TITLE: Closed form for $\int_0^e\mathrm{Li}_2(\ln{x})\,dx$?
QUESTION [18 upvotes]: Inspired by this question and this answer, I decided to investigate the family of integrals 
$$I(k)=\int_0^e\mathrm{Li}_k(\ln{x})\,dx,\tag{1}$$
where $\mathrm{Li}_k(z)$ represents the polylogarithm of order $k$ and argument $z$. $I(1)$ evaluates to $e\gamma$, but $I(2)$ has resisted my efforts (which can be seen here).

Neither ISC nor WolframAlpha could provide a closed form for its numerical value--however, I've conjectured a possible analytic form. $$\eqalign{\int_0^e\mathrm{Li}_2(\ln{x})\,dx&\stackrel?=\,_3F_3(1,1,1;2,2,2;1)+\frac{\pi^2(2e-5)}{12}+\frac{\gamma^2}{2}-\gamma\,\mathrm{Ei}(1)\\&=0.578255559804073275225659054377625577...\tag{2}}$$ 

Brevan Ellefsen has computed that my conjecture is accurate to at least 150 digits. Brevan also gave the alternate form 
$$\frac{\pi^2e}{6}+\gamma G_{1,2}^{2,0}\left(-1\left|\begin{array}{c}1\\0,0\\\end{array}\right.\right)+G_{2,3}^{3,0}\left(-1\left|\begin{array}{c}1,1\\0,0,0\\\end{array}\right.\right).\tag{2.1}$$

Is there a closed form for $I(2)$ that doesn't involve Meijer G or hypergeometrics? The simplicity of the following two equations seems to suggest that there might be. $(3.1)$ follows directly from $(3)$, which I've proven here. 
  $$\begin{align}
\sum_{k=1}^\infty I(k)&=e\tag{3}\\\sum_{k=2}^\infty I(k)&=e(1-\gamma)\tag{3.1}\end{align}$$


PROGRESS UPDATE: Using this equation, I've turned $_3F_3(1,1,1;2,2,2;1)$ into
$$\lim_{c\to 1}\left(\frac{\mathrm{Ei}(1)-\gamma}{c-1}+\frac{1-e}{(c-1)^2}+\frac{(-1)^{-c}\,\Gamma(c-1)}{c-1}+\frac{(-1)^{1-c}\,\Gamma(c,-1)}{(c-1)^2}\right),\tag{4}\label{4}$$ but I don't know how to proceed from there. EDIT: This limit leads nowhere. See below. 
SECOND PROGRESS UPDATE: After some studying of the properties of the Meijer G function, I've finally cracked the limit; however, the result is an underwhelming $_3F_3(1,1,1;2,2,2;1)$. Before I evaluate the limit, I'd first like to state the following intermediate result:

Lemma $(4.1)$: For $z\in\mathbb{C}$, 
  $$G_{2,3}^{3,0}\left(z\left|\begin{array}{c}1,1\\0,0,0\\\end{array}\right.\right)=\gamma\ln{z}+\frac12\ln^2(z)-z\,_3F_3(1,1,1;2,2,2;-z)+\frac{\gamma^2}2+\frac{\pi^2}{12}.\tag{4.1}\label{4.1}$$

