Evaluation of $\int_0^1 \frac{\log^2(1+x)}{x} \ dx$

Question

One of the ways to approach it lies in the area of the dilogarithm, but is it possible to evaluate it
by other means of the real analysis (without using dilogarithm)?

$$\int_0^1 \frac{\log^2(1+x)}{x} \ dx$$

EDIT: maybe you're aware of some easy way to do that. I'd appreciate it!
Some words on the generalization case (by means of the real analysis again)?

$$F(n)=\int_0^1 \frac{\log^n(1+x)}{x} \ dx, \space n\in \mathbb{N}$$

Answer 1

Squaring the series for $\log(1+x)$ yields $$ \log(1+x)^2=\sum_{k=2}^\infty\sum_{j=1}^{k-1}\frac{(-1)^kx^k}{j(k-j)} $$ Dividing by $x$ and integrating gives $$ \begin{align} \int_0^1\frac{\log(1+x)^2}{x}\mathrm{d}x &=\sum_{k=2}^\infty\sum_{j=1}^{k-1}\frac{(-1)^k}{jk(k-j)}\\ &=\sum_{j=1}^\infty\sum_{k=j+1}^\infty\frac{(-1)^k}{jk(k-j)}\\ &=\sum_{j=1}^\infty\sum_{k=1}^\infty\frac{(-1)^{j+k}}{jk(j+k)}\\[9pt] &=\frac{\zeta(3)}{4} \end{align} $$ Using $(5)$ from this answer: $$ \sum_{n=1}^\infty\frac{(-1)^n}{n^2}H_n =-\frac34\zeta(3)+\frac12\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{(-1)^{n+k}}{(n+k)kn} $$ and $(6)$ from the same answer: $$ -\frac58\zeta(3) =\sum_{n=1}^\infty\frac{(-1)^n}{n^2}H_n $$ we get $$ \sum_{j=1}^\infty\sum_{k=1}^\infty\frac{(-1)^{j+k}}{jk(j+k)} =\frac{\zeta(3)}{4} $$

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{1}{\ln^{2}\pars{1 + x} \over x}\,\dd x:\ {\large ?}}$

\begin{align}&\color{#c00000}{% \int_{0}^{1}{\ln^{2}\pars{1 + x} \over x}\,\dd x} =\int_{1}^{2}{\ln^{2}\pars{x} \over x - 1}\,\dd x =\int_{1}^{1/2}{\ln^{2}\pars{1/x} \over 1/x - 1}\,\pars{-\,{\dd x \over x^{2}}} =\int_{1/2}^{1}{\ln^{2}\pars{x} \over x\pars{1 - x}}\,\dd x \\[3mm]&=\int_{1/2}^{1}{\ln^{2}\pars{x} \over x}\,\dd x + \int_{1/2}^{1}{\ln^{2}\pars{x} \over 1 - x}\,\dd x ={1 \over 3}\,\ln^{3}\pars{2} +\sum_{n = 1}^{\infty}\color{#00f}{\int_{1/2}^{1}\ln^{2}\pars{x}x^{n - 1}\,\dd x} \qquad\qquad\pars{1} \end{align}

$$ \color{#00f}{\int_{1/2}^{1}\ln^{2}\pars{x}x^{n - 1}\,\dd x} =\lim_{\mu\ \to\ n - 1}\partiald[2]{}{\mu}\int_{1/2}^{1}x^{\mu}\,\dd x =\lim_{\mu\ \to\ n - 1}\partiald[2]{}{\mu} \bracks{{1 - \pars{1/2}^{\mu + 1} \over \mu + 1}} $$

$$ \color{#00f}{\int_{1/2}^{1}\ln^{2}\pars{x}x^{n - 1}\,\dd x} =-2\,{\pars{1/2}^{n} \over n^{3}}+ {2 \over n^{3}} -\ln^{2}\pars{2}\,{\pars{1/2}^{n} \over n} -2\ln\pars{2}\,{\pars{1/2}^{n} \over n^{2}} $$

By replacing in $\pars{1}$: \begin{align}&\color{#c00000}{% \int_{0}^{1}{\ln^{2}\pars{1 + x} \over x}\,\dd x} \\[3mm]&={1 \over 3}\,\ln^{3}\pars{2} -2{\rm Li}_{3}\pars{\half} +2\zeta\pars{3} - \ln^{2}\pars{2}{\rm Li}_{1}\pars{\half} -2\ln\pars{2}{\rm Li}_{2}\pars{\half}\tag{2} \end{align}

You'll find values for the PolyLogarithm Function $\ds{{\rm Li}_{s}\pars{\half}\,,\ \pars{~s = 1,2,3~}\,,\ }$ in this page: \begin{align} {\rm Li}_{1}\pars{\half} &= \ln\pars{2} \\[3mm] {\rm Li}_{2}\pars{\half} &= {\pi^{2} \over 12} - \half\,\ln^{2}\pars{2} \\[3mm] {\rm Li}_{3}\pars{\half} &= {1 \over 6}\,\ln^{3}\pars{2}- {\pi^{2} \over 12}\,\ln\pars{2} +{7 \over 8}\,\zeta\pars{3} \end{align}

