# Compute $\int_{0}^{1}\frac{\ln(x) \ln^2 (1-x)}{x} dx$

Compute $$\int_{0}^{1}\frac{\ln(x) \ln^2 (1-x)}{x} dx$$

I'm looking for some nice proofs at this problem. One idea would be to use Taylor expansion and then integrating term by term. What else can we do? Thanks.

• Concerning Euler's early work you may see this article from Sandifer's excellent 'How Euler did it'. – Raymond Manzoni Sep 7 '12 at 8:04
• @Raymond Manzoni: it's a nice paper! – user 1591719 Sep 7 '12 at 8:09
• I'm really glad that such questions exist! If they weren't around then we'd need to invent them. This way one may see the real beauty of calculus (my opinion) :-) – user 1591719 Sep 7 '12 at 8:34

This answer is from my old calculation.

First, assume we are well aware of the following famous result.

$$\zeta(2) =\frac{\pi^{2}}{6}, \quad \zeta(4) =\frac{\pi^{4}}{90}$$

Next, by a simple calculation we obtain

$$H_{n} := \sum_{k=1}^{n} \frac{1}{k} =\int_{0}^{1}\frac{1-t^{n}}{1-t}\, dt.$$

and

$$\frac{\log (1-x)}{1-x}\ =\ -\sum_{n=1}^{\infty}H_{n}x^{n}.$$

Finally, define the polylogarithm as

$$\mathrm{Li}_{s}(x) :=\sum_{n=1}^{\infty} \frac{x^n}{n^s},$$

so that it satisfies the recurrence relation

$$\mathrm{Li}_{1}(x) =-\log (1-x) , \quad \mathrm{Li}_{s+1}(x) =\int_{0}^{x}\frac{\mathrm{Li}_{s}(t)}{t}\, dt$$

and the identity

$$\mathrm{Li}_{s}(1) =\zeta(s).$$

The the all-in-one straight calculation goes as follows:

\begin{align*} \int_{0}^{1}\frac{\log x\log^{2}(1-x)}{x}\, dx & = \int_{0}^{1}\frac{\log (1-x)\log^{2}x}{1-x}\, dx = -\sum_{n=1}^{\infty}H_{n}\int_{0}^{1}x^{n}\log^{2}x\, dx\\ & = -2\sum_{n=1}^{\infty}\frac{H_{n}}{(n+1)^{3}}\\ & = 2\sum_{n=1}^{\infty}\left[\frac{1}{(n+1)^{4}}-\frac{H_{n+1}}{(n+1)^{3}}\right] = 2\sum_{n=0}^{\infty}\left[\frac{1}{(n+1)^{4}}-\frac{H_{n+1}}{(n+1)^{3}}\right]\\ & = 2\zeta(4)-2\sum_{n=1}^{\infty}\frac{H_{n}}{n^{3}}\\ & = 2\zeta(4)-2\sum_{n=1}^{\infty}\frac{1}{n^{3}}\int_{0}^{1}\frac{1-t^{n}}{1-t}\, dt = 2\zeta(4)-2\int_{0}^{1}\frac{\zeta(3)-\mathrm{Li}_{3}(t)}{1-t}\, dt\\ & = 2\zeta(4)+\left[2 (\zeta(3)-\mathrm{Li}_{3}(t))\log (1-t)\right]_{0}^{1}+2\int_{0}^{1}\frac{\mathrm{Li}_{2}(t)\log (1-t)}{t}\, dt\\ & = 2\zeta(4)-2\int_{0}^{1}\mathrm{Li}_{2}(t)\frac{d\mathrm{Li}_{2}(t)}{dt}\, dt\\ & = 2\zeta(4)-\left[\mathrm{Li}_{2}^{2}(t)\right]_{0}^{1} = 2\zeta(4)-\zeta(2)^{2} = \frac{\pi^{4}}{45}-\frac{\pi^{4}}{36} = -\frac{\pi^{4}}{180}\\ & = -\frac{1}{2}\zeta(4). \end{align*}

• An amazing answer! – user 1591719 Sep 7 '12 at 7:49
• it would be so nice to have many answers / different approaching ways for questions like this one. :-) – user 1591719 Sep 7 '12 at 8:01
• I just noticed that also $\int_{0}^{1}\frac{\ln^2(x) \ln (1-x)}{x} dx$ has a nice result that involves $\zeta(4)$. – user 1591719 Sep 10 '12 at 8:06
• @Chris'ssister: you may find many integrals of this kind in my link (around (417)) to your other thread $\sum_{n=1}^{\infty}\dfrac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)\cdot (2n+1)}$ (see too (437) !). – Raymond Manzoni Sep 10 '12 at 8:20
• @Raymond Manzoni: thanks! That is great link. I'm trying to learn tricks about solving very difficult integrals and Euler sums. – user 1591719 Sep 10 '12 at 8:21

The Taylor expansion approach gives you $-2 \sum_{k=1}^\infty H_k/(k+1)^3$ where $H_k = \sum_{n=1}^k 1/n$. Wolfram Alpha says this is $-\pi^4/180$, but I don't know how it gets that.

