# About the integral $\int_{0}^{1}\frac{\log(x)\log^2(1+x)}{x}\,dx$

I came across the following Integral and have been completely stumped by it.

$$\large\int_{0}^{1}\dfrac{\log(x)\log^2(1+x)}{x}dx$$

I'm extremely sorry, but the only thing I noticed was that the limits of the Integral were similar to the Beta function. I also got a hint that solving it would require the Polylogarithm, Gamma and the Riemann Zeta Functions. Would it be possible to solve this without using Complex Methods (I haven't learnt them yet) unless absolutely necessary? Any help on this Integral would be greatly appreciated. Many, many thanks in advance!

EDIT: From the comments given below byDavid H Sir, and Alex S Sir, the Integral becomes: $$\int_0^1 \dfrac{\ln^2(1+x)\ln(x)}{x}dx$$

Just an observation: This is strikingly similar to the Beta Function. Also, if we consider $$\int^1_0 (1+x)^ax^bdx$$ and differentiate twice with respect to $a$ and once with respect to $b$ and set $a=b=0$, we get the above Integral (except the $x$ in the denominator).

I'm not sure, but I think taking series representations of the Integrals would be of help, especially since the closed form includes the Riemann Zeta and the Polylogarithm functions.

• Using basic properties of logarithms, the exponential in the numerator reduces to $\ln{(x)}\ln^2{(1+x)}$. – David H Jun 27 '15 at 15:30
• @MakeaDifference now take $e$ to the power of your result. Also, you dropped the $1+$ in the squared logarithm – Alex S Jun 27 '15 at 15:37
• @AlexS Sir, sorry for forgetting to add it. Sir, I just was referring to the exponent. Of course you are right Sir. – Make a Difference Jun 27 '15 at 15:38
• The closed form is to complicated to be answered in both this format and an understandable way. – Zach466920 Jun 27 '15 at 15:39
• @MakeaDifference, is this the full text of the problem ? – Victor Jun 27 '15 at 15:40

$$I=\int_{0}^{1}\frac{\log(x)\log^2(1+x)}{x}\,dx = -\int_{0}^{+\infty} t \log(1+e^{-t})\,dt$$ is an integral that already appeared on MSE. It can be tackled in many ways, for instance by expanding $\log^2(1+x)$ as its Taylor series:

$$\log^2(1+x) = 2\sum_{n\geq 1}\frac{(-1)^{n+1} H_n}{n+1} x^{n+1}$$ from which: $$I = 2\sum_{n\geq 1}\frac{(-1)^{n} H_n}{(n+1)^3}$$ follows. Evaluation of such series is a fashion topic here on MSE: Shobhit mentioned the last series (Euler sum) in his answer to a related question.

• Sorry Sir, but despite trying several different codes for this search, I was unable to find the Integral. Sir, would you remember where the relevant post is located? – Make a Difference Jun 27 '15 at 16:21
• @MakeaDifference: it is a slightly different integral, but leads to the same problem: math.stackexchange.com/questions/465444/… – Jack D'Aurizio Jun 27 '15 at 16:26
• Sir, I thought of evaluating $$\int_0^1 \dfrac{\ln^2(1+x)\ln(x)}{x}dx$$ by rewriting it as $$\int_0^1\sum_{n=1}^\infty\dfrac{x^{2n}}{n^2} \dfrac{\ln(x)}{x}dx$$ Sir, can we exchange the order of Integration and Summation for this problem ie rewrite it as $$\sum_{n=1}^\infty \int\dfrac{x^{2n}\ln(x)}{n^2x}dx?$$ – Make a Difference Jun 27 '15 at 16:28
• @MakeaDifference: $$\log^2(1+x)\color{red}{\neq}\sum_{n=1}^{+\infty}\frac{x^{2n}}{n^2}=\text{Li}_2(x^2).$$ – Jack D'Aurizio Jun 27 '15 at 16:30
• Sir, I saw that $$\ln(1+x)=\sum_{n=1}^\infty \dfrac{(-1)^{n-1}x^n}{n}$$ I thus thought $$\large\ln^2(1+x)=\sum_{n=1}^\infty \dfrac{(((-1))^2)^{n-1}x^{2n}}{n^2}$$ Sir, could you please tell me where I went wrong? – Make a Difference Jun 27 '15 at 16:33

