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

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)?

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

Solution 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 $$ \sum_{j=1}^\infty\sum_{k=1}^\infty\frac{(-1)^{j+k}}{jk(j+k)} =\frac{\zeta(3)}{4} $$

Solution 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}

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

Solution 3:

The following new solution to the classical harmonic series result, $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n}{n^2}=\frac{5}{8}\zeta(3)$, is proposed by Cornel Ioan Valean, using the powerful identity, $$\sum _{k=1}^{\infty } \frac{1}{2k(2k+2n-1)}=\frac{1}{2(2n-1)}\left(2H_{2n}-H_n-2\log(2)\right),\tag1$$ found and proved in $(6.289)$ in the book (Almost) Impossible Integrals, Sums, and Series.

If we multiply both sides of $(1)$ by $1/(2n-1)$, consider the sum from $n=1$ to $\infty$ and then reindex, we have for the right-hand side that $$\sum_{n=1}^{\infty} \frac{H_{2n}}{(2n-1)^2}-\frac{1}{2}\sum_{n=1}^{\infty} \frac{H_n}{(2n-1)^2}-\log(2)\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$$ $$=-\frac{3}{4}\log(2)\zeta(2)+\sum_{n=1}^{\infty} \frac{H_{2n-1}}{(2n-1)^2}-\frac{1}{2}\sum_{n=1}^{\infty} \frac{H_n}{(2n+1)^2}$$ $$=-\frac{7}{8}\zeta(3)+\frac{1}{2}\sum_{n=1}^{\infty}\frac{H_n}{n^2}+\frac{1}{2}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n}{n^2}=\frac{1}{8}\zeta(3)+\frac{1}{2}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n}{n^2}.\tag2$$

On the other hand, based on $(1)$, we have for the left-hand side that $$\sum _{n=1}^{\infty}\left(\sum _{k=1}^{\infty } \frac{1}{2k(2k+2n-1)(2n-1)}\right)=\sum _{k=1}^{\infty}\left(\sum _{n=1}^{\infty } \frac{1}{2k(2k+2n-1)(2n-1)}\right)$$ $$=\frac{1}{4}\sum _{k=1}^{\infty}\frac{1}{k^2}\sum_{n=1}^k \frac{1}{2n-1}=\frac{1}{4}\sum _{k=1}^{\infty}\frac{1}{k^2}\left(H_{2k}-\frac{1}{2}H_k\right)=\sum _{k=1}^{\infty}\frac{H_{2k}}{(2k)^2}-\frac{1}{8}\sum _{k=1}^{\infty}\frac{H_k}{k^2}$$ $$=\frac{1}{4}\sum _{k=1}^{\infty}\frac{1}{k^2}\sum_{n=1}^k \frac{1}{2n-1}=\frac{1}{4}\sum _{k=1}^{\infty}\frac{1}{k^2}\left(H_{2k}-\frac{1}{2}H_k\right)=\sum _{k=1}^{\infty}\frac{H_{2k}}{(2k)^2}-\frac{1}{8}\sum _{k=1}^{\infty}\frac{H_k}{k^2}$$ $$=\frac{3}{8}\sum _{k=1}^{\infty}\frac{H_k}{k^2}-\frac{1}{2}\sum _{k=1}^{\infty}(-1)^{k-1}\frac{H_k}{k^2}=\frac{3}{4}\zeta(3)-\frac{1}{2}\sum _{n=1}^{\infty}(-1)^{n-1}\frac{H_n}{n^2}.\tag3$$

By combining $(2)$ and $(3)$, we obtain that

$$\sum _{n=1}^{\infty}(-1)^{n-1}\frac{H_n}{n^2}=\frac{5}{8}\zeta(3).$$

In the calculations we needed particular cases of the generalizations, \begin{equation*} 2\sum_{k=1}^\infty \frac{H_k}{k^n}=(n+2)\zeta(n+1)-\sum_{k=1}^{n-2} \zeta(n-k) \zeta(k+1), \ n\ge2, \end{equation*} and \begin{equation*} \sum _{k=1}^{\infty}\frac{H_k}{(2k+1)^{2m}}=2m\left(1-\frac{1}{2^{2m+1}}\right)\zeta(2m+1)-2\log(2)\left(1-\frac{1}{2^{2m}}\right)\zeta(2m) \end{equation*} \begin{equation*} -\frac{1}{2^{2m}}\sum_{i=1}^{m-1}(1-2^{i+1})(1-2^{2m-i})\zeta(1+i)\zeta(2m-i), \end{equation*} proved in https://math.stackexchange.com/q/3268851. Cornel's solution to the case, $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n}{n^4}=\frac{59}{32}\zeta(5)-\frac{1}{2}\zeta(2)\zeta(3)$, may be found in https://math.stackexchange.com/q/3269815, and the present technique may be easily extended to calculate the generalization, $\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1} \frac{H_n}{n^{2m}}$.

Since the given integral easily reduces to the calculations of $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n}{n^2}$, the solution is finalized.

Solution 4:

Different approach:

We have

$$\ln^2(1+x)=2\sum_{n=1}^\infty\frac{H_n}{n+1}(-x)^{n+1}$$

Divide by $x$ then integrate to get

\begin{align} \int_0^1\frac{\ln^2(1+x)}{x}\ dx&=2\sum_{n=1}^\infty\frac{(-1)^{n+1}H_n}{n+1}\int_0^1x^n\ dx\\ &=2\sum_{n=1}^\infty\frac{(-1)^{n+1}H_n}{(n+1)^2}\\ &=2\sum_{n=1}^\infty\frac{(-1)^{n}H_{n-1}}{n^2}\\ &=2\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{n^2}-2\sum_{n=1}^\infty\frac{(-1)^{n}}{n^3}\\ &=2\left(-\frac58\zeta(3)\right)-2\operatorname{Li}_3(-1)\\ &=-\frac54\zeta(3)-2\left(-\frac34\zeta(3)\right)\\ &=\boxed{\frac14\zeta(3)} \end{align}

Note:

We have the generating identity

$$\sum_{n=1}^\infty x^n\frac{H_n}{n^2}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$

and by setting $x=-1$ and considering only the real parts we have

$$\Re\sum_{n=1}^\infty (-1)^n\frac{H_n}{n^2}=\operatorname{Li}_3(-1)-\Re\operatorname{Li}_3(2)+\Re\ln2\operatorname{Li}_2(2)+\frac12\underbrace{\Re\ln(-1)\ln^22}_{0}+\zeta(3)\tag{1}$$

Using the trilogarithmic identity

$$\operatorname{Li}_3(x)+\operatorname{Li}_3(1-x)+\operatorname{Li}_3\left(\frac{x-1}{x}\right)=\frac16\ln^3x+\zeta(2)\ln x-\frac12\ln^2x\ln(1-x)+\zeta(3)$$

set $x=-1$ and take the real parts to have

$$ \boxed{\Re\operatorname{Li}_3(2)=\frac78\zeta(3)+\frac32\ln2\zeta(2)}$$

also Landen's identity gives

$$ \boxed{\Re\operatorname{Li}_2(2)=\frac32\zeta(2)}$$

Plugging the boxed results along with $\operatorname{Li}_3(-1)=-\frac34\zeta(3)$ in (1) we have

$$\Re\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^2}=-\frac58\zeta(3)$$

Solution 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

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

or $$(x-y)^n=(-1)^n(y-x)^n=(-1)^n \sum_{k=0}^n{n\choose k}y^{n-k}(-x)^k=\sum_{k=0}^n{n\choose k}(-y)^{n-k}x^k\tag2$$

we get

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

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

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

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

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

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