My proof for this can be found here. Now I return to the limit $\eqref{4}$.
Consider the following: $\frac{1}{c-1}=\frac{c-1}{(c-1)^2}$, $(c-1)\Gamma(c-1)=\Gamma(c)$, and $(-1)^{1-c}=-(-1)^{-c}$. Based on these algebraic identities, the limit can be written as 
$$\lim_{c\to 1}\frac{(c-1)(\mathrm{Ei}(1)-\gamma)+1-e+(-1)^{-c}(\Gamma(c)-\Gamma(c,-1))}{(c-1)^2}.\tag{4.2}$$ 
In this form, the limit is $\frac{0}0$. Using l'Hospital twice, we obtain 
$$\lim_{c\to 1}{(-1)^{-c}\left(-G_{3,4}^{4,0}\left(-1\left|\begin{array}{c}1,1,1\\0,0,0,c\\\end{array}\right.\right)+\Gamma{(c)}\left(\frac{\psi_0{(c)}^2}2-i\pi\psi_0(c)+\frac{\psi_1(c)}2-\frac{\pi^2}2\right)\right)}$$ $$\begin{align}&=G_{3,4}^{4,0}\left(-1\left|\begin{array}{c}1,1,1\\0,0,0,1\\\end{array}\right.\right)-\frac{\psi_0(1)^2}2+i\pi\psi_0(1)-\frac{\psi_1(1)}2+\frac{\pi^2}2\\&=G_{2,3}^{3,0}\left(-1\left|\begin{array}{c}1,1\\0,0,0\\\end{array} \right.\right)-\frac{\gamma^2}{2}-i\gamma\pi+\frac{5\pi^2}{12}.\tag{4.2a}\end{align}$$ 
Using Lemma $\eqref{4.1}$, we know that 
$$G_{2,3}^{3,0}\left(-1\left|\begin{array}{c}1,1\\0,0,0\\\end{array}\right.\right)={}_3F_3(1,1,1;2,2,2;1)+\frac{\gamma^2}2+i\gamma\pi-\frac{5\pi^2}{12},\tag{4.3}$$ which can be rewritten as 
$$G_{2,3}^{3,0}\left(-1\left|\begin{array}{c}1,1\\0,0,0\\\end{array}\right.\right)-\frac{\gamma^2}{2}-i\gamma\pi+\frac{5\pi^2}{12}={}_3F_3(1,1,1;2,2,2;1).\tag{4.3a}$$ 

Thus, 
  $$\eqalign{&\lim_{c\to 1}\left(\frac{\mathrm{Ei}(1)-\gamma}{c-1}+\frac{1-e}{(c-1)^2}+\frac{(-1)^{-c}\,\Gamma(c-1)}{c-1}+\frac{(-1)^{1-c}\,\Gamma(c,-1)}{(c-1)^2}\right)\\=&{}_3F_3(1,1,1;2,2,2;1).\tag{4.4}}$$

REPLY [3 votes]: Lemma 1: For $\lambda \in \mathbb{R}^{+}$ $$\sum\limits_{n=1}^{\infty} \dfrac{\lambda^n}{n!}\mathcal{H}_{n} = e^{\lambda}\left(\ln \lambda + \gamma - \operatorname{Ei}(-\lambda)\right) \tag{1} \label{lemma1}$$
  where $\mathcal{H}_{n}$ — $n$-th harmonic number and $\operatorname{Ei}(\cdot)$ — exponential integral.