• adding the $k+1$ term to $H_k$ we get $\ 2\zeta(4)-2\sum_{k=1}^\infty \frac {H_k}{k^4}$. Euler found (24) of mathworld $2\sum_{k=1}^\infty \frac {H_k}{k^m}=(m+2)\zeta(m+1)-\sum_{n=1}^{m-2}\zeta(m-n)\zeta(n+1)$. I didn't verify this... – Raymond Manzoni Sep 7 '12 at 6:55
• @Raymond Manzoni: do you know some free paper with nice proofs for these sums? – user 1591719 Sep 7 '12 at 7:04
• @Chris'ssister: I don't have my references here but free papers about Euler sums (or MZV multiple zeta values) are numerous (Borwein, Bailey, Broadhurst and many physicists). Start perhaps with Wolfram Euler sums (references look fine!). – Raymond Manzoni Sep 7 '12 at 7:08
• @Raymond Manzoni: OK. Thanks. – user 1591719 Sep 7 '12 at 7:13
• @Chris'ssister: many results were obtained in an experimental way (LLL PSLQ) in the early nineties as seen in this paper of Bailey. Compendium of results (click on one of the names for other papers). It's a rather interesting subject ! – Raymond Manzoni Sep 7 '12 at 7:31
• thanks for the references. (+1) – user 1591719 Sep 7 '12 at 7:35


Note that

\begin{align} {\Gamma\pars{\mu + 1} \over \Gamma\pars{1 + \mu - n}} & = {\Gamma\pars{\mu + 1} \over \pi/\braces{\Gamma\pars{n - \mu}\sin\pars{\pi\bracks{n - \mu}}}} = {\pars{-1}^{n + 1} \over \pi}\,\sin\pars{\pi\mu} \Gamma\pars{\mu + 1}\Gamma\pars{n - \mu} \\[5mm] & = \pars{-1}^{n + 1}\,\mu \braces{\Gamma\pars{1}\Gamma\pars{n} + \bracks{\Gamma'\pars{1}\Gamma\pars{n} - \Gamma\pars{1}\Gamma'\pars{n}}\mu} + \,\mrm{O}\pars{\mu^{3}} \\[5mm] & = \pars{-1}^{n + 1}\,\Gamma\pars{n}\mu + \pars{-1}^{n + 1}\bracks{-\gamma\pars{n - 1}! -\pars{n - 1}!\,\Psi\pars{n}}\color{#f00}{\mu^{2}} + \,\mrm{O}\pars{\mu^{3}} \\[5mm] & = \pars{-1}^{n + 1}\,\Gamma\pars{n}\mu - \pars{-1}^{n + 1}\pars{n - 1}!\,H_{n - 1}\,\color{#f00}{\mu^{2}} + \,\mrm{O}\pars{\mu^{3}} \end{align}

such that $\ds{\left.\partiald[2]{}{\mu}{\Gamma\pars{\mu + 1} \over \Gamma\pars{1 + \mu - n}}\right\vert_{\ \mu\ =\ 0^{+}} = -2\pars{-1}^{n + 1}\pars{n - 1}!\,H_{n - 1}}$

Expression \eqref{1} becomes \begin{align} &\int_{0}^{1}{\ln\pars{x}\ln^{2}\pars{1 - x} \over x}\,\dd x = -\sum_{n = 1}^{\infty}{1 \over n!}\, \bracks{-2\pars{-1}^{n + 1}\pars{n - 1}!\,H_{n - 1}}\, {\pars{-1}^{n} \over n^{2}} = -2\sum_{n = 1}^{\infty}{H_{n - 1} \over n^{3}} \\[5mm] & = -2\sum_{n = 1}^{\infty}{H_{n} \over n^{3}} + 2\sum_{n = 1}^{\infty}{1 \over n^{4}} = -2\pars{\pi^{4} \over 72} + 2\zeta\pars{4} = -\,{5 \over 2}\,\zeta\pars{4} + 2\zeta\pars{4} = \bbx{-\,{1 \over 2}\,\zeta\pars{4}} \end{align}

Note that $\ds{\sum_{n = 1}^{\infty}{H_{n} \over n^{3}} = {\pi^{4} \over 72} = {5 \over 4}\,\zeta\pars{4}}$ is a well known identity. See expression $\pars{19}$ in this page.

