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Question

He estado tratando de evaluar esta integral y mediante el uso de wolfram alpha sé que el valor es$$I=-\int_0^1 \ln(1+x)\ln(1-x)dx=\frac{\pi^2}{6}+2\ln(2)-\ln^2(2)-2$$

Mi Intento:

Empiezo por parametizing la integral como $$I(a)=\int_0^1 -\ln(1+x)\ln(1-ax)dx$$ donde $I=I(1)$. Entonces me diferenciar a conseguir $$I'(a)=\int_0^1 \frac{ax\ln(1+x)}{1-ax}dx=\int_0^1 ax\ln(1+x)\sum_{n=0}^\infty(ax)^ndx=\sum_{n=1}^\infty a^{n+1}\int_0^1 x^{n+1}\ln(1+x)dx$$

Al evaluar esta integral por integración por partes y series geométricas puedo conseguir $$\int_0^1 x^{n+1}\ln(1+x)dx=\frac{x^{n+2}}{n+2}\ln(1+x)|_0^1-\frac{1}{n+2}\int_0^1 \frac{x^{n+2}}{1+x}dx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\int_0^1 x^{n+2}\sum_{k=0}^\infty(-x)^kdx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\sum_{k=0}^\infty(-1)^k\int_0^1 x^{k+n+2}dx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\sum_{k=0}^\infty\frac{(-1)^k}{k+n+2}=\frac{\ln(2)}{n+2}-\frac{1}{2(n+2)}\left(\psi_0\left(\frac{n}{2}+2\right)-\psi_0\left(\frac{n}{2}+\frac{3}{2}\right)\right)$$ So I arrive at $$I'(a)=\sum_{n=0}^\infty a^{n+1}\left(\frac{\ln(2)}{n+2}-\frac{1}{2(n+2)}\left(\psi_0\left(\frac{n}{2}+2\right)-\psi_0\left(\frac{n}{2}+\frac{3}{2}\right)\right)\right)$$ Re-indexación puedo conseguir $$I'(a)=\frac{\ln(2)}{a}\sum_{n=2}^\infty \frac{a^n}{n}+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n}a^{n-1}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n}a^{n-1}$$Integrating both sides from $0$ to $1$ I recover $I(1)$ $$I(1)=\int_0^1 \frac{\ln(2)}{a}\left(-\ln(1-a)-a\right)da+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$ , A continuación, utilizando la ecuación integral para la Dilogarithm llego a $$I(1)=\ln(2)\int_0^1 -\frac{\ln(1-a)}{a}da-\ln(2)+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$ $$I(1)=\frac{\ln(2)\pi^2}{6}-\ln(2)+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$

En este punto yo no podía seguir más, puesto que no sabía cómo simplificar la Digamma términos en la suma. Creo que usando la función Digamma la relación Armónica de los Números podría ser posible explotar los valores conocidos de la Armónica sumas de dinero para llegar a la respuesta, pero no pude obtener las sumas en una forma en que esto iba a funcionar. Si alguien pudiera ayudarme a seguir más o déjame saber si estoy en el camino correcto me sería de gran aprecio. Gracias de antemano.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove armada]{\displaystyle{#1}}\,} \newcommand{\llaves}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{I \equiv -\int_{0}^{1}\ln\pars{1 + x}\ln\pars{1 - x}\,\dd x = {\pi^{2} \over 6} + 2\ln\pars{2} - \ln^{2}\pars{2} - 2:\ {\LARGE ?}}$.

\begin{align} I & \equiv \bbox[10px,#ffd]{-\int_{0}^{1}\ln\pars{1 + x}\ln\pars{1 - x}\,\dd x} \\[5mm] & = \int_{0}^{1}{\bracks{\ln\pars{1 - x} - \ln\pars{1 + x}}^{\, 2} - \bracks{\ln\pars{1 - x} + \ln\pars{1 + x}}^{\, 2} \over 4}\,\dd x \\[5mm] & = \underbrace{{1 \over 4}\int_{0}^{1}\ln^{2}\pars{1 - x \over 1 + x}\,\dd x}_{\ds{\equiv\ \mc{I}_{1}}}\ -\ \underbrace{{1 \over 4}\int_{0}^{1}\ln^{2}\pars{1 - x^{2}}\,\dd x} _{\ds{\equiv\ \mc{I}_{2}}}\ =\ \mc{I}_{1} - \mc{I}_{2}\label{1}\tag{1} \end{align}

