Inspired by the well known $$\int_0^1\frac{\ln(1-x)\ln x}x\mathrm dx=\zeta(3)$$ and the integral given here (writing $\zeta_r:=\zeta(r)$ for easier reading)$$\int_0^1\frac{\ln^3(1-x)\ln x}x\mathrm dx=12\zeta(5)-\pi^2\zeta(3)=6(2\zeta_5-\zeta_3\zeta_2),$$ I have looked at the integrals $$I_{n,m}=:\int_0^1\frac{\ln^n(1-x)\ln^m x}x\mathrm dx$$ for $m,n\in\mathbb N$.

From the power series of $\ln(1-x)$ and recursion, it is not hard to see that $I_{m,n}$ can be written as a rational combination of products of integer zeta values, with the arguments ($\ge 2$) for each product summing up to $m+n+1$, e.g. $$I_{n,1}=\begin{cases} -n!\Bigl(\dfrac{n-1}4\zeta_{n+2}-\dfrac12\sum\limits_{j=1}^{k-1}\zeta_{2j+1}\zeta_{n+1-2j} \Bigr)\ \ &\text{ for }n=2k, \\ -n!\Bigl(\dfrac{n+1}2\zeta_{n+2}- \sum\limits_{j=1}^{k-1} \zeta_{2j+1} \zeta_{n+1-2j}\Bigr)\ \ &\text{ for }n=2k-1. \end{cases}$$

So far, so good. Now the intriguing (numerical) discovery is that there are several families of pairs of those integrals that have rational ratios. For instance, $$\frac{I_{n+2,n-2}}{I_{n-1,n+1}}=\frac{n+2}{n-1}\quad \text{or}\quad\frac{I_{n+1,n-1}}{I_{n,n}}=\frac{n+1}{n}\quad \text{or}\quad\frac{I_{n,2}}{I_{3,n-1}}=\frac{n}{3},$$ e.g. $$I_{5,2}=\color{red}{10}\color{blue}{(61\zeta_8-72\zeta_5\zeta_3+12\zeta_3^2\zeta_2)} $$ and $$I_{3,4}=\color{red}6\color{blue}{(61\zeta_8-72\zeta_5\zeta_3+12\zeta_3^2\zeta_2)}. $$ Note that each such pair has the same proportion as the respective first arguments.
I don't see how that could be possibly proven by recursion, so there must be some deeper connection between those integrals.

How to explain that?

  • $\begingroup$ Your sentence-ending punctuation after $\text{“} I_{n,1}=\cdots \text{”}$ was bizarre, and I did a bit of copyediting on it. I've seen various mathematicians do that before, and I wonder why. $\endgroup$ Jul 5, 2022 at 15:21
  • 1
    $\begingroup$ @MichaelHardy Easy to tell why: it already looks sooo much better than if the full stop comes after the '$$'. 🤣 What a good idea to include it even deeper, i.e. inside the '\cases' as well :) $\endgroup$
    – Wolfgang
    Jul 6, 2022 at 7:59

1 Answer 1


For $n\geq 1$ and $m\geq 0$, an application of integration by parts ($u=\log^n(1-x)$, $dv=\log^m(x)\,dx/x$) followed by the substitution $x\mapsto 1-x$ shows that $$ \frac{I_{n,m}}{I_{m+1,n-1}}=\frac{n}{m+1}. $$ All of your examples are special cases of this identity.

  • $\begingroup$ Yep... so easy! So much for "thinking about a certain approach without considering other possible methods" ... ☺ $\endgroup$
    – Wolfgang
    Apr 22, 2019 at 16:47

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