Closed form for $\int_{-\infty}^0\operatorname{Ei}^3x\,dx$

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 $$\int_{-\infty}^0\operatorname{Ei}^3x\,dx\ ?\tag7$$

Solution 1:

$$\begin{align}\int_0^\infty\operatorname{Ei}^3(-x)\,dx&=-3\operatorname{Li}_2\left(\frac14\right)-6\ln^22.\\\\\int_0^\infty\operatorname{Ei}^4(-x)\,dx&=24\operatorname{Li}_3\left(\frac14\right)-48\operatorname{Li}_2\left(\frac13\right)\ln2-13\,\zeta(3)\\&-32\ln^32+48\ln^22\,\ln3-24\ln2\,\ln^23+6\pi^2\ln2.\end{align}$$

Solution 2:

First notice that $$\int_{-\infty}^{0} \operatorname{Ei}^{3}(x) \, \mathrm{d}x = -\int_{0}^{\infty} \operatorname{E}_{1}^{3}(x) \, \mathrm{d}x \, , $$

where $$\operatorname{E}_{1}(x) =\int_{x}^{\infty} \frac{e^{-t}}{t} \, \mathrm{d}t = \int_{1}^{\infty} \frac{e^{-xu}}{u} \, \mathrm{d}u. $$

Then integrating by parts, we get

$$ - \int_{0}^{\infty} \operatorname{E}_{1}^{3}(x) \, \mathrm{d}x = -x \operatorname{E}_{1}^{3}(x) \Big|^{\infty}_{0} - 3 \int^{\infty}_{0} \operatorname{E}_{1}^{2}(x) e^{-x} \, \mathrm{d}x. $$

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

(See here.)

And

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

where

$$ \begin{align} \int \frac{\log(1+ \frac{z}{2})}{1+z} \, \mathrm{d}z &= \int \frac{\log (1+u)}{u} \, \mathrm{d}u - \log (2) \int\frac{1}{u} \, \mathrm{d}u \\ &= - \operatorname{Li}_{2}(-u) - \log(2) \log u +C \\ &= - \operatorname{Li}_{2}(-z-1) - \log(2) \log(z+1) + C. \end{align}$$

Therefore,

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

The above limit can be evaluated by hand by using the inversion formula for the dilogarithm and then expanding the result at infinity.

(See the second edit for a simpler approach.)

EDIT:

As Lucian pointed out in the comments, we can apply the inversion formula to get

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

Using the identity $$\operatorname{Li} (x) +\operatorname{Li}_{2}(-x) = \frac{1}{2} \, \operatorname{Li}_{2}(x^{2}), \quad |x| <1 ,$$ along with the known value of $\operatorname{Li}_{2}\left(\frac{1}{2} \right),$ the result can also be expressed as $$-3 \operatorname{Li}_{2} \left(\frac{1}{4} \right) - 6 \log^{2}(2). $$

SECOND EDIT:

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

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

where

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

and

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

Therefore.

$$\begin{align} \int_{-\infty}^{0} \operatorname{Ei}^{3}(x) \, dx &= -3 \operatorname{Li}_{2}(-3)-6 \log(3) \log(2) - \frac{\pi^{2}}{2} \\ &= 3 \operatorname{Li}_{2} \left(-\frac{1}{3} \right) + \frac{3}{2} \, \log^{2} (3) -6 \log(3) \log(2) \\&=-3\operatorname{Li}_{2} \left(\frac{1}{4} \right)- \frac{3}{2} \, \log^{2} \left(\frac{4}{3} \right) + \frac{3}{2} \, \log^{2} (3) -6 \log(3) \log(2) \tag{1}\\&= - 3\operatorname{Li}_{2} \left(\frac{1}{4} \right)- 6 \log^{2}(2). \end{align} $$

(Also see Ali Shather's comment below.)

$(1)$ Landen's Identity