Performing integration by parts by taking $u=\ln^2(1+x)$ and $\mathrm dv=\dfrac{\ln x}{x}\ \mathrm dx$, then \begin{align} I&=\int_0^1\frac{\ln x\ln^2(1+x)}{x}\ \mathrm dx\\ &=\ln x\ln^2(1+x)\Bigg|_0^1-\int_0^1\frac{\ln^2{x}\ln(1+x)}{1+x}\,\mathrm dx\\ &=-\int_0^1\sum_{k=1}^\infty (-1)^{k-1}H_{k}\,x^k\ln^2x\,\mathrm dx\tag{1}\\ &=-\sum_{k=1}^\infty (-1)^{k-1}H_{k}\int_0^1x^k\ln^2x\,\mathrm dx\\ &=-2\sum_{k=1}^\infty (-1)^{k-1}\frac{H_{k}}{(k+1)^3}\tag{2}\\ &=-2\sum_{k=1}^\infty (-1)^{k-1}\left[\frac{H_{k+1}}{(k+1)^3}-\frac{1}{(k+1)^4}\right]\tag{3}\\ &=2\sum_{k=1}^\infty (-1)^{k-1}\left[\frac{H_{k}}{k^3}-\frac{1}{k^4}\right]\\ &=\frac{11\pi^4}{180}-4\text{Li}_4 \left(\frac{1}{2} \right)-\frac{7\zeta(3)\ln 2}{2}+\frac{\pi^2\ln^2 2}{6}-\frac{\ln^4 2}{6}-2\eta(4)\tag{4}\\ &=\frac{11\pi^4}{180}-4\text{Li}_4 \left(\frac{1}{2} \right)-\frac{7\zeta(3)\ln 2}{2}+\frac{\pi^2\ln^2 2}{6}-\frac{\ln^4 2}{6}-\frac{7\zeta(4)}{4}\tag{5}\\ &=\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large\frac{\pi^4}{24}-4\text{Li}_4 \left(\frac{1}{2} \right)-\frac{7\zeta(3)\ln 2}{2}+\frac{\pi^2\ln^2 2}{6}-\frac{\ln^4 2}{6}}} \end{align}

Explanation :

$(1)$ Use generating function $\displaystyle\sum_{k=1}^\infty (-1)^{k-1}H_{k}\,x^k=\frac{\ln(1+x)}{1+x}$

$(2)$ Use formula $\displaystyle\int_0^1 x^k \ln^n x\ \mathrm dx=\frac{(-1)^n n!}{(k+1)^{n+1}}\quad,\ n\in\mathbb{Z}_{n\ge0}$

$(3)$ Use property $\displaystyle H_{k}=H_{k+1}-\frac{1}{k+1}$

$(4)$ Use the result $\displaystyle \sum_{k=1}^\infty (-1)^{k-1} \frac{H_k}{k^3} = \frac{11\pi^4}{360}-2\text{Li}_4 \left(\frac{1}{2} \right)-\frac{7\zeta(3)\ln 2}{4}+\frac{\pi^2\ln^2 2}{12}-\frac{\ln^4 2}{12}$

$(5)$ Use property of Dirichlet eta function $\displaystyle \eta(s)=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^s}=\left(1-2^{1-s}\right)\zeta(s)$

• Very nice and insightful answer! I appreciate this kind of comprehensible information! An additional ref to (4) could be helpful. +1 – Markus Scheuer Jun 28 '15 at 8:21
• @MarkusScheuer Thanks. I already added the reference link from the beginning. Just click the result in the explanation section $(4)$. – Venus Jun 28 '15 at 9:41
• Ah, I see! Great! :-) – Markus Scheuer Jun 28 '15 at 15:26

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \ln^{3}\pars{x \over x + 1} & = \ln^{3}\pars{x} - 3\ln^{2}\pars{x}\ln\pars{x + 1} + 3\color{#66f}{\ln\pars{x}\ln^{2}\pars{x + 1}} - \ln^{3}\pars{x + 1} \end{align} such that \begin{align} \color{#66f}{\ln\pars{x}\ln^{2}\pars{x + 1}} & = \ln^{2}\pars{x}\ln\pars{x + 1} + {1 \over 3}\,\ln^{3}\pars{x + 1} + {1 \over 3}\bracks{\ln^{3}\pars{x \over x + 1} - \ln^{3}\pars{x}} \end{align}

