# Ways to prove $\int_0^1 \frac{\ln^2(1+x)}{x}dx = \frac{\zeta(3)}{4}$?

I am wondering if we can show in a simple way that $$I=\int_0^1 \frac{\ln^2(1+x)}{x}dx = \int_1^2 \frac{\ln^2(t)}{t-1}dt = \frac{\zeta(3)}{4}.$$

Because the end result is very simple, I suspect that there might be a fast way to prove this. Can you prove it without using polylog identities? Complex analysis is allowed. It may be easier to show the equivalent identity $$\sum_{k=1}^\infty \frac{(-1)^k H_k}{k^2} = -\frac{5 \zeta (3)}{8}$$ I know you can do that one with the generating function of the harmonic numbers, but that gives a nasty expression with polylogs which I would like to avoid.

• I am aware that a simple indefinite integral in terms of polylogs exists, but I would like to avoid them if possible. Commented Jan 5, 2015 at 13:51
• For elementary solution involving just series manipulations you can check @robjohn's proof of $A(1,2)$ here
– r9m
Commented Jan 5, 2015 at 13:52
• @rqm, That one is quite nice. I wonder if we can we also calculate the integral without using a series representation, or is that too much to ask? Commented Jan 5, 2015 at 14:02
• In this answer, it is shown that $$\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}H_n=\frac58\zeta(3)$$ Ah, I see that r9m has mentioned this.
– robjohn
Commented Jan 5, 2015 at 14:14

Let us denote $I_{\pm}=\displaystyle \int_{0}^1\frac{\ln^2(1\pm x)}{x}dx$. We will express $I_+$ in terms of $I_-$, which is itself related to the standard integral representation of $\zeta(z)$ by the change of variables $x=1-e^{-t}$: $$I_-=\int_0^{\infty}\frac{t^2dt}{e^{t}-1}=2\zeta(3).$$ Indeed, we have \begin{align} \int_0^1\frac{\ln^2\frac{1+x}{1-x}}{x}dx=\int_0^{\infty}\frac{16t^2}{2\sinh 2t}dt&=\int_0^{\infty}16t^2\left(\frac{1}{e^{2t}-1}-\frac{1}{e^{4t}-1}\right)dt=\frac74 I_- \tag{1} \end{align} where the first equality is obtained by setting $x=\tanh t$. Also, it is easy to show ($x^2\to x$) that $$\int_{0}^1\frac{\ln^2(1-x^2)}{x}dx=\frac12I_-. \tag{2}$$ Summing (1) and (2), one finds that $2I_+ +2I_-=\left(\frac74+\frac12\right)I_-$, and hence $\displaystyle I_+=\frac{I_-}{8}=\frac{\zeta(3)}4$.

• Exactly what I was looking for, thanks! Commented Jan 5, 2015 at 19:06
• @user111187 You are welcome! Commented Jan 5, 2015 at 20:00
• (+1) It might be nice to explicitly note that $$\log\left(\frac{1+x}{1-x}\right)^2+\log\left(1-x^2\right)^2=2\log(1+x)^2+2\log(1-x)^2$$
– robjohn
Commented Jan 6, 2015 at 1:40

Using this answer which shows that $$\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}H_n=\frac58\zeta(3)$$ and the series $$\frac{\log(1+x)}{1+x}=\sum_{k=1}^\infty(-1)^{k-1}H_kx^k$$ we get \begin{align} \int_0^1\frac{\log(1+x)^2}{x}\mathrm{d}x &=\int_0^1\log(1+x)^2\,\mathrm{d}\log(x)\\ &=-2\int_0^1\frac{\log(1+x)\log(x)}{1+x}\,\mathrm{d}x\\ &=-2\int_0^1\sum_{k=1}^\infty(-1)^{k-1}H_kx^k\log(x)\,\mathrm{d}x\\ &=2\sum_{k=1}^\infty\frac{(-1)^{k-1}H_k}{(k+1)^2}\\ &=2\sum_{k=1}^\infty\frac{(-1)^{k-1}H_{k+1}}{(k+1)^2}-2\sum_{k=1}^\infty\frac{(-1)^{k-1}}{(k+1)^3}\\ &=2\left(\frac34\zeta(3)-\sum_{k=1}^\infty\frac{(-1)^{k-1}H_k}{k^2}\right)\\ &=\frac{\zeta(3)}4 \end{align}