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$\ds{\Large\mc{I}_{1}:\ ?}$

\begin{align} \mc{I}_{1} & \equiv {1 \over 4}\int_{0}^{1}\ln^{2}\pars{1 - x \over 1 + x}\,\dd x \,\,\,\stackrel{\pars{1 - x}/\pars{1 + x}\ =\ t}{=}\,\,\, {1 \over 2}\int_{0}^{1}{\ln^{2}\pars{t} \over \pars{1 + t}^{2}} \,\dd t \\[5mm] & = {1 \over 2}\sum_{n = 0}^{\infty}\underbrace{-2 \choose n} _{\ds{\pars{n + 1}\pars{-1}^{n}}}\ \underbrace{\int_{0}^{1}\ln^{2}\pars{t}t^{n}\,\dd t} _{\ds{2 \over \pars{n +1}^{3}}}\ =\ \sum_{n = 1}^{\infty}{\pars{-1}^{n + 1}\over n^{2}} = \bbx{\pi^{2} \over 12}\label{2}\tag{2} \end{align}

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$\ds{\Large\mc{I}_{2}:\ ?}$

\begin{align} \mc{I}_{2} & \equiv {1 \over 4}\int_{0}^{1}\ln^{2}\pars{1 - x^{2}}\,\dd x \,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\, {1 \over 8}\int_{0}^{1}x^{-1/2}\ln^{2}\pars{1 - x}\,\dd x \\[5mm] & = \left.{1 \over 8}\,\partiald[2]{}{\mu}\int_{0}^{1}x^{-1/2} \pars{1 - x}^{\mu}\,\dd x\,\right\vert_{\ \mu\ =\ 0} = {1 \over 8}\,\partiald[2]{}{\mu}\bracks{\Gamma\pars{1/2}\Gamma\pars{\mu + 1} \over \Gamma\pars{\mu + 3/2}}_{\ \mu\ =\ 0} \\[5mm] & = \bbx{-\,{\pi^{2} \over 12} - 2\ln\pars{2} + \ln^{2}\pars{2} + 2} \label{3}\tag{3} \end{align}

\eqref{1}, \eqref{2} y \eqref{3} ceder el codiciado resultado $\ds{\bbx{{\pi^{2} \over 6} + 2\ln\pars{2} - \ln^{2}\pars{2} - 2}}$.

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$\ds{\LARGE\mbox{Another Approach}}$

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\begin{align} I & \equiv \bbox[10px,#ffd]{-\int_{0}^{1}\ln\pars{1 + x}\ln\pars{1 - x}\,\dd x} \,\,\,\stackrel{x + 1\ \mapsto\ x}{=}\,\,\, -\int_{1}^{2}\ln\pars{x}\ln\pars{2 - x}\,\dd x \\[5mm] & \stackrel{x/2\ \mapsto\ x}{=}\,\,\, -2\int_{1/2}^{1}\bracks{\ln\pars{x} + \ln\pars{2}} \bracks{\ln\pars{1 - x} + \ln\pars{2}}\,\dd x \\[8mm] & = -2\int_{1/2}^{1}\ln\pars{x}\ln\pars{1 - x}\,\dd x - 2\ln\pars{2}\ \overbrace{\int_{1/2}^{1}\ln\pars{x}\,\dd x}^{\ds{\ln\pars{2} - 1 \over 2}} \\[2mm] & -2\ln\pars{2}\ \underbrace{\int_{1/2}^{1}\ln\pars{1 - x}\,\dd x} _{\ds{-\,{\ln\pars{2} + 1 \over 2}}} - 2\ln^{2}\pars{2}\int_{1/2}^{1}\,\dd x \\[8mm] & = -2\int_{1/2}^{1}\ln\pars{x}\ln\pars{1 - x}\,\dd x + 2\ln\pars{2} - \ln^{2}\pars{2} \end{align}

