Evaluar

Question

Yo estaba tratando de hacer variaciones de una representación integral para $\log\frac{\pi}{2}$ debido a Jonathan Sondow, cuando me preguntaba sobre si es posible evaluar

$$\int_0^1-\frac{1-x}{(1-x+x^2)\log x}\,dx,\tag{1}$$ Wolfram Alpha de la calculadora en línea me proporcionan una forma cerrada con código

int -(1-x)/((1-x+x^2)log(x)) dx, from x=0 to x=1

Pregunta. Por favor me proporcione sugerencias para saber cómo evaluar anterior esta integral definida como $$\int_0^1-\frac{1-x}{(1-x+x^2)\log x}\,dx=\log \left(\frac{\Gamma(1/6)}{\Gamma(2/3)}\right)-\frac{\log \pi}{2}$$ como dijo Wolfram Alpha. Muchas gracias.

Answer 1

Gracias a la serie de energía representación de $1/(1+x^3)$ y el Teorema de convergencia dominada, la integral dada es\begin{align*} -\int_0^1\frac{1-x^2}{(1+x^3)\log x}\,dx &=\int_0^1\sum_{k=0}^{\infty}(-1)^{k+1}\frac{(1-x^2)x^{3k}}{\log x}\,dx\\ &=\int_0^1\sum_{k=0}^{\infty}\frac{(1-x^2)(x^{6k+3}-x^{6k})}{\log x}\,dx\\ &=\sum_{k=0}^{\infty}\int_0^1\frac{(1-x^2)(x^{6k+3}-x^{6k})}{\log x}\,dx\\&= \sum_{k=0}^{\infty}(\ln(6k+3)-\ln(6k+1)+\ln(6k+4)-\ln(6k+6))\\&= \ln\left(\prod_{k=0}^{\infty}\frac{(6k+3)(6k+4)}{(6k+1)(6k+6)}\right) =\ln\left(\frac{\Gamma(1/6)}{\sqrt{\pi}\Gamma(2/3)}\right) \end{align*} donde $\Gamma(x)$ es la función Gamma y utilizamos el hecho de que $$\int_{0}^{1}\frac{x^n-1}{\log x}\,dx = \int_{0}^{+\infty}\frac{1-e^{-nt}}{t}\,e^{-t}dt = \log(n+1)$ $ ( integral de Frullani se aplica en el último paso).

Answer 2

Yo estaba confundido por el uso implícito de la integral de Frullani, así que creo que vale la pena mencionar: $$\begin{align} \int_0^1\frac{x^n-x^m}{\log(x)}\,\mathrm{d}x &=\int_0^\infty\frac{e^{-mu}-e^{-nu}}{u}e^{-u}\,\mathrm{d}u\\ &=\lim_{\epsilon\to0^+}\int_\epsilon^\infty\frac{e^{-(m+1)u}-e^{-(n+1)u}}{u}\,\mathrm{d}u\\ &=\lim_{\epsilon\to0^+}\int_{(m+1)\epsilon}^{(n+1)\epsilon}\frac{e^{-u}}{u}\,\mathrm{d}u\\[3pt] &=\log\left(\frac{n+1}{m+1}\right) \end {Alinee el} $$ $$\begin{align} -\int_0^1\frac{1-x}{\left(1-x+x^2\right)\log(x)}\,\mathrm{d}x &=-\int_0^1\frac{1-x^2}{\left(1+x^3\right)\log(x)}\,\mathrm{d}x\\ &=-\sum_{k=0}^\infty(-1)^k\frac{x^{3k}-x^{3k+2}}{\log(x)}\,\mathrm{d}x\\ &=\sum_{k=0}^\infty(-1)^k\log\left(\frac{3k+3}{3k+1}\right)\\ &=\sum_{k=0}^\infty\log\left(\frac{(6k+3)(6k+4)}{(6k+1)(6k+6)}\right)\\ &=\log\left(\prod_{k=0}^\infty\frac{\left(k+\frac12\right)\left(k+\frac23\right)}{\left(k+\frac16\right)(k+1)}\right)\\ &=\lim_{n\to\infty}\log\left(\prod_{k=0}^{n-1}\frac{\color{#C00}{\left(k+\frac12\right)}\color{#090}{\left(k+\frac23\right)}}{\color{#00F}{\left(k+\frac16\right)}(k+1)}\right)\\ &=\log\left(\lim_{n\to\infty}\color{#C00}{\frac{\Gamma\left(n+\frac12\right)}{\Gamma\left(\frac12\right)}}\color{#090}{\frac{\Gamma\left(n+\frac23\right)}{\Gamma\left(\frac23\right)}}\color{#00F}{\frac{\Gamma\left(\frac16\right)}{\Gamma\left(n+\frac16\right)}}\frac{\Gamma(1)}{\Gamma(n+1)}\right)\\ &=\log\left(\frac1{\sqrt\pi}\frac{\Gamma\left(\frac16\right)}{\Gamma\left(\frac23\right)}\right)+\log\left(\lim_{n\to\infty}\frac{\Gamma\left(n+\frac12\right)\Gamma\left(n+\frac23\right)}{\Gamma\left(n+\frac16\right)\Gamma(n+1)}\right)\\ &=\log\left(\frac1{\sqrt\pi}\frac{\Gamma\left(\frac16\right)}{\Gamma\left(\frac23\right)}\right) \end {Alinee el} $$ el último paso es por la desigualdad de Gautschi.

