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

simply as

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


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

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.

  • $\begingroup$ The integral then relates two Nielsen polylogs as, $$I = -12\,S_{3,3}(-1) = -\frac{\pi^6}{252}-18\,S_{4,2}(-1)+3\zeta^2(3)$$ $\endgroup$ – Tito Piezas III Jun 3 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 $(-1)^{n+p-1}/n! p!$.

  • $\begingroup$ Thanks. I suspected as much. I'm just wondering why some authors use the 2nd form of the integral, when the 1st form is better known and even has a name (Nielsen polylogs). I guess they have their reasons. $\endgroup$ – Tito Piezas III Jun 4 at 2:06

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