Start from two series representations of lower incomplete gamma function $\gamma(\beta, \lambda)$:
\begin{align}
\gamma(\beta, \lambda) &= e^{-\lambda}\sum\limits_{n=0}^{\infty} \dfrac{\lambda^{n+\beta}}{\beta\left(\beta+1\right)\ldots\left(\beta+n\right)} \\
\gamma(\beta, \lambda) &= \sum\limits_{n=0}^{\infty} (-1)^{n}\dfrac{\lambda^{n+\beta}}{n!\left(\beta+n\right)}
\end{align}
Now take the derivative with respect to $\beta$ at $1$. Since 
\begin{align*}
\dfrac{\mathrm{d}}{\mathrm{d}\beta}\left(\dfrac{1}{\beta\left(\beta+1\right)\ldots\left(\beta+n\right)}\right)_{\beta=1} &= -\left.\dfrac{1}{\beta\left(\beta+1\right)\ldots\left(\beta+n\right)}\left(\dfrac{1}{\beta}+\dfrac{1}{\beta+1}+\ldots+\dfrac{1}{\beta+n}\right)\right\vert_{\ \beta=1} \\
&= -\dfrac{\mathcal{H}_{n+1}}{(n+1)!}
\end{align*}
we have that
\begin{align*}
-e^{-\lambda}\sum\limits_{n=1}^{\infty} \dfrac{\lambda^{n}}{n!}\mathcal{H}_{n}+\ln \lambda\ e^{-\lambda}\sum\limits_{n=1}^{\infty} \dfrac{\lambda^{n}}{n!} &= -\ln \lambda\sum\limits_{n=1}^{\infty} (-1)^{n}\dfrac{\lambda^{n}}{n!} + \sum\limits_{n=1}^{\infty} (-1)^{n}\dfrac{\lambda^{n}}{n!\ n} \\
-e^{-\lambda}\sum\limits_{n=1}^{\infty} \dfrac{\lambda^{n}}{n!}\mathcal{H}_{n}+\ln \lambda \left(1-e^{-\lambda}\right) &= \ln \lambda \left(1-e^{-\lambda}\right) + \sum\limits_{n=1}^{\infty} (-1)^{n}\dfrac{\lambda^{n}}{n!\ n} \\
\sum\limits_{n=1}^{\infty} \dfrac{\lambda^{n}}{n!}\mathcal{H}_{n} &= -e^{\lambda}\sum\limits_{n=1}^{\infty} (-1)^{n}\dfrac{\lambda^{n}}{n!\ n} \\
&= e^{\lambda}\left(\ln \lambda + \gamma - \operatorname{Ei}(-\lambda)\right)
\end{align*}

Let 
$$
\mathfrak{I}(\lambda, \alpha, \beta) = \int\limits_{0}^{\infty} \dfrac{e^{-\lambda x}\,x^{\alpha - 1}}{\left(1+x\right)^{\alpha+\beta}}\,\mathrm{d}x
$$

Lemma 2: Let $\alpha, \lambda \in \mathbb{R}^{+}$ and $\alpha + \beta > 0$. Then
  $$
\mathfrak{I}(\lambda, \alpha, \beta) = \dfrac{\Gamma(\alpha)}{\Gamma(\alpha+\beta)}\lambda^{\beta}\mathfrak{I}(\lambda, \alpha + \beta, -\beta) \tag{2} \label{lemma2}
$$

Using Laplace transform properties we have
\begin{align*}
\mathfrak{I}(\lambda, \alpha, \beta) &= \int\limits_{0}^{\infty} \dfrac{e^{-\lambda x}x^{\alpha - 1}}{\left(1+x\right)^{\alpha+\beta}}\,\mathrm{d}x \\
&= \int\limits_{0}^{\infty} e^{-\lambda t}t^{\alpha-1}\,\mathcal{L}\left\{\dfrac{e^{-x}x^{\alpha+\beta-1}}{\Gamma(\alpha+\beta)}\right\}(t)\,\mathrm{d}t \\
&= \int\limits_{0}^{\infty} \dfrac{\Gamma(\alpha)}{\left(\lambda+x\right)^{\alpha}}\dfrac{e^{-x}x^{\alpha+\beta-1}}{\Gamma(\alpha+\beta)} \,\mathrm{d}x \\
&= \dfrac{\Gamma(\alpha)}{\Gamma(\alpha+\beta)}\lambda^{\beta}\int\limits_{0}^{\infty} \dfrac{e^{-\lambda x}x^{\alpha+\beta-1}}{\left(1+x\right)^{\alpha}}\,\mathrm{d}x \\
&= \dfrac{\Gamma(\alpha)}{\Gamma(\alpha+\beta)}\lambda^{\beta}\mathfrak{I}(\lambda, \alpha + \beta, -\beta)
\end{align*}


Claim 1: For $\lambda > 0$ and $\beta \in \mathbb{R}$
$$
\mathfrak{I}(\lambda, 1, \beta) = e^{\lambda}\lambda^{\beta}\Gamma(-\beta, \lambda) \tag{3} \label{claim1}
$$
  where $\Gamma(\cdot, \cdot)$ — upper incomplete gamma function.

