Simplify the Laplace Transform for $E_{i}(-y)^{2}$

Question

I want to simplify the Laplace transform expression of $E_{i}(-y)^{2}$, where $E_{i}(y)$ is the exponential integral defined by $E_{i}(y) = -\int\limits_{-y}^{\infty} \frac{e^{-t}}{t} dt$.

Question: Can someone please show a simpler expression than what I derived, as shown below?

I know that the Laplace Transform $\mathcal{L}_{p} (E_{i}(-y)) = -\frac{log(p+1)}{p}, ROC \in Re(p) \gt 0$

For $E_{i}(-y)^{2}$, I used the formula for laplace transform of product from Wikipedia.

$\mathcal{L}_{p} (E_{i}(-y)^{2}) = \frac{1}{2\pi i} \int\limits_{c-i \infty}^{c + i \infty} \left(-\frac{log(1+\sigma)}{\sigma}\right) \left(-\frac{log(1+p-\sigma)}{p-\sigma}\right) d\sigma$, where $c$ is a vertical line right of origin (to be in ROC of one of these).

I closed the contour on the left half of the complex plane with a keyhole contour, with the hole at $(-1,0)$. If I am right, the integrals for big semicircle and the small circle at the hole vanish. Then I am left with two lines in the contour: one from $-\infty$ to $-1$ (slightly above the x-axis) and one from $-1$ to $-\infty$ slightly below x-axis.

Doing some math (not entirely sure of its correctness), I got the integral to be

$$-\int\limits_{0}^{\infty} \frac{log(p+2+x)}{(p+1+x)(x+1)} dx$$

When I entered this in Wolfram, it gave me this complicated result:

As you can see, the tool didn't simplify it. I am not very familiar with Polylog functions. Is there a simplification possible?

Answer 1

As is often the case, I must apologize for the length of the derivation below. In this case I needed to take great care that everything worked for complex parameters, which isn't my strong-suit.

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The Laplace transform $\mathcal{L}$ of a function $f:\mathbb{R}_{>0}\rightarrow\mathbb{R}$ is given by the improper integral

$$\mathcal{L}_{t}{\left[f{\left(t\right)}\right]}{\left(s\right)}=\int_{0}^{\infty}\mathrm{d}t\,e^{-st}f{\left(t\right)},$$

for all $s\in\mathbb{C}$ such that the integral converges.

Our goal is to obtain as simple an expression as possible for the Laplace transform of $\operatorname{Ei}{\left(-y\right)}^{2}$, where the exponential integral $\operatorname{Ei}{\left(z\right)}$ is defined for complex argument by

$$\operatorname{Ei}{\left(z\right)}:=-\int_{-z}^{\infty}\mathrm{d}t\,\frac{e^{-t}}{t};~~~\small{z\in\mathbb{C}\setminus[0,\infty)}.$$

This will be accomplished in terms of polylogarithms.

Recall that for nonnegative order $n$, the polylogarithm of a complex argument may be defined iteratively by

$$\operatorname{Li}_{n+1}{\left(z\right)}:=\int_{0}^{z}\mathrm{d}t\,\frac{\operatorname{Li}_{n}{\left(t\right)}}{t};~~~\small{n\in\mathbb{Z}_{\ge0}\land z\in\mathbb{C}\setminus(1,\infty)\land\left(n\ge1\lor z\neq1\right)},$$

$$\operatorname{Li}_{0}{\left(z\right)}:=\frac{z}{1-z};~~~\small{z\in\mathbb{C}\land z\neq1}.$$

Integrating by parts $n$ times yields the following integral representation for the polylogarithm of positive order:

$$\operatorname{Li}_{n+1}{\left(z\right)}=\frac{(-1)^{n}}{n!}\int_{0}^{1}\mathrm{d}t\,\frac{z\ln^{n}{\left(t\right)}}{1-zt};~~~\small{n\in\mathbb{Z}_{\ge0}\land z\in\mathbb{C}\setminus(1,\infty)\land\left(n\ge1\lor z\neq1\right)}.$$

