# 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)?

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

• do you need a closed form? Otherwise $\log(1+x) \sim x - \frac{x^2}{2} + \frac{x^3}{3}$ is a good approximation – Alex May 15 '14 at 13:17
• Yes, I need closed forms. – user 1357113 May 15 '14 at 13:44
• Wolfram Alpha thinks the first two answers (for $n=1$ and $n=2$) are ${1\over2}\zeta(2)$ and ${1\over4}\zeta(3)$. After that it gets dilogarithmically messy. – Barry Cipra May 15 '14 at 13:54
• An antiderivative of $\frac{\log^{n}(1+x)}{x}$ in terms of polylogarithms can found by repeatedly integrating by parts. And if you're interested, joriki evaluated the case $n=2$ using contour integration. math.stackexchange.com/questions/316745/… – Random Variable May 15 '14 at 14:09
• @RandomVariable Yes, indeed (I missed a tricky series while doing things in a hurry). – user 1357113 May 15 '14 at 14:20

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

• Thanks, that's the way. (+1) – user 1357113 May 15 '14 at 14:23


\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}

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

• Good job there! :-) (+1) – user 1357113 Jun 30 '14 at 19:38

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.