Encontrar el Valor de la Integral

Question

Encontrar el Valor de $$\begin{align}I=\int_{0}^{1}\frac{\ln(x)\,dx}{1-x^2}\end{align}$$

He intentado como esta: Tenemos $$\begin{align}2I=\int_{0}^{1}\frac{\ln(x^2)\,dx}{1-x^2}=\int_{0}^{1}\frac{\ln(1-(1-x^2))\,dx}{1-x^2}\end{align}$$ Lo

$$2I=\begin{align}\int_{0}^{1}\frac{-(1-x^2)-\frac{(1-x^2)^2}{2}-\frac{(1-x^2)^3}{3}-\cdots }{1-x^2}\end{align}=$$

Necesito ayuda desde aquí..

Answer 1

El uso de la expansión de la serie de $\dfrac{1}{1-x^2}$ i.e

$\displaystyle \frac{1}{1-x^2}=\sum_{k=0}^{\infty} x^{2k}\,dx$

Por lo tanto, $$I=\int_0^1 \ln(x)\left(\sum_{k=0}^{\infty} x^{2k}\right)\,dx=\sum_{k=0}^{\infty} \int_0^1 x^{2k}\ln x\,dx=-\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}$$ donde la última integral se puede evaluar mediante la sustitución de $\ln x=-t$.

Desde entonces, $$\sum_{k=1}^{\infty} \frac{1}{k^2}=\sum_{k=1}^{\infty} \frac{1}{(2k)^2}+\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} \Rightarrow \sum_{k=0}^{\infty} \frac{1}{(2k+1)^2}=\frac{\pi^2}{6}-\frac{1}{4}\frac{\pi^2}{6}$$ $$\Rightarrow \sum_{k=0}^{\infty} \frac{1}{(2k+1)^2}=\frac{3\pi^2}{24}=\frac{\pi^2}{8}$$

Por lo tanto, $$\boxed{I=-\dfrac{\pi^2}{8}}$$

Answer 2

Vamos $$ I(\alpha)=\int_{0}^{1}\frac{x^{\alpha}dx}{1-x^2}. $$ A continuación,$I'(0)=I$. Ahora \begin{eqnarray*} I(\alpha)=\lim_{a\to 1^-}\int_{0}^{a}\sum_{n=0}^\infty x^{\alpha+2n}dx=\lim_{a\to 1^-}\sum_{n=0}^\infty \frac{1}{\alpha+2n+1}a^{\alpha+2n+1} \end{eqnarray*} y por lo tanto \begin{eqnarray*} I'(\alpha)&=&\lim_{a\to 1^-}\sum_{n=0}^\infty \left(\frac{-1}{(\alpha+2n+1)^2}a^{\alpha+2n+1}+\frac{1}{\alpha+2n+1}a^{\alpha+2n+1}\ln a\right)\\ &=&-\sum_{n=0}^\infty\frac{1}{(\alpha+2n+1)^2} \end{eqnarray*} Así $$ I=I'(0)=-\sum_{n=0}^\infty\frac{1}{(2n+1)^2}=-\frac{\pi^2}{8}. $$

Answer 3

$\newcommand{\+}{^{\daga}} \newcommand{\ángulos}[1]{\left\langle\, nº 1 \,\right\rangle} \newcommand{\llaves}[1]{\left\lbrace\, nº 1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, nº 1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, nº 1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\piso}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\mitad}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\a la derecha\vert\,} \newcommand{\cy}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left (\, nº 1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\vphantom{\large Un}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, nº 1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} I&=\color{#66f}{\Large\int_{0}^{1}{\ln\pars{x} \over 1 - x^{2}}\,\dd x} =\half\int_{0}^{1}{\ln\pars{x} \over 1 - x}\,\dd x +\half\int_{0}^{1}{\ln\pars{x} \over 1 + x}\,\dd x \\[3mm]&=\half\int_{0}^{1}{\ln\pars{x} \over 1 - x}\,\dd x -\half\int_{0}^{-1}{\ln\pars{-x} \over 1 - x}\,\dd x =\half\int_{0}^{1}{\ln\pars{1 - x} \over x}\,\dd x -\half\int_{0}^{-1}{\ln\pars{1 - x} \over x}\,\dd x \\[3mm]&=-\half\sum_{n = 1}^{\infty}{1 \over n^{2}} +\half\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^{2}} =-\sum_{n = 1}^{\infty}{1 \over \pars{2n - 1}^{2}} =-\sum_{n = 1}^{\infty}{1 \over n^{2}} +\sum_{n = 1}^{\infty}{1 \over \pars{2n}^{2}} \\[3mm]&=-\,{3 \over 4}\ \underbrace{\sum_{n = 1}^{\infty}{1 \over n^{2}}} _{\ds{\color{#c00000}{\zeta\pars{2} = {\pi^{2} \over 6}}}} =-\,{3 \over 4}\,{\pi^{2} \over 6} = \color{#66f}{\Large -\,{\pi^{2} \over 8}} \end{align}

desde \begin{align} \int_{0}^{a}{\ln\pars{1 - x} \over x}\,\dd x\, &= \int_{0}^{a}{1 \over x}\pars{-\sum_{n = 1}^{\infty}{x^{n} \over n}}\,\dd x =-\sum_{n = 1}^{\infty}{a^{n} \over n^{2}}\,,\qquad \verts{a}\ \leq\ 1 \end{align}

Answer 4

La integración por partes,

$$ \begin{align} \int \frac{\ln (x)}{1-x^{2}} \ dx &= \ln(x) \ \text{arctanh} (x) - \int \frac{\text{arctanh}(x)}{x} \ dx \\ &= \ln (x) \ \text{arctanh} (x) - \frac{1}{2} \int \frac{\ln (1+x)}{x} \ dx + \frac{1}{2} \int \frac{\ln (1-x)}{x} \ dx \\ &= \ln(x) \ \text{arctanh}(x) + \frac{1}{2} \text{Li}_{2}(-x) - \frac{1}{2} \text{Li}_{2}(x) + C \end{align}$$

donde $\text{Li}_{2}(x)$ es el dilogarithm función.

Entonces

$$ \begin{align} \int_{0}^{1} \frac{\ln (x)}{1-x^{2}} \ dx &= \frac{1}{2} \Big( \text{Li}_{2}(-1) - \text{Li}_{2}(1) \Big) \\ &= \frac{1}{2} \Big( - \frac{\zeta(2)}{2} - \zeta(2) \Big) \\ &= \frac{1}{2} \left(- \frac{\pi^{2}}{12}-\frac{\pi^{2}}{6} \right) \\ &= -\frac{\pi^{2}}{8} . \end{align}$$