With these identities and result $\pars{2}$: \begin{align}&\color{#c00000}{% \int_{0}^{1}{\ln^{2}\pars{1 + x} \over x}\,\dd x} \\[3mm]&=\color{#00f}{{1 \over 3}\,\ln^{3}\pars{2}} +\ \overbrace{\bracks{\color{#00f}{-\,{1 \over 3}\,\ln^{3}\pars{2}} + \color{magenta}{{\pi^{2} \over 6}\,\ln\pars{2}} {\large -{7 \over 4}\,\zeta\pars{3}}}}^{\ds{-2{\rm Li}_{3}\pars{\half}}}\ +\ {\large 2\zeta\pars{3}} \\[3mm]&+\ \underbrace{\bracks{\color{#990099}{-\ln^{3}\pars{2}}}} _{\ds{-\ln^{2}\pars{2}{\rm Li}_{1}\pars{\half}}}\ +\ \underbrace{\bracks{\color{magenta}{-\,{\pi^{2} \over 6}\,\ln\pars{2}} +\color{#990099}{\ln^{3}\pars{2}}}}_{\ds{-2\ln\pars{2}{\rm Li}_{2}\pars{\half}}}\ =\ \pars{2 - {7 \over 4}}\zeta\pars{3} \end{align}

$$ \color{#66f}{\large% \int_{0}^{1}{\ln^{2}\pars{1 + x} \over x}\,\dd x = {\zeta\pars{3} \over 4}} \approx 0.3005 $$

Answer 3

On the path of Felix Marin, \begin{align}J&=\int_0^1 \frac{\ln(1+x)^2}{x}\\ &\overset{y=\frac{1}{1+x}}=\int_{\frac{1}{2}}^1 \frac{\ln^2 x}{x(1-x)}\,dx\\ &=\int_{\frac{1}{2}}^1 \frac{\ln^2 x}{x}\,dx+\int_{\frac{1}{2}}^1 \frac{\ln^2 x}{1-x}\,dx\\ &=\frac{1}{3}\left(\ln^3 (1)-\ln^3\left(\frac{1}{2}\right)\right)+\int_0^1 \frac{\ln^2 x}{1-x}\,dx-\int_0^{\frac{1}{2}} \frac{\ln^2 x}{1-x}\,dx\\ &=\frac{1}{3}\ln^3 2+\int_0^1 \frac{\ln^2 x}{1-x}\,dx-\int_0^{\frac{1}{2}} \frac{\ln^2 x}{1-x}\,dx\\ &\overset{y=\frac{x}{1-x},\text{the 2nd integral}}=\frac{1}{3}\ln^3 2+\int_0^1 \frac{\ln^2 x}{1-x}\,dx-\int_0^1\frac{\ln^2\left(\frac{x}{1+x}\right)}{1+x}\,dx\\ &=\frac{1}{3}\ln^3 2+\int_0^1 \frac{\ln^2 x}{1-x}\,dx-\int_0^1\frac{\ln^2 x}{1+x}\,dx-\int_0^1\frac{\ln^2 (1+x)}{1+x}\,dx+2\int_0^1\frac{\ln(1+x)\ln x}{1+x}\,dx\\ &\overset{IBP}=\frac{1}{3}\ln^3 2+\int_0^1 \frac{\ln^2 x}{1-x}\,dx-\int_0^1\frac{\ln^2 x}{1+x}\,dx-\int_0^1\frac{\ln^2 (1+x)}{1+x}\,dx-J\\ &=\frac{1}{3}\ln^3 2+\int_0^1 \frac{2x\ln^2 x}{1-x}\,dx-\frac{1}{3}\ln^3 2-J\\ &\overset{y=x^2}=\frac{1}{4}\int_0^1 \frac{\ln^2 x}{1-x}\,dx-J\\ J&=\frac{1}{8}\int_0^1 \frac{\ln^2 x}{1-x}\,dx\\ &=\frac{1}{8}\times 2\zeta(3)\\ &=\boxed{\frac{1}{4}\zeta(3)} \end{align} NB: i assume that, \begin{align}\int_0^1 \frac{\ln^2 x}{1-x}\,dx=2\zeta(3)\end{align} (proof: Taylor expansion)

Answer 4

Here is a solution by finding the closed form of $\int \frac{\ln^2(1-x)}{x}dx$ then letting $x\mapsto -x$:

$$\int \frac{\ln^2(1-x)}{x}dx=\int \frac{\ln(1-x)\ln(1-x)}{x}dx\overset{IBP}{=}-\operatorname{Li}_2(x)\ln(1-x)-\int\frac{\operatorname{Li}_2(x)}{1-x}dx$$

For the last integral, set $1-x=y$ then use the reflection formula: $$\operatorname{Li}_2(1-y)=\zeta(2)-\ln(y)\ln(1-y)-\operatorname{Li}_2(y)$$