$\ds{\Huge\left. b\right)}$ \begin{align} &\int_{0}^{1}{\ln\pars{x}\ln^{2}\pars{1 - x} \over x}\,\dd x = {1 \over 3}\int_{0}^{1}\bracks{% \ln^{3}\pars{x \over 1 - x} - \ln^{3}\pars{x}}\,{\dd x \over x} \\[5mm] + &\ {1 \over 3}\int_{0}^{1}{\ln^{3}\pars{1 - x} \over x}\,\dd x\ +\ \underbrace{\int_{0}^{1}{\ln^{2}\pars{x}\ln\pars{1 - x} \over x}\,\dd x} _{\ds{-\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{x}\,\dd x}}\label{2}\tag{2} \end{align}

Note that

\begin{align} {1 \over 3}\int_{0}^{1}{\ln^{3}\pars{1 - x} \over x}\,\dd x = {1 \over 3}\int_{0}^{1}{\ln^{3}\pars{x} \over 1 - x}\,\dd x = \int_{0}^{1}\ln\pars{1 - x}{\ln^{2}\pars{x} \over x}\,\dd x = -\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{x}\,\dd x \end{align} such that \eqref{2} is reduced to \begin{align} \int_{0}^{1}{\ln\pars{x}\ln^{2}\pars{1 - x} \over x}\,\dd x & = \underbrace{{1 \over 3}\int_{0}^{1}\bracks{% \ln^{3}\pars{x \over 1 - x} - \ln^{3}\pars{x}}\,{\dd x \over x}} _{\ds{\mc{I}_{1}}} \\[5mm] & -\ \underbrace{2\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{x}\,\dd x} _{\ds{\mc{I}_{2}}}\ =\ \mc{I}_{1} - \,\mc{I}_{2}\label{3}\tag{3} \end{align}

Hereafter, I'll evaluate $\ds{\,\mc{I}_{1}}$ and $\ds{\,\mc{I}_{2}}$.

$\ds{\large\quad\mc{I}_{1}:\ ?}$. \begin{align} \mc{I}_{1} & \equiv {1 \over 3}\int_{0}^{1}\bracks{% \ln^{3}\pars{x \over 1 - x} - \ln^{3}\pars{x}}\,{\dd x \over x} \\[5mm] & = \lim_{\epsilon \to 0^{+}}\bracks{% {1 \over 3}\int_{\epsilon}^{1}\ln^{3}\pars{x \over 1 - x}\,{\dd x \over x} - {1 \over 3}\int_{\epsilon}^{1}{\ln^{3}\pars{x} \over x}\,\dd x} \end{align} In the RHS first integral, lets make the change $\ds{{x \over 1 - x} \mapsto x}$: \begin{align} \mc{I}_{1} & = \lim_{\epsilon \to 0^{+}}\bracks{% {1 \over 3}\int_{\epsilon/\pars{1 - \epsilon}}^{\infty} {\ln^{3}\pars{x} \over x\pars{1 + x}}\,\dd x - {1 \over 3}\int_{\epsilon}^{1}{\ln^{3}\pars{x} \over x}\,\dd x} \\[5mm] & = \lim_{\epsilon \to 0^{+}}\left[% -\,{1 \over 3}\int_{\epsilon}^{\epsilon/\pars{1 - \epsilon}} {\ln^{3}\pars{x} \over x\pars{1 + x}}\,\dd x + {1 \over 3}\int_{\epsilon}^{\infty}{\ln^{3}\pars{x} \over x\pars{1 + x}} \,\dd x\right. \\[5mm] & \left.\phantom{= \lim_{\epsilon \to 0^{+}}\left[\right]}- {1 \over 3}\int_{\epsilon}^{1}{\ln^{3}\pars{x} \over x}\,\dd x\right] \end{align} The RHS first integral vanishes out when $\ds{\epsilon \to 0^{+}}$: \begin{align} \mc{I}_{1} & = {1 \over 3}\int_{0}^{1}\ln^{3}\pars{x} \bracks{{1 \over x\pars{1 + x}} - {1 \over x}}\,\dd x + {1 \over 3}\int_{1}^{\infty}{\ln^{3}\pars{x} \over x\pars{1 + x}}\,\dd x \\[5mm] & = -\,{1 \over 3}\int_{0}^{1}{\ln^{3}\pars{x} \over 1 + x}\,\dd x + {1 \over 3}\int_{1}^{\infty}{\ln^{3}\pars{x} \over x\pars{1 + x}}\,\dd x \end{align} In the second integral, lets $\ds{x \mapsto 1/x}$: \begin{align} \mc{I}_{1} & = -\,{1 \over 3}\int_{0}^{1}{\ln^{3}\pars{x} \over 1 + x}\,\dd x + {1 \over 3}\int_{1}^{0}{\ln^{3}\pars{1/x} \over \pars{1/x}\pars{1 + 1/x}} \pars{-\,{1 \over x^{2}}}\,\dd x \\[5mm] & = -\,{2 \over 3}\int_{0}^{1}{\ln^{3}\pars{x} \over 1 + x}\,\dd x = {2 \over 3}\int_{0}^{-1}{\ln^{3}\pars{-x} \over 1 - x}\,\dd x = 2\int_{0}^{-1}\ln\pars{1 - x}\,{\ln^{2}\pars{-x} \over x}\,\dd x \\[5mm] & = -2\int_{0}^{-1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{-x}\,\dd x = 4\int_{0}^{-1}\mrm{Li}_{3}'\pars{x}\ln\pars{-x}\,\dd x = -4\int_{0}^{-1}\mrm{Li}_{4}'\pars{x}\,\dd x \\[5mm] & = -4\,\underbrace{\mrm{Li}_{4}\pars{-1}}_{\ds{-\,{7 \over 8}\,\zeta\pars{4}}} \implies \bbx{\,\mc{I}_{1} = {7 \over 2}\,\zeta\pars{4}}\label{4}\tag{4} \end{align}
$\ds{\large\quad\mc{I}_{2}:\ ?}$. \begin{align} \mc{I}_{2} & \equiv 2\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{x}\,\dd x = -4\int_{0}^{1}\mrm{Li}_{3}'\pars{x}\ln\pars{x}\,\dd x = 4\int_{0}^{1}\mrm{Li}_{4}\pars{x}\,\dd x = 4\mrm{Li}_{4}\pars{1} \\[5mm] & \implies \bbx{\,\mc{I}_{2} = 4\zeta\pars{4}}\label{5}\tag{5} \end{align}
With \eqref{3}, \eqref{4} and \eqref{5}: \begin{align} &\int_{0}^{1}{\ln\pars{x}\ln^{2}\pars{1 - x} \over x}\,\dd x = {7 \over 2}\,\zeta\pars{4} - 4\zeta\pars{4} \end{align} $$\bbox[#ffe,15px,border:1px dotted navy]{\ds{% \int_{0}^{1}{\ln\pars{x}\ln^{2}\pars{1 - x} \over x}\,\dd x = -\,{1 \over 2}\,\zeta\pars{4}}}$$