Then, \begin{align} &\int_{0}^{1}{\color{#66f}{\ln\pars{x}\ln^{2}\pars{1 + x}} \over x}\,\dd x = \sum_{i = 1}^{3}\mc{I}_{i} \\[5mm] &\mbox{where}\quad \left\{\begin{array}{rcl} \ds{\mc{I}_{1}} & \ds{\equiv} & \ds{\phantom{1 \over 3}\int_{0}^{1}{\ln^{2}\pars{x}\ln\pars{1 + x} \over x} \,\dd x} \\[3mm] \ds{\mc{I}_{2}} & \ds{\equiv} & \ds{{1 \over 3}\int_{0}^{1}{\ln^{3}\pars{1 + x} \over x}\,\dd x} \\[3mm] \ds{\mc{I}_{3}} & \ds{\equiv} & \ds{{1 \over 3}\int_{0}^{1}\bracks{\ln^{3}\pars{x \over x + 1} - \ln^{3}\pars{x}}{\dd x \over x}} \end{array}\right.\label{1}\tag{1} \end{align}
$\bbox[15px,#ffe,border:1px dotted navy]{\ds{\mc{I}_{1} =\ {\large ?}}}$. \begin{align} \mc{I}_{1} & \equiv \int_{0}^{1}{\ln^{2}\pars{x}\ln\pars{1 + x} \over x}\,\dd x = \int_{0}^{-1}{\ln^{2}\pars{-x}\ln\pars{1 - x} \over x}\,\dd x \\[5mm] & = -\int_{0}^{-1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{-x}\,\dd x = 2\int_{0}^{-1}{\mrm{Li}_{2}\pars{x} \over x}\,\ln\pars{-x}\,\dd x \\[5mm] & = 2\int_{0}^{-1}\mrm{Li}_{3}'\pars{x}\,\ln\pars{-x}\,\dd x = -2\int_{0}^{-1}{\mrm{Li}_{3}\pars{x} \over x}\,\dd x = -2\int_{0}^{-1}\mrm{Li}_{4}'\pars{x}\,\dd x \\[5mm] & = -2\,\mrm{Li}_{4}\pars{-1} = \bbox[15px,#ffe,border:1px dotted navy]{\ds{7\pi^{4} \over 360}} \label{I1}\tag{I1} \end{align}
$\bbox[15px,#ffe,border:1px dotted navy]{\ds{\mc{I}_{2} =\ {\large ?}}}$. \begin{align} \mc{I}_{2} & \equiv {1 \over 3}\int_{0}^{1}{\ln^{3}\pars{1 + x} \over x}\,\dd x = {1 \over 3}\int_{1}^{2}{\ln^{3}\pars{x} \over x - 1}\,\dd x = {1 \over 3}\int_{1}^{1/2}{\ln^{3}\pars{1/x} \over 1/x - 1} \pars{-\,{1 \over x^{2}}}\,\dd x \\[5mm] = &\ -\,{1 \over 3}\int_{1/2}^{1}{\ln^{3}\pars{x} \over x\pars{1 - x}}\,\dd x = -\,{1 \over 3}\int_{1/2}^{1}{\ln^{3}\pars{x} \over x}\,\dd x - {1 \over 3}\int_{1/2}^{1}{\ln^{3}\pars{x} \over 1 - x}\,\dd x \\[5mm] & = {1 \over 12}\,\ln^{4}\pars{2} - {1 \over 3}\bracks{\ln^{4}\pars{2} + 3\int_{1/2}^{1}{\ln\pars{1 - x} \over x}\,\ln^{2}\pars{x}\,\dd x} \\[5mm] & = -\,{1 \over 4}\,\ln^{4}\pars{2} + \int_{1/2}^{1}\mrm{Li}_{2}'\pars{x}\,\ln^{2}\pars{x}\,\dd x \\[5mm] & = -\,{1 \over 4}\,\ln^{4}\pars{2} - \,\mrm{Li}_{2}\pars{1 \over 2}\ln^{2}\pars{2} -2\int_{1/2}^{1}{\mrm{Li}_{2}\pars{x} \over x}\,\ln\pars{x}\,\dd x \\[5mm] & = -\,{1 \over 4}\,\ln^{4}\pars{2} - \,\mrm{Li}_{2}\pars{1 \over 2}\ln^{2}\pars{2} -2\int_{1/2}^{1}\mrm{Li}_{3}'\pars{x}\,\ln\pars{x}\,\dd x \\[5mm] & = -\,{1 \over 4}\,\ln^{4}\pars{2} - \,\mrm{Li}_{2}\pars{1 \over 2}\ln^{2}\pars{2} - 2\,\mrm{Li}_{3}\pars{1 \over 2}\ln\pars{2} + 2\int_{1/2}^{1}{\mrm{Li}_{3}\pars{x} \over x}\,\dd x \\[5mm] & = -\,{1 \over 4}\,\ln^{4}\pars{2} - \,\mrm{Li}_{2}\pars{1 \over 2}\ln^{2}\pars{2} - 2\,\mrm{Li}_{3}\pars{1 \over 2}\ln\pars{2} + 2\int_{1/2}^{1}\mrm{Li}_{4}'\pars{x}\,\dd