\begin{align*}
\mathfrak{I}(\lambda, 1, \beta) &= \int\limits_{0}^{\infty} \dfrac{e^{-\lambda x}}{\left(1+x\right)^{1+\beta}}\,\mathrm{d}x \\
&= e^{\lambda}\int\limits_{1}^{\infty} \dfrac{e^{-\lambda x}}{x^{1+\beta}}\,\mathrm{d}x \\
&= e^{\lambda}\lambda^{\beta}\int\limits_{\lambda}^{\infty} \dfrac{e^{-x}}{x^{1+\beta}}\,\mathrm{d}x \\
&= e^{\lambda}\lambda^{\beta}\Gamma(-\beta, \lambda)
\end{align*}

After applying \eqref{lemma2} to \eqref{claim1} we have that
$$
\int\limits_{0}^{\infty} \dfrac{e^{-\lambda x}x^{\beta}}{1+x}\,\mathrm{d}x = e^{\lambda}\Gamma(1+\beta)\Gamma(-\beta, \lambda) \tag{4} \label{consec1}
$$
Taking the derivative with respect to the variable $\beta$ at point $(\lambda, \beta) = (1, 0)$ leads us to
\begin{align*}
\int\limits_{0}^{\infty} \dfrac{e^{-x}\ln x}{1+x}\,\mathrm{d}x &= e\dfrac{\partial}{\partial \beta}\Big(\Gamma(1+\beta)\Gamma(-\beta, 1)\Big)\Bigg\vert_{\beta=0} \\
&= -e\gamma \int\limits_{1}^{\infty} \dfrac{e^{-x}}{x}\,\mathrm{d}x -e \int\limits_{1}^{\infty} \dfrac{e^{-x}\ln x}{x}\,\mathrm{d}x \\
&= e\gamma\operatorname{Ei}(-1)-\dfrac{1}{2}e\int\limits_{1}^{\infty} e^{-x}\ln^2 x\,\mathrm{d}x \\
&= e\gamma\operatorname{Ei}(-1)-\dfrac{1}{2}e\dfrac{\mathrm{d}^2}{\mathrm{d}x^2}\Gamma(1)+\dfrac{1}{2}e\int\limits_{0}^{1} e^{-x}\ln^2 x\,\mathrm{d}x \\
\end{align*}

$$
\int\limits_{0}^{\infty} \dfrac{e^{-x}\ln x}{1+x}\,\mathrm{d}x = e\gamma\operatorname{Ei}(-1)-\dfrac{1}{2}e\left(\gamma^2+\dfrac{1}{6}\pi^2\right)-e\sum\limits_{n=1}^{\infty} \dfrac{(-1)^n}{n!\,n^2} \tag{5} \label{first}
$$