In this answer I will make use of a Maclaurin series expansion for the term $\ln^2 (1 - x)$, which I show here to be $$\ln^2 (1 - x) = 2 \sum_{n = 2}^\infty \frac{H_{n - 1} x^n}{n},$$ and the well-known Euler sum of $$\sum_{n = 1}^\infty \frac{H_n}{n^3} = \frac{1}{2} \zeta^2 (2),$$ several proofs for which can be found here.

From the above Maclaurin series expansion for $\ln^2 (1 - x)$ the integral can be written as \begin{align*} \int_0^1 \frac{\ln (x) \ln^2 (1 - x)}{x} \, dx = 2 \sum_{n = 2}^\infty \int_0^1 x^{n - 1} \ln x \, dx. \end{align*} The integral that appears to the right can be readily found by parts. The result is $$\int_0^1 x^{n - 1} \ln x \, dx = -\frac{1}{n^2}.$$ Thus $$\int_0^1 \frac{\ln (x) \ln^2 (1 - x)}{x} \, dx = -2 \sum_{n = 2}^\infty \frac{H_{n - 1}}{n^3}.$$

From properties of harmonic numbers, since $$H_n = H_{n - 1} + \frac{1}{n},$$ the integral becomes $$\int_0^1 \frac{\ln (x) \ln^2 (1 - x)}{x} \, dx = 2 \sum_{n = 2}^\infty \frac{1}{n^4} - 2 \sum_{n = 2}^\infty \frac{H_n}{n^3} = 2 \sum_{n = 1}^\infty \frac{1}{n^4} - 2 \sum_{n = 1}^\infty \frac{H_n}{n^3}.$$ As $$\sum_{n = 1}^\infty \frac{1}{n^4} = \zeta (4) \quad \text{and} \quad \sum_{n = 1}^\infty \frac{H_n}{n^3} = \frac{1}{2} \zeta^2 (2),$$ we have $$\int_0^1 \frac{\ln (x) \ln^2 (1 - x)}{x} \, dx = 2 \zeta (4) - \zeta^2 (2) = - \frac{\pi^4}{180} = -\frac{1}{2} \zeta (4),$$ as expected.