How to evaluate $\int_0^1\frac{\log^2(1+x)}x\mathrm dx$?

The definite integral

$$\int_0^1\frac{\log^2(1+x)}x\mathrm dx=\frac{\zeta(3)}4$$

arose in my answer to this question. I couldn't find it treated anywhere online. I eventually found two ways to evaluate the integral, and I'm posting them as answers, but they both seem like a complicated detour for a simple result, so I'm posting this question not only to record my answers but also to ask whether there's a more elegant derivation of the result.

Note that either using the method described in this blog post or substituting the power series for $\log(1+x)$ and using

$$\frac1k\frac1{s-k}=\frac1s\left(\frac1k+\frac1{s-k}\right)\;$$

yields

$$\int_0^1\frac{\log^2(1+x)}x\mathrm dx=2\sum_{n=1}^\infty\frac{(-1)^{n+1}H_n}{(n+1)^2}\;.$$

However, since the corresponding identity without the alternating sign is used to obtain the sum by evaluating the integral and not vice versa, I'm not sure that this constitutes progress.

The similar integral

$$\int_0^1\frac{\log^2(1-x)}x\mathrm dx=2\zeta(3)$$

is evaluated in this blog post using the substitution $u=-\log(1-x)$:

\begin{align} \int_0^1\frac{\log^2(1-x)}x\mathrm dx &= \int_0^\infty\frac{u^2}{1-\mathrm e^{-u}}\mathrm e^{-u}\,\mathrm du \\ &= \int_0^\infty u^2\sum_{n=1}^\infty\mathrm e^{-nu}\mathrm du \\ &= \sum_{n=1}^\infty\int_0^\infty u^2\mathrm e^{-nu}\mathrm du \\ &= \sum_{n=1}^\infty\frac2{n^3} \\ &= 2\zeta(3)\;. \end{align}

Analogously substituting $u=\log(1+x)$ in the present integral leads to an integral up to $\log2$ that can be evaluated in terms of polylogarithms evaluated at $\frac12$:

\begin{align} &\int_0^{\log2}\frac{\log^2(1+x)}x\mathrm dx \\ =& \int_0^{\log2}\frac{u^2}{\mathrm e^u-1}\mathrm e^u\,\mathrm du \\ =& \int_0^{\log2}\frac{u^2}{1-\mathrm e^{-u}}\mathrm du \\ =& \int_0^{\log2} u^2\sum_{n=0}^\infty\mathrm e^{-nu}\mathrm du \\ =& \sum_{n=0}^\infty\int_0^{\log2} u^2\mathrm e^{-nu}\mathrm du \\ =& \sum_{n=0}^\infty\int_0^{\log2} u^2\mathrm e^{-nu}\mathrm du \\ =& \frac13\log^32+\sum_{n=1}^\infty\frac1n\left(-2^{-n}\log^22+2\int_0^{\log2} u\mathrm e^{-nu}\mathrm du\right) \\ =& \frac13\log^32+\sum_{n=1}^\infty\left(-\frac1n2^{-n}\log^22+\frac2{n^2}\left(-2^{-n}\log2+\int_0^{\log2}\mathrm e^{-nu}\mathrm du\right)\right) \\ =& \frac13\log^32+\sum_{n=1}^\infty\left(-\frac1n2^{-n}\log^22-\frac2{n^2}2^{-n}\log2-\frac2{n^3}\left(2^{-n}-1\right)\right) \\ =& \def\Li{\operatorname{Li}} \frac13\log^32-\Li_1\left(\frac12\right)\log^22-2\Li_2\left(\frac12\right)\log2-2\Li_3\left(\frac12\right)+2\zeta(3) \\ =& \frac13\log^32-\log2\log^22-2\left(\frac{\pi^2}{12}-\frac12\log^22\right)\log2-2\left(\frac16\log^32-\frac{\pi^2}{12}\log2+\frac78\zeta(3)\right)+2\zeta(3) \\ =& \frac{\zeta(3)}4\;. \end{align}

Not only is this a rather complicated derivation of a much simpler result; it also looks as if the polylogarithm values may have been obtained using the present integral in the first place.