Let 
$$
f(\lambda) = \int\limits_{0}^{\infty} e^{-x}\ln^2 \left(\lambda x + x^2\right)\,\mathrm{d}x
$$
Then
\begin{align*}
f'(\lambda) &= 2\int\limits_{0}^{\infty} e^{-x}\dfrac{\ln \left(\lambda x + x^2\right)}{\lambda + x}\,\mathrm{d}x \\
&= 2\int\limits_{0}^{\infty} \dfrac{e^{-\lambda x}}{1+x}\Big(\ln \left(1+x\right)+\ln x + 2 \ln \lambda\Big)\,\mathrm{d}x \\
&= 4\ln \lambda \int\limits_{0}^{\infty} \dfrac{e^{-\lambda x}}{1+x}\,\mathrm{d}x + 2\int\limits_{0}^{\infty} \dfrac{e^{-\lambda x}}{1+x}\ln \left(1+x\right)\,\mathrm{d}x + 2\int\limits_{0}^{\infty} \dfrac{e^{-\lambda x}}{1+x}\ln x\,\mathrm{d}x \\
&= -4\ln \lambda e^{\lambda}\operatorname{Ei}(-\lambda) - 2\dfrac{\partial}{\partial \beta}\left.\left(\int\limits_{0}^{\infty} \dfrac{e^{-\lambda x}}{\left(1+x\right)^{1+\beta}}\,\mathrm{d}x\right)\right\vert_{\beta=0} + 2\dfrac{\partial}{\partial \beta}\left.\left(\int\limits_{0}^{\infty} \dfrac{e^{-\lambda x}x^{\beta}}{1+x}\,\mathrm{d}x\right)\right\vert_{\beta=0} \\
&= -4\ln \lambda e^{\lambda}\operatorname{Ei}(-\lambda) -2e^{\lambda}\dfrac{\partial}{\partial \beta}\Big(\lambda^{\beta}\Gamma(-\beta,\lambda)\Big)\Bigg\vert_{\beta=0} + 2e^{\lambda}\dfrac{\partial}{\partial \beta}\Big(\Gamma(1+\beta)\Gamma(-\beta,\lambda)\Big)\Bigg\vert_{\beta=0} \\
&= -2\ln \lambda e^{\lambda}\operatorname{Ei}(-\lambda)+2\gamma e^{\lambda}\operatorname{Ei}(-\lambda)
\end{align*}
Integrating over $[0,1]$ leads us to
\begin{align*}
f(1) &= f(0) -2\int\limits_{0}^{1} \ln \lambda e^{\lambda}\operatorname{Ei}(-\lambda)\,\mathrm{d}\lambda + 2\gamma\Big(e^{\lambda}\operatorname{Ei}(-\lambda)-\ln \lambda\Big)\Bigg\vert_{0}^{1} \\
&= 2\gamma e\operatorname{Ei}(-1)+2\gamma^2+\dfrac{2}{3}\pi^2-2\int\limits_{0}^{1} \ln \lambda e^{\lambda}\operatorname{Ei}(-\lambda)\,\mathrm{d}\lambda
\end{align*}
For integral in last formula we use \eqref{lemma1}:
\begin{align*}
\int\limits_{0}^{1} \ln \lambda e^{\lambda}\operatorname{Ei}(-\lambda)\,\mathrm{d}\lambda &= \int\limits_{0}^{1} \ln^2 \lambda e^{\lambda}\,\mathrm{d}\lambda +\gamma \int\limits_{0}^{1} \ln \lambda e^{\lambda}\,\mathrm{d}\lambda + \sum\limits_{n=2}^{\infty} \dfrac{\mathcal{H}_{n-1}}{n!\ n} \\
&= \sum\limits_{n=1}^{\infty} \dfrac{1}{n!\ n^2} + \gamma^2-\gamma\operatorname{Ei}(1) + \sum\limits_{n=1}^{\infty} \dfrac{\mathcal{H}_{n}}{n!\ n} 
\end{align*}
Combining all together we have

$$
\int\limits_{0}^{\infty} e^{-x}\ln^2 \left(x+x^2\right)\,\mathrm{d}x = 2\gamma\Big(e\operatorname{Ei}(-1)+\operatorname{Ei}(1)\Big)+\dfrac{2}{3}\pi^2-2\sum\limits_{n=1}^{\infty} \dfrac{1}{n!\ n^2}-2\sum\limits_{n=1}^{\infty} \dfrac{\mathcal{H}_{n}}{n!\ n} \tag{6} \label{second}
$$

From another side
\begin{align*}
\int\limits_{0}^{\infty} e^{-x}\ln^2 \left(x+x^2\right)\,\mathrm{d}x &= \int\limits_{0}^{\infty} e^{-x}\Big(\ln^2 \left(1+x\right) + \ln^2 x + 2\ln \left(1+x\right)\ln x\Big)\,\mathrm{d}x \\
&= \gamma^2+\dfrac{1}{6}\pi^2+e\int\limits_{1}^{\infty}e^{-x}\ln^2 x\,\mathrm{d}x + 2\int\limits_{0}^{\infty} e^{-x}\ln \left(1+x\right)\ln x\,\mathrm{d}x \\
&= \left(1+e\right)\left(\gamma^2+\dfrac{1}{6}\pi^2\right)+2e\sum\limits_{n=1}^{\infty} \dfrac{(-1)^n}{n!\ n^2} + 2\int\limits_{0}^{\infty} e^{-x}\ln \left(1+x\right)\ln x\,\mathrm{d}x
\end{align*}
 Express last integral