Also recall the dilogarithm obeys the following well-known functional relations:

$$\frac12\operatorname{Li}_{2}{\left(z^{2}\right)}=\operatorname{Li}_{2}{\left(z\right)}+\operatorname{Li}_{2}{\left(-z\right)};~~~\small{z\in\mathbb{C}\setminus(-\infty,-1)\cup(1,\infty)},$$

$$\operatorname{Li}_{2}{\left(z\right)}+\operatorname{Li}_{2}{\left(\frac{1}{z}\right)}+\frac12\ln^{2}{\left(-z\right)}=2\operatorname{Li}_{2}{\left(-1\right)}=-\operatorname{Li}_{2}{\left(1\right)};~~~\small{z\in\mathbb{C}\setminus[0,\infty)},$$

$$\operatorname{Li}_{2}{\left(z\right)}+\operatorname{Li}_{2}{\left(\frac{z}{z-1}\right)}+\frac12\ln^{2}{\left(1-z\right)}=0;~~~\small{z\in\mathbb{C}\setminus[1,\infty)}.$$

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We will also make use of the following integration formula:

$$\int_{0}^{1}\mathrm{d}t\,\frac{b\ln{\left(1-at\right)}}{1-bt}=\operatorname{Li}_{2}{\left(a\right)}+\operatorname{Li}_{2}{\left(\frac{b}{b-1}\right)}-\operatorname{Li}_{2}{\left(\frac{a-b}{1-b}\right)};~~~\small{a\in\mathbb{C}\setminus(1,\infty)\land b\in\mathbb{C}\setminus[1,\infty)}.$$

Proof:

$$\begin{align} \int_{0}^{1}\mathrm{d}t\,\frac{b\ln{\left(1-at\right)}}{1-bt} &=-\int_{0}^{1}\mathrm{d}t\,\frac{b}{1-bt}\int_{0}^{1}\mathrm{d}x\,\frac{at}{1-axt}\\ &=-\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}x\,\frac{abt}{\left(1-axt\right)\left(1-bt\right)}\\ &=-\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}t\,\frac{abt}{\left(1-axt\right)\left(1-bt\right)}\\ &=-\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}t\,\frac{ab}{\left(ax-b\right)}\left[\frac{1}{1-axt}-\frac{1}{1-bt}\right]\\ &=-\int_{0}^{1}\mathrm{d}x\,\frac{1}{x\left(ax-b\right)}\left[b\int_{0}^{1}\mathrm{d}t\,\frac{ax}{1-axt}-ax\int_{0}^{1}\mathrm{d}t\,\frac{b}{1-bt}\right]\\ &=-\int_{0}^{1}\mathrm{d}x\,\frac{1}{x\left(ax-b\right)}\left[-b\ln{\left(1-ax\right)}+ax\ln{\left(1-b\right)}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{b\ln{\left(1-ax\right)}}{x\left(ax-b\right)}-\frac{a\ln{\left(1-b\right)}}{ax-b}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\left[-\frac{\ln{\left(1-ax\right)}}{x}+\frac{a\ln{\left(1-ax\right)}}{ax-b}-\frac{a\ln{\left(1-b\right)}}{ax-b}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\left[-\frac{\ln{\left(1-ax\right)}}{x}+\frac{a\ln{\left(\frac{1-ax}{1-b}\right)}}{ax-b}\right]\\ &=-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-ax\right)}}{x}+\int_{0}^{1}\mathrm{d}x\,\frac{a\ln{\left(\frac{1-ax}{1-b}\right)}}{ax-b}\\ &=\operatorname{Li}_{2}{\left(a\right)}+\int_{0}^{1}\mathrm{d}x\,\frac{a\ln{\left(\frac{1-ax}{1-b}\right)}}{ax-b}\\ &=\operatorname{Li}_{2}{\left(a\right)}+\int_{0}^{a}\mathrm{d}y\,\frac{\ln{\left(\frac{1-y}{1-b}\right)}}{y-b};~~~\small{\left[ax=y\right]}\\ &=\operatorname{Li}_{2}{\left(a\right)}+\int_{1-a}^{1}\mathrm{d}t\,\frac{\ln{\left(\frac{t}{1-b}\right)}}{1-t-b};~~~\small{\left[1-y=t\right]}\\ &=\operatorname{Li}_{2}{\left(a\right)}+\int_{\frac{1-a}{1-b}}^{\frac{1}{1-b}}\mathrm{d}u\,\frac{\ln{\left(u\right)}}{1-u};~~~\small{\left[\frac{t}{1-b}=u\right]}\\ &=\operatorname{Li}_{2}{\left(a\right)}+\int_{\frac{a-b}{1-b}}^{\frac{b}{b-1}}\mathrm{d}v\,\frac{(-1)\ln{\left(1-v\right)}}{v};~~~\small{\left[1-u=v\right]}\\ &=\operatorname{Li}_{2}{\left(a\right)}+\operatorname{Li}_{2}{\left(\frac{b}{b-1}\right)}-\operatorname{Li}_{2}{\left(\frac{a-b}{1-b}\right)}.\\ \end{align}$$