• Nice! I think that your calculation can be generalized: $$\int_0^1 \frac{\log^n(1+x)}{x}dx=\frac{\log^{n+1}(1+z)}{1+n}-n! \sum_{j=1}^n \frac{\log^{n-j+1}(1+z)}{(n-j+1)!}\text{Li}_j \left(\frac{1}{1+z} \right)+n! \zeta(n+1)-n! \text{Li}_{n+1} \left( \frac{1}{1+z}\right)$$ – Shobhit Bhatnagar Nov 21 '13 at 6:45

We can combine the present integral with the similar integral

$$\int_0^1\frac{\log^2(1-x)}x\mathrm dx=2\zeta(3)$$

\begin{align} \int_0^1\frac{\log^2(1+x)}x\mathrm dx-\int_0^1\frac{\log^2(1-x)}x\mathrm dx &= \int_0^1\frac{\log^2(1+x)}x\mathrm dx+\int_{-1}^0\frac{\log^2(1+x)}x\mathrm dx \\ &= \int_{-1}^1\frac{\log^2(1+x)}x\mathrm dx\;. \end{align}

Then we can complete the contour of integration by a semicircle in the upper half complex plane:

\begin{align} \int_{-1}^1\frac{\log^2(1+x)}x\mathrm dx &= \oint\frac{\log^2(1+z)}z\mathrm dz-\int\frac{\log^2\left(1+\def\e{\mathrm e^{\mathrm i\phi}}\e\right)}{\e}\mathrm d\e \\ &= -\mathrm i\int_0^\pi\log^2\left(1+\e\right)\mathrm d\phi\;, \end{align}

where the integral over the closed contour vanishes since there are no poles inside the contour.

We know that the imaginary part of this expression vanishes, since it sums to zero with a real integral, so we only have to evaluate the real part:

\begin{align} -\mathrm i\int_0^\pi\log^2\left(1+\e\right)\mathrm d\phi &= \Re\left(-\mathrm i\int_0^\pi\log^2\left(1+\e\right)\mathrm d\phi\right) \\ &= \Re\left(-\mathrm i\int_0^\pi\left(\log\left|1+\e\right|+\mathrm i\arg\left(1+\e\right)\right)^2\mathrm d\phi\right) \\ &= 2\int_0^\pi\log\left|1+\e\right|\arg\left(1+\e\right)\mathrm d\phi \\ &= 2\int_0^\pi\frac12\log\left(2+2\cos\phi\right)\frac\phi2\mathrm d\phi \\ &= \frac12\int_0^\pi\log\left(\left(\mathrm e^{\mathrm i\phi/2}+\mathrm e^{-\mathrm i\phi/2}\right)^2\right)\phi\mathrm d\phi \\ &= \int_0^\pi\log\left(\mathrm e^{\mathrm i\phi/2}+\mathrm e^{-\mathrm i\phi/2}\right)\phi\,\mathrm d\phi \\ &= \Re\int_0^\pi\left(\frac{\mathrm i\phi}2+\log\left(1+\mathrm e^{-\mathrm i\phi}\right)\right)\phi\,\mathrm d\phi \\ &= \Re\int_0^\pi\sum_{n=1}^\infty\frac{(-1)^{n+1}\mathrm e^{-\mathrm in\phi}}n\phi\,\mathrm d\phi \\ &= \sum_{n=1}^\infty\frac{-1+(-1)^n}{n^3} \\ &= -\zeta(3)-\eta(3) \\ &= -\zeta(3)-\frac34\zeta(3) \\ &= -\frac74\zeta(3)\;. \end{align}

The desired integral is the sum of the two results:

\begin{align} \int_0^1\frac{\log^2(1+x)}x\mathrm dx &= \int_0^1\frac{\log^2(1-x)}x\mathrm dx-\mathrm i\int_0^\pi\log^2\left(1+\e\right)\mathrm d\phi \\ &= 2\zeta(3)-\frac74\zeta(3) \\ &= \frac{\zeta(3)}4 \;. \end{align}

This raises the question whether there's a deeper reason for both of these seemingly quite different integrals to evaluate to a multiple of $\zeta(3)$.

