TITLE: Conjectured closed form for $\int_0^1\frac{\operatorname{li}^4(x)}{x^4}\,dx$
QUESTION [11 upvotes]: It has been conjectured  by Kirill that
$$\int_0^1\frac{\operatorname{li}^4(x)}{x^4}\,dx=\int_0^\infty\operatorname{Ei}^4(-x)\,e^{3x}\,dx\stackrel{\color{gray}?}=\frac{26\pi^4}{135}+\frac49\left(\pi^2\ln^22-\ln^42\right)-\frac{32}{3}\operatorname{Li}_4\!\left(\tfrac12\right).$$
Can we prove this conjecture?

REPLY [3 votes]: Although I'm 7 years late, here is an alternative direct method:
First, we shall introduce a Laplace transform that will assist us:
$$e^{2x} \operatorname{Ei}^2 (-x) = \int_{0}^{\infty} \frac{2e^{-x t}\ln (1+t)}{2+t} \, dt$$
This can be proven by multiplying the following representation with itself:
$$-e^x \operatorname{Ei} (-x) = \int_{0}^{\infty} \frac{e^{-x t}}{t+1} \, dt$$
In a similar fashion
$$e^{4 x} \operatorname{Ei}^4 (-x) = \int_{0}^{\infty} \int_{0}^{\infty} \frac{4 e^{-x(t+u)} \ln (1+t) \ln (1+u)}{(2+t)(2+u)}\, du \, dt$$
$$\begin{align} \implies I = \int_{0}^{\infty} e^{3x} \operatorname{Ei}^4 (-x) \, dx &= \int_{0}^{\infty} \int_{0}^{\infty} \int_{0}^{\infty} \frac{4 e^{-x(t+u+1)} \ln (1+t) \ln (1+u)}{(2+t)(2+u)} \, dx \, du \, dt
\\ &= \int_{0}^{\infty} \frac{4 \ln (1+t)}{2+t} \int_{0}^{\infty} \frac{\ln (1+u)}{(2+u)(1+t+u)} \, du \, dt \\ &= \int_{0}^{\infty} \frac{-4\ln (1+t)}{(2+t)(t-1)} \left(\frac{\pi^2}{12} + \operatorname{Li}_2 (-t) \right) \, dt\end{align}$$
Enforce the substitution $t = e^x - 1 \> \implies dt = e^x \, dx$:
$$I = -4 \int_{0}^{\infty} \frac{x e^x}{(e^x+1)(e^x-2)}\left(\frac{\pi^2}{12}+\operatorname{Li}_2 (1-e^x)\right) \, dx$$
Define
$$A=-\frac{4}{3} \int_{0}^{\infty} \frac{x}{e^x+1} \left(\frac{\pi^2}{12}+\operatorname{Li}_2 (1-e^x)\right) \, dx$$
$$B=-\frac{8}{3} \int_{0}^{\infty} \frac{x}{e^x-2} \left(\frac{\pi^2}{12}+\operatorname{Li}_2 (1-e^x)\right) \, dx$$
such that $I=A+B$
Using
$$(1-2^{1-s}) \zeta (s) \Gamma (s)=\int_{0}^{\infty} \frac{x^{s-1}}{e^x+1} \, dx$$
$$\operatorname{Li}_2 (x) = \int_{0}^{\infty} \frac{x t}{e^t-x} \, dt$$
We have
$$\implies A=-\frac{\pi^4}{108} - \frac{4}{3} \int_{0}^{\infty} \frac{x \operatorname{Li}_2 (1-e^x)}{e^x+1} \, dx$$
$$\implies A = -\frac{\pi^4}{108} - \frac{4}{3} \int_{0}^{\infty} \frac{x \operatorname{Li}_2 (1-e^x)}{1-e^x} \, dx - \frac{8}{3} \int_{0}^{\infty} \frac{x}{e^x-2} \left(\frac{\pi^2}{12}+\operatorname{Li}_2 (1-e^x)\right) \, dx \\= -\frac{\pi^4}{108} - \frac{4}{3} \underbrace{\int_{0}^{\infty} \frac{x \operatorname{Li}_2 (1-e^x)}{1-e^x} \, dx}_J + B$$
Using dilogarithm identities
$$J = \int_{0}^{\infty} \frac{x \operatorname{Li}_2 (1-e^x)}{1-e^x} \, dx = \int_{0}^{\infty} \frac{x (\operatorname{Li}_2 (1-e^{-x})+ x^2/2)}{e^x-1} \, dx$$
Using $$\zeta(s) \Gamma (s) = \int_{0}^{\infty} \frac{x^{s-1}}{e^x - 1} \, dx$$