x \\[5mm] & = -\,{1 \over 4}\,\ln^{4}\pars{2} - \,\mrm{Li}_{2}\pars{1 \over 2}\ln^{2}\pars{2} - 2\,\mrm{Li}_{3}\pars{1 \over 2}\ln\pars{2} + 2\mrm{Li}_{4}\pars{1} - 2\mrm{Li}_{4}\pars{1 \over 2} \end{align} By using the well know values of $\ds{\,\mrm{Li}_{2}\pars{1/2}}$ and $\ds{\,\mrm{Li}_{3}\pars{1/2}}$ and $\ds{\,\mrm{Li}_{4}\pars{1} = \zeta\pars{4} = \pi^{4}/90}$: $$\mc{I}_{2} = \bbox[15px,#ffe,border:1px dotted navy]{\ds{{\pi^{4} \over 45} + {\pi^{2}\ln^{2}\pars{2} \over 12} - {\ln^{4}\pars{2} \over 12} - {7\ln\pars{2} \over 4}\,\zeta\pars{3} - 2\,\mrm{Li}_{4}\pars{1 \over 2}}}\label{I2}\tag{I2}$$
$\bbox[15px,#ffe,border:1px dotted navy]{\ds{\mc{I}_{3} =\ {\large ?}}}$. \begin{align} \mc{I}_{3} & \equiv {1 \over 3}\int_{0}^{1}\bracks{\ln^{3}\pars{x \over x + 1} - \ln^{3}\pars{x}}{\dd x \over x} = {1 \over 3}\,\lim_{\epsilon \to 0^{+}}\bracks{% \int_{\epsilon}^{1}\ln^{3}\pars{x \over x + 1}\,{\dd x \over x} - \int_{\epsilon}^{1}{\ln^{3}\pars{x} \over x}\,\dd x} \end{align} In the RHS first integral I'll make the change of variables $\ds{x/\pars{1 + x} \mapsto x}$ such that \begin{align} \mc{I}_{3} & = {1 \over 3}\,\lim_{\epsilon \to 0^{+}}\bracks{% \int_{\epsilon/\pars{1 + \epsilon}}^{1/2}{\ln^{3}\pars{x} \over x - x^{2}} \,\dd x - \int_{\epsilon}^{1}{\ln^{3}\pars{x} \over x}\,\dd x} \\[5mm] & = {1 \over 3}\,\ \underbrace{\lim_{\epsilon \to 0^{+}} \int_{\epsilon/\pars{1 + \epsilon}}^{\epsilon} {\ln^{3}\pars{x} \over x - x^{2}}\,\dd x}_{\ds{=\ 0}}\ +\ {1 \over 3}\int_{0}^{1/2}\bracks{% {\ln^{3}\pars{x} \over x - x^{2}} - {\ln^{3}\pars{x} \over x}}\dd x - {1 \over 3}\int_{1/2}^{1}{\ln^{3}\pars{x} \over x}\,\dd x \\[5mm] & = {1 \over 3}\int_{0}^{1/2}{\ln^{3}\pars{x} \over 1 - x}\,\dd x\ -\ \underbrace{{1 \over 3}\int_{1/2}^{1}{\ln^{3}\pars{x} \over x}\,\dd x} _{\ds{-\,{1 \over 12}\,\ln^{4}\pars{2}}} \end{align} The remaining integral can be evaluated by successive integration by parts: A procedure I already exploited in the $\ds{\,\mc{I}_{2}}$-evaluation $\pars{~\mbox{see}\ \eqref{I2}~}$. Indeed, they are rather similar. Therefore, $$\mc{I}_{3} = \bbox[15px,#ffe,border:1px dotted navy]{\ds{% {\pi^{2}\ln^{2}\pars{2} \over 12} - {\ln^{4}\pars{2} \over 12} - {7\ln\pars{2} \over 4}\,\zeta\pars{3} - 2\,\mrm{Li}_{4}\pars{1 \over 2}}}\label{I3}\tag{I3}$$
By replacing $\ds{\,\mc{I}_{1},\,\mc{I}_{2}\ \mbox{and}\ \,\mc{I}_{3}}$ $\pars{~\mbox{see expressions}\ \eqref{I1},\eqref{I2}\ \mbox{and}\ \eqref{I3}~}$ in \eqref{1}: $$\int_{0}^{1}{\ln\pars{x}\ln^{2}\pars{1 + x} \over x}\,\dd x = \bbox[15px,#ffe,border:1px dotted navy]{\ds{{\pi^{4} \over 24} + {\pi^{2}\ln^{2}\pars{2} \over 6} - {\ln^{4}\pars{2} \over 6} - {7\ln\pars{2} \over 2}\,\zeta\pars{3} - 4\,\mrm{Li}_{4}\pars{1 \over 2}}}$$