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \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}}} \\[3mm]&=\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 \\[3mm]&={1 \over 3}\,\ln^{3}\pars{2} +\color{#66f}{\sum_{n = 0}^{\infty}\int_{1/2}^{1}\ln^{2}\pars{x}x^{n}\,\dd x} \end{align}

\begin{align}&\color{#66f}{\sum_{n = 0}^{\infty}\int_{1/2}^{1}\ln^{2}\pars{x}x^{n} \,\dd x} =\left.\partiald[2]{}{\mu}\sum_{n = 1}^{\infty}\int_{1/2}^{1}x^{\mu - 1} \,\dd x\,\right\vert_{\,\mu\ =\ n} =\left.\partiald[2]{}{\mu}\sum_{n = 1}^{\infty} {1 - 2^{-\mu} \over \mu}\,\right\vert_{\,\mu\ =\ n} \\[3mm]&=2\sum_{n = 1}^{\infty}{1 \over n^{3}} -2\sum_{n = 1}^{\infty}{\pars{1/2}^{n} \over n^{3}} -2\ln\pars{2}\sum_{n = 1}^{\infty}{\pars{1/2}^{n} \over n^{2}} -\ln^{2}\pars{2}\sum_{n = 1}^{\infty}{\pars{1/2}^{n} \over n} \\[3mm]&=2\zeta\pars{3} - 2{\rm Li}_{3}\pars{\half} -2\ln\pars{2}{\rm Li}_{2}\pars{\half} -\ln^{2}\pars{2}{\rm Li}_{1}\pars{\half} \end{align}

From this link \begin{align} {\rm Li}_{1}\pars{\half} &= \ln\pars{2} \\[1mm] {\rm Li}_{2}\pars{\half} &= {\pi^{2} \over 12} - \half\,\ln^{2}\pars{2} \\[1mm] {\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}

Let,

$\displaystyle A=\int_0^1 \dfrac{\ln\left(1+x\right)^2}{x}dx$

$\displaystyle B=\int_0^1 \dfrac{\ln\left(1-x\right)^2}{x}dx$

Perform the change of variable $y=1-x$,

$\displaystyle B=\int_0^1 \dfrac{\ln\left(x\right)^2}{1-x}dx$

Using Taylor expansion,

$\displaystyle B=2\zeta(3)$

$\displaystyle C=\int_0^1 \dfrac{\ln\left(\dfrac{1-x}{1+x}\right)^2}{x}dx$

Perform the change of variable $y=\dfrac{1-x}{1+x}$,

$\displaystyle C=2\int_0^1 \dfrac{\ln\left(x\right)^2}{1-x^2}dx$

Using Taylor expansion,

$C=\dfrac{7}{2}\zeta(3)$

$\displaystyle D=\int_0^1 \dfrac{\ln\left(1-x^2\right)^2}{x}dx$

Perform the change of variable $y=x^2$,

\begin{align}\displaystyle D&=\dfrac{1}{2}\int_0^1 \dfrac{\ln\left(1-x\right)^2}{x}dx\\ &=\dfrac{1}{2}B\\ &=\zeta(3) \end{align}

Since $(a+b)^2+(a-b)^2=2a^2+2b^2$ therefore,

$C+D=2A+2B$

Therefore,

\begin{align}A&=\dfrac{1}{2}\left(C+D-2B\right)\\ &=\dfrac{1}{2}\left(\dfrac{7}{2}\zeta(3)+\zeta(3)-4\zeta(3)\right)\\ &=\boxed{\dfrac{1}{4}\zeta(3)} \end{align}

I wrote this to answer a question which was deleted (before I posted) because the answers to this question answered that question. \begin{align} \int_0^1\frac{\log(1+x)^2}x\,\mathrm{d}x &=-2\int_0^1\frac{\log(1+x)\log(x)}{1+x}\,\mathrm{d}x\tag1\\ &=-2\sum_{k=0}^\infty(-1)^kH_k\int_0^1x^k\log(x)\,\mathrm{d}x\tag2\\ &=-2\sum_{k=0}^\infty(-1)^k\frac{H_k}{(k+1)^2}\tag3\\ &=-2\sum_{k=0}^\infty(-1)^k\left(\frac{H_{k+1}}{(k+1)^2}-\frac1{(k+1)^3}\right)\tag4\\[3pt] &=-2\left(\frac58\zeta(3)-\frac34\zeta(3)\right)\tag5\\[6pt] &=\frac14\zeta(3)\tag6 \end{align} Explanation:
$$(1)$$: integration by parts
$$(2)$$: $$\frac{\log(1+x)}{1+x}=\sum\limits_{k=0}^\infty(-1)^{k-1}H_kx^k$$
$$(3)$$: $$H_{k+1}=H_k+\frac1{k+1}$$
$$(4)$$: $$\int_0^1x^k\log(x)\,\mathrm{d}x=-\frac1{(k+1)^2}$$
$$(5)$$: $$(7)$$ from this answer
$$(6)$$: simplify