$$\implies J = \frac{\pi^4}{30} + \int_{0}^{\infty} x \left(\frac{\pi^2}{6} - \operatorname{Li}_2 (1-e^{-x})\right) \, dx + \int_{0}^{\infty} x^2 \ln (1-e^{-x}) + \frac{x}{2} \ln^2 (1-e^{-x})\, dx$$
With a few applications of integration by parts we conclude $$J = \frac{17 \pi^4}{360} \implies A = -\frac{13 \pi^4}{180}+B$$
$$I = \frac{29\pi^4}{540} - \frac{8}{3} \int_{0}^{\infty} \frac{x\operatorname{Li}_2 (1-e^x)}{e^x+1} \, dx$$
Applying integration by parts we have
$$\begin{align} I &= \frac{29\pi^4}{540} - \frac{8}{3} \int_{0}^{\infty}\frac{x}{e^{-x}-1}\left(\operatorname{Li}_2(-e^{-x})-x\ln(1+e^{-x})\right) \, dx \\
&=\frac{29\pi^4}{540}-\frac{8}{3}\left(\frac{\pi^4}{144}-\frac{7}{2}\ln(2)\zeta(3)\right)-\frac{8}{3}\int_{0}^{\infty}\frac{x \operatorname{Li}_2(-e^{-x})}{1-e^{-x}} \, dx\\ &= \frac{19 \pi^4}{540} + \frac{28}{3} \ln (2) \zeta (3)-\frac{8}{3}\underbrace{\int_{0}^{\infty}\frac{x \operatorname{Li}_2(-e^{-x})}{1-e^{-x}} \, dx}_K \end{align}$$
Once again using
$$\operatorname{Li}_2 (x) = \int_{0}^{\infty} \frac{x t}{e^t-x} \, dt$$
We have
$$K = \int_{0}^{\infty}\frac{x \operatorname{Li}_2(-e^{-x})}{1-e^{-x}} \, dx=-\frac{\pi^2}{3}\int_{0}^{\infty}\frac{t^2}{1+e^t} \, dt+\int_{0}^{\infty}\frac{t^2\ln(1+e^{-t})}{1+e^t} \, dt+\frac{1}{2}\int_{0}^{\infty}\frac{t\ln^2(1+e^{-t})}{1+e^t}\, dt+\int_{0}^{\infty}\frac{t\operatorname{Li}_2\left(\frac{1}{1+e^{-t}}\right)}{1+e^t} \, dt$$
It is not too much work to determine
$$\implies K =-\frac{7\pi^4}{120}-\frac{\pi^2}{8}\ln^2(2)+\frac{\ln^4(2)}{8}+3\operatorname{Li}_4 \left(\frac{1}{2}\right)+\frac{21}{8}\ln(2)\zeta(3)-\int_{0}^{\infty}x\left(\frac{\pi^2}{12}+\operatorname{Li}_2\left(\frac{1}{e^{-x}-1}\right)\right)\, dx$$
$$-\int_{0}^{\infty}x\left(\frac{\pi^2}{12}+\operatorname{Li}_2\left(\frac{1}{e^{-x}-1}\right)\right)\, dx = \int_{0}^{\infty}\frac{x^2}{2(e^x-1)}\left(\ln\left(1-\frac{e^{-x}}{2}\right)-\ln(1-e^{-x})+\ln(2)\right)\, dx$$
I will state a proposition that we shall now use that I will omit the proof of as it is somewhat tedious:

Proposition 1
$$f(b) = \int_{0}^{\infty}\frac{x^2\ln(1+be^{-x})}{2(e^x-1)}dx=-\frac{1}{2}\text{Li}_2(-b)^2-\ln(1+b)\text{Li}_3(-b)+\text{Li}_4(-b)+\ln(1+b)\zeta(3)$$

Putting everything together we have then
$$\implies I = \frac{19 \pi^4}{540} + \frac{28}{3} \ln (2) \zeta (3) -\frac{8}{3}\left(-\frac{7\pi^4}{120}-\frac{\pi^2}{8}\ln^2(2)+\frac{\ln^4(2)}{8}+3\operatorname{Li}_4\left(\frac{1}{2}\right)+\frac{21}{8}\ln(2)\zeta(3)+\ln(2)\zeta(3)+f\left(-\frac{1}{2}\right)-f(-1)\right)$$
Note that $f(-1)$ should be realised as $\lim_{b \, \to \, -1} f(b)$.
Upon simplification, this gives the desired result
$$I=\frac{26\pi^4}{135}+\frac49\left(\pi^2\ln^22-\ln^42\right)-\frac{32}{3}\operatorname{Li}_4\!\left(\tfrac12\right)$$
$\square$