Relating $\int_0^1\frac{(\ln x)^{n-1}(\ln(1-z\,x))^p}{x}dx$ and $\int_0^1\frac{(\ln x)^{n}(\ln(1-z\,x))^{p-1}}{1-z\,x}dx$

This post, after a complicated analysis, evaluates the integral $$I=\int_0^1\frac{\ln^2(x)\,\ln^3(1+x)}xdx$$

simply as

$$I =-\frac{\pi^6}{252}-18\zeta(\bar{5},1)+3\zeta^2(3)\tag1$$

where,

$$\zeta(\bar{5},1)=\frac{1}{24}\int^1_0\frac{\ln^4{x}\ln(1+x)}{1+x}{\rm d}x$$

More succinctly,

$$I = -12\,S_{3,3}(-1)\tag2$$

with Nielsen generalized polylogarithm $$S_{n,p}(z)$$.

Question: How do we show that $$\zeta(\bar{5},1)$$ is also a Nielsen generalized polylogarithm in disguise? More generally, for $$-1\leq z\leq1$$, how to show

\begin{aligned}S_{n,p}(z) &= C_1\int_0^1\frac{(\ln x)^{n-1}\big(\ln(1-z\,x)\big)^p}{x}dx\\ &\overset{?}= C_2\int_0^1\frac{(\ln x)^{n}\;\big(\ln(1-z\,x)\big)^{p-1}}{1-z\,x}dx\end{aligned}\tag3

where, $$C_1 = \frac{(-1)^{n+p-1}}{(n-1)!\,p!},\qquad C_2 = \frac{(-1)^{n+p-1}}{n!\,(p-1)!}\color{red}z$$

If true, this implies,

$$\zeta(\bar{5},1) \overset{\color{red}?}= S_{4,2}(-1)\tag4$$

Edit: It turns out the notation $$\zeta(\bar{5},1)$$ is a multiple zeta function so,

$$\zeta(\bar{a},1)=\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^a}\,(-1)^{n+1} = S_{a-1,2}(-1)$$

with harmonic numbers $$H_n$$, hence $$(4)$$ indeed is true and is just the case $$a=5$$. However, $$(3)$$ still needs to be proved in general.

• The integral then relates two Nielsen polylogs as, $$I = -12\,S_{3,3}(-1) = -\frac{\pi^6}{252}-18\,S_{4,2}(-1)+3\zeta^2(3)$$ – Tito Piezas III Jun 3 '19 at 8:32

Without knowing much about the context for this problem, the relationship between the two integrals seems to be pretty direct from integration by parts. For $$n, p \geq 1$$, writing $$u(x) = (\ln x)^n$$ and $$v(x) = (\ln(1 - zx))^p$$, we have $$\frac{du}{dx} = \frac{n(\ln x)^{n-1}}{x} \qquad \text{and} \qquad \frac{dv}{dx} = -\frac{pz (\ln (1 - zx))^{p-1}}{1 - zx}$$ hence using integration by parts: \begin{align*} n\int_0^1 \frac{(\ln x)^{n-1} (\ln(1 - zx))^p}{x} \,dx &= \int_0^1 \frac{du}{dx} v \,dx \\ &= (uv)|_0^1 - \int_0^1 u \frac{dv}{dx} \,dx \\ &= pz\int_0^1 \frac{(\ln x)^n (\ln (1 - zx))^{p-1}}{1 - zx} \,dx \end{align*} where the last equation holds since $$u(x)v(x) \to 0$$ as $$x \to 0$$ or $$x \to 1$$.
This is $$(3)$$, up to the factor $$(-1)^{n+p-1}/n! p!$$.