Answer 3

Pensé que podría ser instructivo para presentar un enfoque que no depende de la integral de Frullani.

Para ello, en primer lugar observamos que $\frac{x-1}{\log(x)}=\int_0^1 x^s\,ds$.

Por lo tanto, podemos escribir

$$\begin{align} \int_0^1 \frac{1+x}{1+x^3}\,\frac{x-1}{\log(x)}\,dx&=\int_0^1 \left(\int_0^1 \frac{x^s+x^{s+1}}{1+x^3} \right)\,ds\\\\ &=\int_0^1 \sum_{n=0}^\infty (-1)^n\int_0^1 (x^{s+3n}+x^{s+3n+1})\,dx\,ds\\\\ &=\int_0^1 \sum_{n=0}^\infty (-1)^n \left(\frac{1}{s+3n+1}+\frac{1}{s+3n+2}\right)\,ds\\\\ &=\sum_{n=0}^\infty (-1)^n \log\left(\frac{3n+3}{3n+1}\right)\\\\ &=\sum_{n=0}^\infty \log\left(\frac{(6n+3)(6n+4)}{(6n+1)(6n+6)}\right)\tag 1\\\\ &=\log\left(\frac{\Gamma(1/6)}{\sqrt{\pi}\Gamma(2/3)}\right)\tag2 \end {Alinee el} $$

donde ir de $(1)$ $(2)$ confiamos en el análisis publicado en solución de Rob en el presente.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove armada]{\displaystyle{#1}}\,} \newcommand{\llaves}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &-\int_{0}^{1}{1 - x \over \pars{1 - x + x^{2}}\ln\pars{x}}\,\dd x = \int_{0}^{1}\ \overbrace{{1 + x \over 1 + x^{3}}} ^{\ds{1 \over 1 - x + x^{2}}}\,\,\, \overbrace{{x - 1 \over \ln\pars{x}}}^{\ds{\int_{0}^{1}x^{t}\,\dd t}}\ \,\dd x = \int_{0}^{1}\int_{0}^{1}{x^{t} + x^{t + 1} \over 1 + x^{3}}\,\dd x\,\dd t \\[5mm] & = \int_{0}^{1}\int_{0}^{1}{x^{t} + x^{t + 1} - x^{t + 3} - x^{t + 4} \over 1 - x^{6}}\,\dd x\,\dd t \\[5mm] & \stackrel{x^{6}\ \mapsto\ x}{=}\,\,\, {1 \over 6}\int_{0}^{1}\int_{0}^{1} {x^{t/6 - 5/6} + x^{t/6 - 2/3} - x^{t/6 - 1/3} - x^{t/6 - 1/6} \over 1 - x}\,\dd x\,\dd t \\[5mm] & = {1 \over 6}\int_{0}^{1}\!\!\pars{% -\int_{0}^{1}\!\!{1 - x^{t/6 - 5/6} \over 1 - x}\,\dd x - \int_{0}^{1}\!\!{1 - x^{t/6 - 2/3} \over 1 - x}\,\dd x + \int_{0}^{1}\!\!{1 - x^{t/6 - 1/3} \over 1 - x}\,\dd x + \int_{0}^{1}\!\!{1 - x^{t/6 - 1/6} \over 1 - x}\,\dd x} \\[5mm] & = {1 \over 6}\ \overbrace{\int_{0}^{1}\bracks{% -\Psi\pars{{t \over 6} + {1 \over 6}} - \Psi\pars{{t \over 6} + {1 \over 3}} + \Psi\pars{{t \over 6} + {2 \over 3}} + \Psi\pars{{t \over 6} + {5 \over 6}}}\dd t} ^{\ds{\Psi:\ Digamma\ Function}} \\[5mm] & = \left.\ln\pars{\Gamma\pars{t/6 + 2/3}\Gamma\pars{t/6 + 5/6} \over \Gamma\pars{t/6 + 1/6}\Gamma\pars{t/6 + 1/3}}\right\vert_{\ 0}^{\ 1}\qquad \pars{~\Gamma:\ Gamma\ Function.\ \Psi\pars{z} \stackrel{\mrm{def.}}{=} \totald{\ln\pars{\Gamma\pars{z}}}{z}~} \\[5mm] & = \ln\pars{{\Gamma\pars{5/6}\Gamma\pars{1} \over \Gamma\pars{1/3}\Gamma\pars{1/2}}\, {\Gamma\pars{1/6}\Gamma\pars{1/3} \over \Gamma\pars{2/3}\Gamma\pars{5/6}}} \\[5mm] & = \ln\pars{{1 \over \Gamma\pars{1/2}}\, {\Gamma\pars{1/6} \over \Gamma\pars{2/3}}}\qquad \pars{\ds{\mbox{Note that}\ \Gamma\pars{1} = 1\ \mbox{and}\ \Gamma\pars{1 \over 2} = \root{\pi}}} \\[5mm] & = \bbx{\ln\pars{{1 \over \root{\pi}}\, {\Gamma\pars{1/6} \over \Gamma\pars{2/3}}}} \approx 0.8412 \end{align}

Tenga en cuenta que $\ds{\left.\vphantom{\Large A}\Psi\pars{z}\right\vert_{\ \Re\pars{z}\ >\ 0} = -\gamma + \int_{0}^{1}{1 - t^{z - 1} \over 1 - t}\,\dd t}$. $\ds{\gamma}$ is the Euler-Mascheroni Constant. See $\ds{\color{#000}{\mathbf{6.3.22}}}$ en a & S de la Tabla.