\begin{align*}
\int\limits_{0}^{\infty} e^{-x}\ln \left(1+x\right)\ln x\,\mathrm{d}x &= \gamma\Big(e\operatorname{Ei}(-1)+\operatorname{Ei}(1)\Big) -\dfrac{1}{2}\gamma^2+\dfrac{1}{4}\pi^2-\dfrac{1}{2}e\left(\gamma^2+\dfrac{1}{6}\pi^2\right) \\
&-e\sum\limits_{n=1}^{\infty} \dfrac{(-1)^n}{n!\ n^2}-\sum\limits_{n=1}^{\infty} \dfrac{1}{n!\ n^2}-\sum\limits_{n=1}^{\infty} \dfrac{\mathcal{H}_{n}}{n!\ n}
\end{align*}

Now use integration by parts and already obtained result \eqref{first}:
\begin{align*}
\int\limits_{0}^{\infty} e^{-x}\ln \left(1+x\right)\ln x\,\mathrm{d}x &= \int\limits_{0}^{\infty} e^{-x}\dfrac{\ln x}{1+x}\,\mathrm{d}x + \int\limits_{0}^{\infty} e^{-x}\dfrac{\ln \left(1+x\right)}{x}\,\mathrm{d}x \\
&= e\gamma\operatorname{Ei}(-1)-\dfrac{1}{2}e\left(\gamma^2+\dfrac{1}{12}\pi^2\right)-e\sum\limits_{n=1}^{\infty} \dfrac{(-1)^n}{n!\ n^2}+ \int\limits_{0}^{\infty} e^{-x}\dfrac{\ln \left(1+x\right)}{x}\,\mathrm{d}x
\end{align*}
So

$$ 
\int\limits_{0}^{\infty} e^{-x}\dfrac{\ln \left(1+x\right)}{x}\,\mathrm{d}x = -\dfrac{1}{2}\gamma^2+\dfrac{1}{4}\pi^2+\gamma\operatorname{Ei}(1)-\sum\limits_{n=1}^{\infty} \dfrac{1}{n!\ n^2}-\sum\limits_{n=1}^{\infty} \dfrac{\mathcal{H}_{n}}{n!\ n} \tag{7} \label{final}
$$


Back to original problem
With integration by parts and substitution original integral can be converted to
\begin{align*}
\int\limits_{0}^{e} \operatorname{Li}_2\left(\ln x\right)\,\mathrm{d}x &= \dfrac{1}{6}e\pi^2-\int\limits_{0}^{\infty} e^{-x}\dfrac{\ln \left(1+x\right)}{x}\,\mathrm{d}x+\int\limits_{0}^{1} e^{x}\dfrac{\ln \left(1-x\right)}{x}\,\mathrm{d}x \\
&= \dfrac{1}{6}e\pi^2 + \dfrac{1}{2}\gamma^2-\dfrac{1}{4}\pi^2-\gamma\operatorname{Ei}(1)+\sum\limits_{n=1}^{\infty} \dfrac{1}{n!\ n^2}+\sum\limits_{n=1}^{\infty} \dfrac{\mathcal{H}_{n}}{n!\ n} \\
&+\int\limits_{0}^{1} \dfrac{\ln\left(1-x\right)}{x}\,\mathrm{d}x+\sum\limits_{n=1}^{\infty}\dfrac{1}{n!}\int\limits_{0}^{1} x^{n-1}\ln\left(1-x\right)\,\mathrm{d}x \\
&= \dfrac{1}{6}e\pi^2 + \dfrac{1}{2}\gamma^2-\dfrac{5}{12}\pi^2-\gamma\operatorname{Ei}(1)+\sum\limits_{n=1}^{\infty} \dfrac{1}{n!\ n^2}
\end{align*}