A definite integral involving logarithmic functions: $\int_{0}^{1} \frac{\ln(x) \ln (1+x)}{1-x} \, \mathrm{d} x$

Question

I'm trying to show $$ \int_0^1 \frac{\ln x \cdot \ln(1+x)}{1-x}dx=-\frac{1}{4}\pi^2 \ln(2)+\zeta(3). $$ I am unsure how to approach this integral as I do not know how to use a power series representation for the integrand. I cannot use the generating function $$ \frac{\ln(1-x)}{1-x}=-\sum_{n=1}^\infty H_n x^n $$ since I do not have this in my integrand so it is not so easy to approach. Thanks for the help!

Answer 1

EDIT:

I modified my answer so that it no longer involves dealing with polylogarithms.

There is also a very nice evaluation HERE.

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The generating function of the alternating harmonic numbers (i.e., $\bar{H}_{n} = \sum_{k=1}^{n} \frac{(-1)^{k-1}}{k})$ is $$\frac{\ln(1+x)}{1-x} = \sum_{n=1}^{\infty} \bar{H}_{n} x^{n}. $$

This can be derived using the Cauchy product.

(This was mentioned in a partial answer that was deleted shortly after it was posted.)

Using this generating function, we find that

$$ \begin{align} \int_{0}^{1} \frac{\ln (x) \ln(1+x)}{1-x} \, dx &= \int_{0}^{1} \ln(x) \sum_{n=1}^{\infty} \bar{H}_{n} \, x^{n} \, dx \\ &= \sum_{n=1}^{\infty} \bar{H}_{n} \int_{0}^{1} \ln(x) x^{n} \, dx \\ &= -\sum_{n=1}^{\infty} \frac{\bar{H}_{n}}{(n+1)^{2}} \\&= -\sum _{n=1}^{\infty} \frac{\bar{H}_{n+1}}{(n+1)^{2}} + \sum_{n=1}^{\infty} \frac{(-1)^{n}}{(n+1)^{3}} \\&= \left( -\sum_{n=1}^{\infty}\frac{\bar{H}_{n}}{n^{2}} +1 \right) + \left( \frac{ 3\zeta(3)}{4} -1 \right) \\ &= -\sum_{n=1}^{\infty}\frac{\bar{H}_{n}}{n^{2}} + \frac{3\zeta(3)}{4}. \end{align}$$

In general, $$2 \sum_{n=1}^{\infty} \frac{\bar{H}_{n}}{n^{q}} = 2 \zeta(q) \ln(2)-q \zeta(q+1) + 2 \eta(q+1) + \sum_{k=1}^{q} \eta(k)\eta(q-k+1) , $$ where $\eta(z)$ is the Dirichlet eta function.

(See Theorem 7.1 in this paper for details about how to derive this formula using contour integration.)

Therefore, $$\sum_{n=1}^{\infty} \frac{\bar{H}_{n}}{n^{2}} = \zeta(2) \ln(2) - \zeta(3) + \eta(3) + \eta(1) \eta(2) = \frac{\pi^{2}}{4} \ln(2) - \frac{\zeta(3)}{4}, $$ and

$$\int_{0}^{1} \frac{\ln (x) \ln(1+x)}{1-x} \, dx = \frac{\zeta(3)}{4} - \frac{\pi^{2}}{4} \log(2) + \frac{3\zeta(3)}{4} = - \frac{\pi^{2}}{4} \ln(2) + \zeta(3).$$

Answer 2

Use Feynman’s trick \begin{align} J(a) &= \int_0^1\frac{\ln x}{1+x}\left[ \frac{ \ln(1-ax)}x+\ln(1-x) \right]dx\\ J’(a) &=\int_0^1\frac{-\ln x\ dx}{(1+x)(1-ax)}= \frac{-1}{1+a}\int_0^1 \left(\frac{\ln x}{1+x}+ \frac{a\ln x}{1-ax} \right)dx \\ &= \frac{\zeta(2)}{2(1+a)}- \frac1{1+a}\int_0^a\frac{\ln y}{1-y}dy - \frac{\ln a\ln(1-a)}{1+a}\\ \end{align} Note that $$J(1) = \int_0^1 \frac{\ln x\ln(1-x)}{x}dx \overset{ibp} = \frac12\int_0^1 \frac{\ln^2x}{1-x}dx= \zeta(3) \tag1 $$ On the other hand \begin{align} J(1)& =J(0)+ \int_0^1 J’(a)da = \frac{\zeta(2)}2\ln2-\int_0^1 d[\ln(1+a)]\int_0^a \frac{\ln y}{1-y}dy \\ &\overset{ibp}= \frac{3}2\ln2\zeta(2)+ \int_0^1 \frac{\ln a \ln(1+a)}{1-a}da \tag2 \\\end{align}

Equate (1) and (2) to obtain

$$\int_0^1 \frac{\ln a \ln(1+a)}{1-a}da= -\frac{3}2\ln2\zeta(2)+\zeta(3)= -\frac{\pi^2}{4}\ln2+\zeta(3) $$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x} \over 1 - x}\,\dd x = -\,{1 \over 4}\,\pi^{2}\ln\pars{2} + \zeta\pars{3}} \approx -0.5082:\ {\Large ?}}$.