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Suppose $s\in\mathbb{C}\land\Re{(s)}\ge-2$. We begin by using the integral representation of the exponential integral to rewrite the Laplace transform as a triple integral, and then change the order of integration:

$$\begin{align} \mathcal{L}_{y}{\left[\operatorname{Ei}{\left(-y\right)}^{2}\right]}{\left(s\right)} &=\int_{0}^{\infty}\mathrm{d}y\,e^{-sy}\operatorname{Ei}{\left(-y\right)}^{2}\\ &=\int_{0}^{\infty}\mathrm{d}y\,e^{-sy}\left[-\int_{y}^{\infty}\mathrm{d}x\,\frac{e^{-x}}{x}\right]^{2}\\ &=\int_{0}^{\infty}\mathrm{d}y\,e^{-sy}\left[\int_{1}^{\infty}\mathrm{d}t\,\frac{e^{-yt}}{t}\right]^{2};~~~\small{[x=yt]}\\ &=\int_{0}^{\infty}\mathrm{d}y\,e^{-sy}\int_{1}^{\infty}\mathrm{d}t\int_{1}^{\infty}\mathrm{d}u\,\frac{e^{-yt}}{t}\cdot\frac{e^{-yu}}{u}\\ &=\int_{0}^{\infty}\mathrm{d}y\int_{1}^{\infty}\mathrm{d}t\int_{1}^{\infty}\mathrm{d}u\,\frac{e^{-sy}e^{-yt}e^{-yu}}{tu}\\ &=\int_{0}^{\infty}\mathrm{d}y\int_{1}^{\infty}\mathrm{d}t\int_{1}^{\infty}\mathrm{d}u\,\frac{e^{-sy-yt-yu}}{tu}\\ &=\int_{0}^{\infty}\mathrm{d}y\int_{1}^{\infty}\mathrm{d}t\int_{1}^{\infty}\mathrm{d}u\,\frac{e^{-(s+t+u)y}}{tu}\\ &=\int_{1}^{\infty}\mathrm{d}t\int_{1}^{\infty}\mathrm{d}u\int_{0}^{\infty}\mathrm{d}y\,\frac{e^{-(s+t+u)y}}{tu}\\ &=\int_{1}^{\infty}\mathrm{d}t\int_{1}^{\infty}\mathrm{d}u\,\frac{1}{tu(s+t+u)}.\\ \end{align}$$

The inner integral in the double integral above is elementary, so we can reduce the Laplace transform to the following single-variable integral:

$$\begin{align} \mathcal{L}_{y}{\left[\operatorname{Ei}{\left(-y\right)}^{2}\right]}{\left(s\right)} &=\int_{1}^{\infty}\mathrm{d}t\int_{1}^{\infty}\mathrm{d}u\,\frac{1}{tu(s+t+u)}\\ &=\int_{1}^{\infty}\mathrm{d}t\int_{1}^{0}\mathrm{d}y\,\frac{(-1)}{y^{2}}\cdot\frac{1}{ty^{-1}(s+t+y^{-1})};~~~\small{\left[u=y^{-1}\right]}\\ &=\int_{1}^{\infty}\mathrm{d}t\int_{0}^{1}\mathrm{d}y\,\frac{1}{t\left[\left(s+t\right)y+1\right]}\\ &=\int_{1}^{\infty}\mathrm{d}t\,\frac{1}{t\left(s+t\right)}\int_{0}^{1}\mathrm{d}y\,\frac{\left(s+t\right)}{\left[1+\left(s+t\right)y\right]}\\ &=\int_{1}^{\infty}\mathrm{d}t\,\frac{1}{t\left(s+t\right)}\int_{0}^{1}\mathrm{d}y\,\frac{d}{dy}\ln{\left(1+(s+t)y\right)}\\ &=\int_{1}^{\infty}\mathrm{d}t\,\frac{\ln{\left(1+s+t\right)}}{t\left(s+t\right)}.\\ \end{align}$$

(Note: This is equivalent to the integral you found up to a single sign error out front.)