We obtain that

$$\int\frac{\operatorname{Li}_2(x)}{1-x}dx=-\int\frac{\operatorname{Li}_2(1-y)}{y}dy$$

$$=-\zeta(2)\int\frac{dy}y+\int\frac{\ln(y)\ln(1-y)}{y}dy+\int\frac{\operatorname{Li}_2(y)}{y}dy$$

$$=-\zeta(2)\ln(y)+\left[-\operatorname{Li}_2(y)\ln(y)+\int\frac{\operatorname{Li}_2(y)}{y}dy\right]+\int\frac{\operatorname{Li}_2(y)}{y}dy$$

$$=-\zeta(2)\ln(y)-\operatorname{Li}_2(y)\ln(y)+2\operatorname{Li}_3(y)$$

$$=-\zeta(2)\ln(1-x)-\operatorname{Li}_2(1-x)\ln(1-x)+2\operatorname{Li}_3(1-x)$$

Then

$$\int\frac{\ln^2(1-x)}{x}dx=\ln(1-x)\left[\operatorname{Li}_2(1-x)-\operatorname{Li}_2(x)+\zeta(2)\right]-2\operatorname{Li}_3(1-x)$$

Now consider the integral boundaries $(0,a)$,

$$\int_0^a\frac{\ln^2(1-x)}{x}dx=\ln(1-a)\left[\operatorname{Li}_2(1-a)-\operatorname{Li}_2(a)+\zeta(2)\right]-2\operatorname{Li}_3(1-a)+2\zeta(3)$$

Therefore

$$\int_0^1\frac{\ln^2(1+x)}{x}dx\overset{x\mapsto -x}{=}\int_0^{-1}\frac{\ln^2(1-x)}{x}dx$$

$$=\ln(2)\left[\operatorname{Li}_2(2)-\operatorname{Li}_2(-1)+\zeta(2)\right]-2\operatorname{Li}_3(2)+2\zeta(3)$$

substitute $\Re\operatorname{Li}_2(2)=\frac32\zeta(2)$ and $\Re\operatorname{Li}_3(2)=\frac78\zeta(3)+\frac32\ln2\zeta(2)$, the closed form follows.

Answer 5

You can find a nice generalization for $\int_0^1\frac{\ln^n(1+x)}{x}dx$ in lemma $2.2$ in this article and I am going to type it here with little more details.

Start with subbing $\frac{1}{1+x}=y$

$$I_n=\int_0^1\frac{\ln^n(1+x)}{x}dx=(-1)^n\int_{1/2}^1\frac{\ln^n(y)}{y(1-y)}dy$$

$$=(-1)^n\int_{1/2}^1\frac{\ln^n(y)}{y}dy+(-1)^n\int_{1/2}^1\frac{\ln^n(y)}{1-y}dy$$

$$=(-1)^n\left[(-1)^n\frac{\ln^{n+1}(2)}{n+1}\right]+(-1)^n\int_{0}^1\frac{\ln^n(y)}{1-y}dy-(-1)^n\int_{0}^{1/2}\frac{\ln^n(y)}{1-y}dy$$

$$=(-1)^n\left[(-1)^n\frac{\ln^{n+1}(2)}{n+1}\right]+(-1)^n\left[(-1)^n n!\zeta(n+1)\right]-(-1)^n\int_{0}^{1/2}\frac{\ln^n(y)}{1-y}dy$$

$$=\frac{\ln^{n+1}(2)}{n+1}+n!\zeta(n+1)-(-1)^n\int_{0}^{1/2}\frac{\ln^n(y)}{1-y}dy\tag1$$

By using the binomial theorem

$$(x+y)^n=\sum_{k=0}^n{n\choose k}x^{n-k}y^k$$

we get

$$\int_{0}^{1/2}\frac{\ln^n(y)}{1-y}dy\overset{2y=x}{=}\int_0^1\frac{(\ln(x)-\ln(2))^n}{2-x}dx$$

$$=\sum_{k=0}^n{n\choose k}(-\ln(2))^{n-k}\left(\int_0^1\frac{\ln^k(x)}{2-x}dx\right)$$

$$=\sum_{k=0}^n{n\choose k}(-\ln(2))^{n-k}\left(\sum_{i=1}^\infty\frac1{2^i} \int_0^1 x^{i-1}\ln^k(x)dx\right)$$

$$=\sum_{k=0}^n{n\choose k}(-\ln(2))^{n-k}\left((-1)^k k!\sum_{i=1}^\infty\frac1{2^i i^{k+1}}\right)$$

$$=\sum_{k=0}^n{n\choose k}(-\ln(2))^{n-k}(-1)^k k!\operatorname{Li}_{k+1}\left(\frac12\right)\tag3$$

Plug $(3)$ in $(1)$ we get

$$I_n=\frac{\ln^{n+1}(2)}{n+1}+n!\zeta(n+1)-\sum_{k=0}^n k!{n\choose k}\ln^{n-k}(2)\operatorname{Li}_{k+1}\left(\frac12\right)$$ or $$(-1)^n\int_{1/2}^1\frac{\ln^n(y)}{y(1-y)}dy=\frac{\ln^{n+1}(2)}{n+1}+n!\zeta(n+1)-\sum_{k=0}^n k!{n\choose k}\ln^{n-k}(2)\operatorname{Li}_{k+1}\left(\frac12\right)$$