• I'll add this just for completeness.
– robjohn
Commented Jan 6, 2015 at 1:52

Hmm, i don't know if this answer fulfills the requirement to be a fast way, but it is relatively straightforward:

1.) Use the sub $1+x=e^y$ The integral is now $$\int_0^{\log(2)}\frac{y^2}{1-e^{-y}}dy$$ 2.) By help of geometric series we obtain $$\sum_{n=0}^{\infty}\int_{0}^{\log(2)}y^2e^{-ny}dy$$ 3.) Seperating the $n=0$ term and doing the trivial integrations we obtain $$-\left(\log^2(2)\sum_{n=1}^{\infty}\frac{1}{n2^n}+2\log(2)\sum_{n=1}^{\infty}\frac{1}{n^22^n}+2\sum_{n=1}^{\infty}\frac{1}{n^32^n}-2\sum_{n=1}^{\infty}\frac{1}{n^3}\right)+\frac{1}{3}\log^3(2)$$ 4.) Remembering the definition of Polylog $\text{Li}_s(z)=\sum_{n=1}^{\infty}\frac{z^n}{n^s}$ we can now look up the values $\text{Li}_{1}(1/2),\text{Li}_{2}(1/2),\text{Li}_{3}(1/2)$ in some table and put everything together to obtain (magic!) $$\frac{\zeta(3)}{4}$$

Here is a particularly efficient way to get to your Euler sum.

In this post I show that $$\ln^2 (1 - x) = 2 \sum_{n = 2}^\infty \frac{H_{n - 1} x^n}{n}.$$ Replacing $x$ with $-x$ gives $$\ln^2 (1 + x) = 2 \sum_{n = 2}^\infty \frac{(-1)^n H_{n - 1} x^n}{n}.$$

So if we replace the term $\ln^2 (1 + x)$ with its above Maclaurin series expansion the integral becomes $$\int_0^1 \frac{\ln^2 (1 + x)}{x} \, dx = 2 \sum_{n = 2}^\infty \frac{(-1)^n H_{n - 1}}{n} \int_0^1 x^{n - 1} \, dx = 2 \sum_{n = 2}^\infty \frac{(-1)^n H_{n - 1}}{n^2}.$$

Making use of the following property for harmonic numbers $$H_n = H_{n - 1} + \frac{1}{n},$$ the integral can be expressed as $$\int_0^1 \frac{\ln^2 (1 + x)}{x} \, dx = 2 \sum_{n = 2}^\infty \frac{(-1)^n H_n}{n^2} - 2 \sum_{n = 2}^\infty \frac{(-1)^n}{n^3} = 2 \sum_{n = 1}^\infty \frac{(-1)^n H_n}{n^2} - 2 \sum_{n = 1}^\infty \frac{(-1)^n}{n^3}.$$

For the sums, as you note $$\sum_{n = 1}^\infty \frac{(-1)^n H_n}{n^2} = -\frac{5}{8} \zeta (3),$$ and $$\sum_{n = 1}^\infty \frac{(-1)^n}{n^3} = - \sum_{n = 1}^\infty \frac{(-1)^{n - 1}}{n^3} = - \eta (3) = -(1 - 2^{1-3}) \zeta (3) = -\frac{3}{4} \zeta (3),$$ where $\eta (s)$ is the Dirichlet eta function, one finally has $$\int_0^1 \frac{\ln^2 (1 + x)}{x} \, dx = -\frac{5}{4} \zeta (3) + \frac{3}{2} \zeta (3) = \frac{1}{4} \zeta (3),$$ as expected.

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large\int_{0}^{1}{\ln^{2}\pars{1 + x} \over x}\,\dd x}\ \stackrel{\dsc{1 + x}\ \mapsto\ \dsc{x}}{=}\ \int_{1}^{2}{\ln^{2}\pars{x} \over x - 1}\,\dd x\ \stackrel{\dsc{x}\ \mapsto\ \dsc{1 \over x}}{=}\ \int_{1}^{1/2}{\ln^{2}\pars{1/x} \over 1/x - 1}\,\pars{-\,{\dd x \over x^{2}}} \\[5mm]&=\int_{1/2}^{1}\ {\ln^{2}\pars{x} \over x\pars{1 - x}}\,\dd x =\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 \\[5mm]&={1 \over 3}\,\ln^{3}\pars{2} -\left.\vphantom{\LARGE A}\ln\pars{1 - x}\ln^{2}\pars{x}\right\vert_{1/2}^{1} +\int_{1/2}^{1}\ln\pars{1 - x}\bracks{2\ln\pars{x}\,{1 \over x}}\,\dd x \\[5mm]&=-\,{2 \over 3}\,\ln^{3}\pars{2} - 2\int_{1/2}^{1}\Li{2}'\pars{x}\ln\pars{x}\,\dd x \end{align} where $\Li{\rm s}$ is a PolyLogarithm Function. We already used the identity $\ds{\Li{\rm s}'\pars{t}= {\Li{{\rm s} - 1}\pars{t} \over t}}$ with $\Li{1}\pars{t}=-\ln\pars{1 - t}$.