Then, assuming $\Re{(s)}>-1$, we have

$$\begin{align} s\mathcal{L}_{y}{\left[\operatorname{Ei}{\left(-y\right)}^{2}\right]}{\left(s\right)} &=s\int_{1}^{\infty}\mathrm{d}t\,\frac{\ln{\left(t+1+s\right)}}{t\left(t+s\right)}\\ &=s\int_{1}^{0}\mathrm{d}u\,\frac{(-1)u^{-2}\ln{\left(u^{-1}+1+s\right)}}{u^{-1}\left(u^{-1}+s\right)};~~~\small{\left[t=u^{-1}\right]}\\ &=\int_{0}^{1}\mathrm{d}u\,\frac{s\ln{\left(\frac{1+(1+s)u}{u}\right)}}{1+su}\\ &=\int_{0}^{1}\mathrm{d}u\,\frac{s\ln{\left(1+(1+s)u\right)}}{1+su}-\int_{0}^{1}\mathrm{d}u\,\frac{s\ln{\left(u\right)}}{1+su}\\ &=\operatorname{Li}_{2}{\left(-\frac{1}{1+s}\right)}-\operatorname{Li}_{2}{\left(-1-s\right)}-\operatorname{Li}_{2}{\left(\frac{s}{1+s}\right)}-\operatorname{Li}_{2}{\left(-s\right)}\\ &=\operatorname{Li}_{2}{\left(-\frac{1}{1+s}\right)}-\operatorname{Li}_{2}{\left(-1-s\right)}+\frac12\ln^{2}{\left(1+s\right)}\\ &=2\operatorname{Li}_{2}{\left(-1\right)}-2\operatorname{Li}_{2}{\left(-1-s\right)},\\ \end{align}$$

and thus,

$$\mathcal{L}_{y}{\left[\operatorname{Ei}{\left(-y\right)}^{2}\right]}{\left(s\right)}=\frac{2\operatorname{Li}_{2}{\left(-1\right)}-2\operatorname{Li}_{2}{\left(-1-s\right)}}{s};~~~\small{\Re{(s)}>-1}.\blacksquare$$

I'm quite confident you won't find a simpler expression than this. Cheers!

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Answer 2

Assume that $s>0$.

Integrating by parts and choosing $-\frac{1}{s}\left(e^{-sx}-1 \right)$ for the antiderivative of $e^{-sx}$ so that the boundary term doesn't blow up as $x \to 0^{+}$, we have

$$ \begin{align} \int_{0}^{\infty} e^{-sx} \operatorname{Ei}(-x)^{2} \, \mathrm dx &= \frac{2}{s} \int_{0}^{\infty} \frac{e^{-(1+s)x}-e^{-x}}{x} \, \operatorname{Ei}(-x) \, \mathrm dx \\ &= -\frac{2}{s} \int_{0}^{\infty}\frac{e^{-(1+s)x}-e^{-x}}{x} \int_{1}^{\infty} \frac{e^{-xt}}{t} \, \mathrm dt \, \mathrm dx \\ &= -\frac{2}{s} \int_{1}^{\infty} \frac{1}{t} \int_{0}^{\infty} \frac{e^{-(1+s+t)x}-e^{-(1+t)x}}{x} \, \mathrm dx \, \mathrm dt \\ &\overset{\clubsuit}{=} -\frac{2}{s} \int_{1}^{\infty} \frac{1}{t} \, \ln \left(\frac{1+t}{1+s+t}\right) \, \mathrm dt \\ &\overset{\spadesuit}{=} -\frac{2}{s} \int_{0}^{1} \frac{1}{u} \, \ln \left(\frac{1+u}{1+(1+s)u} \right) \, \mathrm du \\ &= -\frac{2}{s} \left(\int_{0}^{1} \frac{\ln(1+u)}{u} \, \mathrm du - \int_{0}^{1} \frac{\ln \left(1+(1+s)u\right)}{u} \, \mathrm du \right) \\ &= \frac{2}{s} \left(\operatorname{Li}_{2}(-1) -\operatorname{Li}_{2}(-1-s)\right), \end{align}$$ which matches David H's result.

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$\clubsuit$ https://en.wikipedia.org/wiki/Frullani_integral

$\spadesuit$ Let $t = \frac{1}{u}$.

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Let $\operatorname{Li}_{2}()$ be the principal branch of the dilogarithm.

If we assign $-\frac{2}{s} \left(\operatorname{Li}_{s}(-1)-\operatorname{Li}_{2}(-1-s) \right)$ its limit value at $s=0$, then the results holds for $\Re(s) >-2$ by the identity theorem.