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\begin{align} &\bbox[5px,#ffd]{% \int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x} \over 1 - x}\,\dd x} \\[5mm] = &\ {1 \over 2}\int_{0}^{1}{\ln^{2}\pars{x} \over 1 - x}\,\dd x + {1 \over 2}\int_{0}^{1}{\ln^{2}\pars{1 + x} - \ln^{2}\pars{2}\over 1 - x}\,\dd x \\[2mm] & \!\!\!\!\! -\,{1 \over 2}\int_{0}^{1} \bracks{\ln^{2}\pars{x \over 1 + x} - \ln^{2}\pars{2}}\,{\dd x \over 1 - x} \end{align} In the second integral I'll make the change $\ds{{1 + x \over 2} \mapsto x}$ while I'll set $\ds{{x \over x + 1} \mapsto x}$ in the last one: \begin{align} &\bbox[5px,#ffd]{% \int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x} \over 1 - x}\,\dd x} \\[5mm] = &\ {1 \over 2}\ \overbrace{\int_{0}^{1}{\ln^{2}\pars{x} \over 1 - x}\,\dd x} ^{\ds{I_{1}}}\ +\ {1 \over 2}\ \overbrace{\int_{1/2}^{1} {\ln^{2}\pars{2x} - \ln^{2}\pars{2}\over 1 - x}\,\dd x}^{\ds{I_{2}}} \\[2mm] & \!\!\!\!\! -\,{1 \over 2}\bracks{% \underbrace{\int_{0}^{1}{\ln^{2}\pars{x/2} - \ln^{2}\pars{2} \over 1 - x}\,\dd x}_{\ds{I_{3}}}\ -\ \underbrace{\int_{0}^{1/2}{\ln^{2}\pars{x} - \ln^{2}\pars{2} \over 1 - x}\,\dd x}_{\ds{I_{4}}}}\label{1}\tag{1} \end{align} In the last line first integral I did the additional scaling $\ds{x \mapsto {x \over 2}}$.

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In (\ref{1}), all the integrals are reduced to simple ones by integrating twice by parts. Namely, \begin{equation} \left\{\begin{array}{rcl} \ds{I_{1}} & \ds{=} & \ds{\phantom{-\,\,}2\zeta\pars{3}} \\[2mm] \ds{I_{2}} & \ds{=} & \ds{\phantom{-\,\,}{2\ln^{3}\pars{2} \over 3} - {\pi^{2}\ln\pars{2} \over 6} + {\zeta\pars{3} \over 4}} \\[2mm] \ds{I_{3}} & \ds{=} & \ds{\phantom{-\,\,}{\pi^{2}\ln\pars{2} \over 3} + 2\zeta\pars{3}} \\[2mm] \ds{I_{4}} & \ds{=} & \ds{-\,{2\ln^{3}\pars{2} \over 3} + {7\zeta\pars{3} \over 4}} \end{array}\right. \end{equation} Then, \begin{align} &\bbox[5px,#ffd]{% \int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x} \over 1 - x}\,\dd x} = {I_{1} + I_{2} - I_{3} + I_{4} \over 2} \\[5mm] = &\ \bbx{-\,{1 \over 4}\,\pi^{2}\ln\pars{2} + \zeta\pars{3}} \approx -0.5082 \\ & \end{align}

Answer 4

\begin{align}J&=\int_{0}^{1} \frac{\ln(x) \ln (1+x)}{1-x} \, \mathrm{d} x\\&\overset{\text{IBP}}=\left[\left(\int_0^x \frac{\ln t}{1-t}dt\right)\ln(1+x)\right]_0^1-\int_0^1 \frac{1}{1+x}\left(\int_0^x \frac{\ln t}{1-t}dt\right)dx\\&=-\zeta(2)\ln 2-\int_0^1\int_0^1 \frac{x\ln(tx)}{(1+x)(1-tx)}dtdx\\ &=-\zeta(2)\ln 2-\int_0^1\int_0^1 \frac{\ln(tx)}{(1+t)(1-tx)}dtdx+\int_0^1\int_0^1 \frac{\ln(tx)}{(1+x)(1+t)}dtdx\\ &\overset{u(x)=tx}=-\zeta(2)\ln 2-\int_0^1 \frac{1}{t(1+t)}\left(\int_0^t \frac{\ln u}{1-u}du\right)dt+2\ln 2\int_0^1 \frac{\ln x}{1+x}dx\\ &\overset{\text{IBP}}=-\zeta(2)\ln 2-\left[\ln\left(\frac{t}{1+t}\right)\left(\int_0^t \frac{\ln u}{1-u}du\right)\right]_0^1+\int_0^1 \frac{\ln\left(\frac{t}{1+t}\right)\ln t}{1-t}dt+\\&2\ln 2\int_0^1 \frac{\ln x}{1+x}dx\\ &=-2\zeta(2)\ln 2+\int_0^1 \frac{\ln^2 t}{1-t}dt-J+2\ln 2\int_0^1 \frac{\ln x}{1+x}dx\\ &=-2\zeta(2)\ln 2+2\zeta(3)-J+2\ln 2 \times -\frac{1}{2}\zeta(2)\\ &=-3\zeta(2)\ln 2+2\zeta(3)-J\\ &=-3\times \frac{\pi^2}{6}\ln^ 2+2\zeta(3)-J\\ J&=\boxed{\zeta(3)-\frac{1}{4}\pi^2\ln 2} \end{align}

NB: \begin{align}\int_0^1 \frac{\ln x}{1+x}dx&=\int_0^1 \frac{\ln x}{1-x}dx-\int_0^1 \frac{2t\ln t}{1-t^2}dt\\ &\overset{x=t^2}=\int_0^1 \frac{\ln x}{1-x}dx-\frac{1}{2}\int_0^1 \frac{\ln x}{1-x}dx\\ &=\frac{1}{2}\int_0^1 \frac{\ln x}{1-x}dx\\ &=-\frac{1}{2}\zeta(2)\\ \end{align} Moreover, i assume that: \begin{align}\int_0^1 \frac{\ln x}{1-x}dx&=-\zeta(2)\\ \zeta(2)&=\frac{1}{6}\pi^2\\ \int_0^1 \frac{\ln x}{1-x}dx&=2\zeta(3)\\ \end{align}