Then, \begin{align} &\color{#66f}{\large\int_{0}^{1}{\ln^{2}\pars{1 + x} \over x}\,\dd x} =-\,{2 \over 3}\,\ln^{3}\pars{2} - 2\Li{2}\pars{\half}\ln\pars{2} +2\int_{1/2}^{1}\Li{2}\pars{x}\,{1 \over x}\,\dd x \\[5mm]&=-\,{2 \over 3}\,\ln^{3}\pars{2} - 2\Li{2}\pars{\half}\ln\pars{2} +2\int_{1/2}^{1}\Li{3}'\pars{x}\,\dd x \\[5mm]&=-\,{2 \over 3}\,\ln^{3}\pars{2} - 2\Li{2}\pars{\half}\ln\pars{2} +2\ \overbrace{\Li{3}\pars{1}}^{\dsc{\zeta\pars{3}}}\ - 2\Li{3}\pars{\half} \\[1cm]&=-\,{2 \over 3}\,\ln^{3}\pars{2} -2\ \overbrace{\bracks{{1 \over 12}\,\pi^{2} - \half\,\ln^{2}\pars{2}}} ^{\ds{=\ \dsc{\Li{2}\pars{\half}}}}\ \ln\pars{2} + 2\zeta\pars{3} \\[5mm]&\phantom{=}- 2\ \overbrace{% \bracks{{1 \over 6}\,\ln^{3}\pars{2} - {1 \over 12}\,\pi^{2}\ln\pars{2} + {7 \over 8}\,\zeta\pars{3}}}^{\ds{=\ \dsc{\Li{3}\pars{\half}}}} \ =\ \color{#66f}{\large{1 \over 4}\,\zeta\pars{3}} \end{align}

$\ds{\Li{2}\pars{\half}}$ and $\ds{\Li{3}\pars{\half}}$ are well known values ( a few ones !!! ) and they are given elsewhere.

Using the fact that

$$\ln^2(1+x)=\frac12\ln^2\left(\frac{1-x}{1+x}\right)+\frac12\ln^2(1-x^2)-\ln^2(1-x)$$

we have $$\int_0^1\frac{\ln^2(1+x)}{x}dx=\frac12\underbrace{\int_0^1\frac{\ln^2\left(\frac{1-x}{1+x}\right)}{x}dx}_{(1-x)/(1+x)\to x}+\frac12\underbrace{\int_0^1\frac{\ln^2(1-x^2)}{x}dx}_{1-x^2\to x}-\underbrace{\int_0^1\frac{\ln^2(1-x)}{x}dx}_{1-x\to x}$$

$$=\int_0^1\frac{\ln^2(x)}{1-x^2}dx+\frac14\int_0^1\frac{\ln^2(x)}{1-x}dx-\int_0^1\frac{\ln^2(x)}{1-x}dx.$$

Since $$\frac1{1-x^2}-\frac1{1-x}=\frac{x}{1-x^2},$$ we have $$\int_0^1\frac{\ln^2(1+x)}{x}dx=\frac14\int_0^1\frac{\ln^2(x)}{1-x}dx-\underbrace{\int_0^1\frac{x\ln^2(x)}{1-x^2}dx}_{x^2\to x}$$

$$=\frac18\int_0^1\frac{\ln^2(x)}{1-x}dx=\frac14\zeta(3).$$

Note:

$$\int_0^1\frac{\ln^a(x)}{1-x}dx=\sum_{n=1}^\infty\int_0^1 x^{n-1}\ln^a(x)dx=\sum_{n=1}^\infty\frac{(-1)^aa!}{n^{a+1}}=(-1)^aa!\zeta(a+1).$$