int \frac{\log(1+ \frac{z}{2})}{1+z} \ dz &= \int \frac{\log (1+u)}{u} \ du - \log (2) \int\frac{1}{u} \ du \\ &= - \text{Li}_{2}(-u) - \log(2) \log u +C \\ &= - \text{Li}_{2}(-z-1) - \log(2) \log(z+1) + C. \end{align}$$

Therefore,

$$ \begin{align} \int_{-\infty}^{0} \text{Ei}^{3}(x) \ dx &= - 3 \log^{2}(2) - 3 \text{Li}_{2} \left(- \frac{1}{2} \right) + 3 \text{Li}_{2}(-2) + 3 \log^{2}(2) \\ & + 3 \lim_{b \to \infty} \left[\text{Li}_{2} \left(-\frac{b}{2} \right) - \text{Li}_{2}(-b-1) - \log(2) \log(b+1) \right] \\ &= -3 \text{Li}_{2} \left(- \frac{1}{2} \right) + 3 \text{Li}_{2}(-2) - \frac{3 \log^{2}(2)}{2} \\ &\approx -3.68568. \end{align}$$

That limit can be evaluated by hand by using the inversion formula for the dilogarithm.

*See the second edit for a better way to evaluate that log integral.


EDIT:

As Lucian pointed out in the comments, the answer can be simplified using the inversion formula.

Specifically,

$$ \int_{-\infty}^{0} \text{Ei}(x)^{3} \ dx= - \frac{\pi^{2}}{2} - 3 \log^{2}(2) - 6 \text{Li}_{2} \left( -\frac{1}{2}\right).$$

Wolfram Alpha states than an alternative form of the expression on the right is $$-3 \text{Li}_{2} \left(\frac{1}{4} \right) - 6 \log^{2}(2). $$


SECOND EDIT:

A better way to evaluate $$\int_{1}^{\infty} \frac{\log(z+2)}{z(1+z)} \, dz$$ is to make the substitution $u = \frac{1}{z}$.

Then $$ \int_{1}^{\infty} \frac{\log(z+2)}{z(1+z)} \, dz = \int_{0}^{1} \frac{\log(1+2u)}{1+u} \, du - \int_{0}^{1} \frac{\log(u)}{1+u} \, du$$

where

$$ \begin{align} \int_{0}^{1} \frac{\log(1+2u)}{1+u} \, du &= \log(1+2u) \log(2+2u)\Bigg|_{0}^{1} - 2\int_{0}^{1} \frac{\log(2+2u)}{1+2u} \, du \\ &= \log(3) \log(4)- \int_{0}^{2}\frac{\log(2+v)}{1+v} \ dv \\ &= \log(3) \log(4) - \int_{1}^{3} \frac{\log(1+w)}{w} \ dw \\ &=\log(3) \log(4) + \text{Li}_{2}(-3) -\text{Li}_{2}(-1) \\ &= 2\log(3) \log(2) + \text{Li}_{2}(-3) + \frac{\pi^{2}}{12} \end{align}$$

and

$$ \int_{0}^{1} \frac{\log(1+u)}{u} \, du = -\frac{\pi^{2}}{12}.$$

So we get the equivalent answer

$$\int_{-\infty}^{0} \text{Ei}^{3}(x) \, dx = -6 \log(3) \log(2) -3 \text{Li}_{2}(-3) - \frac{\pi^{2}}{2} \approx -3.68568 .$$

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    $\begingroup$ I think you just forgot a dz on the 6th equality you have. Very nice proof as always, +1!! $\endgroup$ – Jeff Faraci Mar 17 '14 at 4:34
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    $\begingroup$ $\text{Li}_2\bigg(-\dfrac12\bigg)-\text{Li}_2(-2)=\zeta(2)+\dfrac{\ln^22}2+2 \cdot\text{Li}_2\bigg(-\dfrac12\bigg)$ $\endgroup$ – Lucian Mar 17 '14 at 5:23
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$$\begin{align}\int_0^\infty\operatorname{Ei}^3(-x)\,dx&=-3\operatorname{Li}_2\left(\frac14\right)-6\ln^22.\\\\\int_0^\infty\operatorname{Ei}^4(-x)\,dx&=24\operatorname{Li}_3\left(\frac14\right)-48\operatorname{Li}_2\left(\frac13\right)\ln2-13\,\zeta(3)\\&-32\ln^32+48\ln^22\,\ln3-24\ln2\,\ln^23+6\pi^2\ln2.\end{align}$$

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    $\begingroup$ This is scary... $\endgroup$ – karvens Oct 1 '14 at 16:29

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