El resto de la integral se evalúa con Euler Reflexión Fórmula. Es decir, \begin{align} I & \equiv \bbox[10px,#ffd]{-\int_{0}^{1}\ln\pars{1 + x} \ln\pars{1 - x}\,\dd x} \\[5mm] & = -2\int_{1/2}^{1}\bracks{{\pi^{2} \over 6} - \mrm{Li}_{2}\pars{x} - \mrm{Li}_{2}\pars{1 - x}}\dd x + 2\ln\pars{2} - \ln^{2}\pars{2} \\ & \pars{~\mrm{Li}_{2}:\ Dilogarithm~} \\[5mm] & = -\,{\pi^{2} \over 6} + 2\int_{0}^{1}\mrm{Li}_{2}\pars{x}\dd x + 2\ln\pars{2} - \ln^{2}\pars{2} \\[5mm] & \stackrel{\mrm{IBP}}{=}\,\,\, -\,{\pi^{2} \over 6} + \braces{2\,\mrm{Li}_{2}\pars{1} - 2\int_{0}^{1}x\bracks{-\,{\ln\pars{1- x} \over x}}}\dd x + 2\ln\pars{2} - \ln^{2}\pars{2} \\[5mm] & = \bbx{{\pi^{2} \over 6} + 2\ln\pars{2} - \ln^{2}\pars{2} - 2} \qquad\qquad \left\{\begin{array}{rcl} \ds{\mrm{Li}_{2}\pars{1}} & \ds{=} & \ds{\pi^{2} \over 6} \\[1mm] \ds{\mrm{Li}_{2}'\pars{x}} & \ds{=} & \ds{-\,{\ln\pars{1 - x} \over x}} \end{array}\right. \end{align}

Answer 2

Otra solución, esta vez utilizando la suma de Euler alterna de Mike Spivey :

\begin{align*} I &=-\int_0^1\ln(1-x)\ln(1+x)\,\mathrm dx \\ &= \int_0^1\ln(1-x)\sum_{n=1}^\infty\frac{(-1)^n}{n}x^n\,\mathrm dx,\qquad\text{Mercator series}\\ &= \sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^1\ln(1-x)x^n\,\mathrm dx\\ &= \sum_{n=1}^\infty\frac{(-1)^n}{n}\cdot\frac{\partial}{\partial\alpha}\left[ \int_0^1(1-x)^\alpha x^n\,\mathrm dx\right ]\Bigg\vert_{\alpha=0}\\ &= \sum_{n=1}^\infty\frac{(-1)^n}{n}\cdot\frac{\partial}{\partial\alpha} \text{B}\left(\alpha+1,\,n+1 \right )\Bigg\vert_{\alpha=0},\qquad\text{beta function}\\ &= \sum_{n=1}^\infty\frac{(-1)^n}{n}\cdot\left[ \frac{\Gamma(\alpha+1)\Gamma(n+1)\left(\psi(\alpha+1)-\psi(\alpha+n+2) \right )}{\Gamma(\alpha+n+2)}\right ]\Bigg\vert_{\alpha=0}\\ &= -\sum_{n=1}^\infty\frac{(-1)^n}{n}\cdot\frac{H_{n+1}}{n+1},\qquad\text{used }\psi(1)=-\gamma\text{ and }\psi(m)=H_{m-1}-\gamma\\ &= -\sum_{n=1}^\infty\frac{(-1)^nH_{n+1}}{n}+\sum_{n=1}^\infty\frac{(-1)^nH_{n+1}}{n+1}\\ &= -\sum_{n=1}^\infty\frac{(-1)^nH_n}{n}-\sum_{n=1}^\infty\frac{(-1)^n}{n(n+1)}+\sum_{n=1}^\infty\frac{(-1)^nH_{n+1}}{n+1}\\ &= -\sum_{n=1}^\infty\frac{(-1)^nH_n}{n}-1+2\ln(2)+\sum_{n=1}^\infty\frac{(-1)^nH_{n+1}}{n+1}\\ &= -2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n}-2+2\ln(2)\\ &= \frac{\pi^2}{6}+2\ln(2)-\ln^2(2)-2,\qquad\text{applied Mike's sum.} \end{align*}

Answer 3

\begin{align}J&=\int_0^1 \ln(1+x)\ln(1-x)\,dx\\ &=\Big[\left(\left(1+x\right)\ln(1+x)-x-2\ln 2+1\right)\ln(1-x)\Big]_0^1+\int_0^1\frac{\left(1+x\right)\ln(1+x)-x-2\ln 2+1}{1-x}\,dx\\ &=\int_0^1\frac{\left(1+x\right)\ln(1+x)-x-2\ln 2+1}{1-x}\,dx\\ \end {align} Realiza el cambio de la variable $y=\dfrac{1-x}{1+x}$ , \begin{align}J&=\int_0^1\frac{\frac{2\ln\left(\frac{2}{1+x}\right)}{1+x}-\frac{1-x}{1+x}-2\ln 2+1}{x(1+x)}\\ &=\int_0^1\frac{\frac{2\ln\left(\frac{2}{1+x}\right)}{1+x}+\frac{2x}{1+x}-2\ln 2}{x(1+x)}\,dx\\ &=\int_0^1\frac{\frac{2\ln\left(\frac{2}{1+x}\right)}{1+x}+\frac{2x}{1+x}-2\ln 2}{x}\,dx-\int_0^1\frac{\frac{2\ln\left(\frac{2}{1+x}\right)}{1+x}+\frac{2x}{1+x}-2\ln 2}{1+x}\,dx\\ &=\int_0^1\frac{\frac{2\ln\left(\frac{2}{1+x}\right)}{1+x}+\frac{2x}{1+x}-2\ln 2}{x}\,dx+2+2\ln^2 2-4\ln 2\\ &=\int_0^1 \left(\frac{2\ln\left(\frac{2}{1+x}\right)}{x(1+x)}-\frac{2\ln 2}{x}\right)\,dx+2+2\ln^2 2-2\ln 2\\ &=\int_0^1 \left(\frac{2\ln\left(\frac{2}{1+x}\right)}{x}-\frac{2\ln\left(\frac{2}{1+x}\right)}{1+x}-\frac{2\ln 2}{x}\right)\,dx+2+2\ln^2 2-2\ln 2\\ &=-2\int_0^1 \frac{\ln(1+x)}{x}\,dx-2\int_0^1 \frac{\ln\left(\frac{2}{1+x}\right)}{1+x}\,dx+2+2\ln^2 2-2\ln 2\\ &=-2\int_0^1 \frac{\ln(1+x)}{x}\,dx+2+\ln^2 2-2\ln 2\\ \end {align} Pero, \begin{align}\int_0^1 \frac{\ln(1+x)}{x}\,dx=\frac{1}{2}\int_0^1 \frac{2x\ln(1-x^2)}{x^2}\,dx-\int_0^1 \frac{\ln(1-x)}{x}\,dx\end {align} En la segunda integral, realiza el cambio de la variable $y=x^2$ , \begin{align}\int_0^1 \frac{\ln(1+x)}{x}\,dx&=-\frac{1}{2}\int_0^1 \frac{\ln(1-x)}{x}\,dx\\ &=-\frac{1}{2}\times -\frac{\pi^2}{6}\\ &=\frac{\pi^2}{12} \end {align} Por lo tanto, \begin{align}\boxed{J=2+2\ln^2 2-2\ln 2-\dfrac{\pi^2}{6}}\end {align}