Evaluar $\int_{0}^{1}{\int_{0}^{1}{\frac{x^{\alpha -1}y^{\beta -1}}{\left( 1+xy \right)\ln xy}dxdy}}$

Question

En AoPS (Art of Problem Solving), la siguiente integral fue publicada por el compañero pprime, pero nadie ha podido acercarse a resolverla. Se han hecho muchos intentos, pero ninguno ha tenido éxito. He venido aquí como último recurso para que algún amable usuario me ayude a evaluar esta integral.

http://www.artofproblemsolving.com/community/c7h1288609_extreme_integration_marathon

Evaluar $$\int_{0}^{1}{\int_{0}^{1}{\frac{x^{\alpha -1}y^{\beta -1}}{\left( 1+xy \right)\ln xy}dxdy}}$$

Una pista: $\int_{0}^{+\infty }{\exp \left( -u\ln \left( xy \right) \right)du}=\frac{1}{\ln x+\ln y}$

Answer 1

Sólo hay que tener en cuenta que con Teorema de Frullani podemos llegar al producto infinito con bastante rapidez. Tenemos que $$I(\alpha,\beta)=\int_{0}^{1}\int_{0}^{1}\frac{x^{\alpha-1}y^{\beta-1}}{\left(1+xy\right)\log\left(xy\right)}dxdy\stackrel{xy=u}{=}\int_{0}^{1}y^{\beta-\alpha-1}\int_{0}^{y}\frac{u^{\alpha-1}}{\left(1+u\right)\log\left(u\right)}dudy $$ $$\stackrel{IBP}{=}\frac{1}{\beta-\alpha}\int_{0}^{1}\frac{y^{\alpha-1}-y^{\beta-1}}{\left(1+y\right)\log\left(y\right)}dy=\frac{1}{\beta-\alpha}\sum_{k\geq0}\left(-1\right)^{k}\int_{0}^{1}\frac{y^{\alpha-1+k}-y^{\beta-1+k}}{\log\left(y\right)}dy$$ $$\stackrel{y=e^{-v}}{=}\frac{1}{\beta-\alpha}\sum_{k\geq0}\left(-1\right)^{k+1}\int_{0}^{\infty}\frac{e^{-v(\alpha+k)}-e^{-v(\beta+k)}}{v}dv $$ y ahora podemos utilizar el teorema de Frullani y obtener $$I(\alpha,\beta)=\frac{1}{\beta-\alpha}\sum_{k\geq0}\left(-1\right)^{k+1}\log\left(\frac{\beta+k}{\alpha+k}\right)=\frac{1}{\beta-\alpha}\log\left(\prod_{k\geq0}\left(\frac{\beta+2k+1}{\alpha+2k+1}\right)\left(\frac{\beta+2k}{\alpha+2k}\right)^{-1}\right) $$ Ahora bien, tenga en cuenta que $$\prod_{k=0}^{N}\left(a+2k\right)=2^{N}\prod_{k=0}^{N}\left(\frac{a}{2}+k\right)=2^{N}a\left(\frac{a}{2}+1\right)_{N} $$ y $$\prod_{k=0}^{N}\left(a+2k+1\right)=2^{N}\prod_{k=0}^{N}\left(k+\frac{a+1}{2}\right)=\left(a+1\right)2^{N}\left(\frac{a+1}{2}+1\right)_{N} $$ donde $\left(x\right)_{N} $ es el Símbolo del martillo pilón . Así que $$\prod_{k=0}^{N}\left(\frac{\beta+2k+1}{\alpha+2k+1}\right)\left(\frac{\beta+2k}{\alpha+2k}\right)^{-1}=\frac{\beta+1}{\alpha+1}\frac{\alpha}{\beta}\frac{\left(\frac{\beta+1}{2}+1\right)_{N}}{\left(\frac{\alpha+1}{2}+1\right)_{N}}\frac{\left(\frac{\alpha}{2}+1\right)_{N}}{\left(\frac{\beta}{2}+1\right)_{N}} $$ y ahora usando el asintótica para el símbolo de Pochhammer obtenemos

$$I(\alpha,\beta)=\color{red}{\frac{1}{\beta-\alpha}\log\left(\frac{\Gamma\left(\frac{\alpha+1}{2}\right)\Gamma\left(\frac{\beta}{2}\right)}{\Gamma\left(\frac{\alpha}{2}\right)\Gamma\left(\frac{\beta+1}{2}\right)}\right)}$$ donde $$\, \beta,\alpha>0,\,\beta\neq \alpha$$

como se quiera. Si $\alpha=\beta=\gamma $ tenemos $$I(\gamma)=\int_{0}^{1}\int_{0}^{1}\frac{\left(xy\right)^{\gamma-1}}{\left(1+xy\right)\log\left(xy\right)}dxdy\stackrel{xy=u}{=}\int_{0}^{1}y^{-1}\int_{0}^{y}\frac{u^{\alpha-1}}{\left(1+u\right)\log\left(u\right)}dudy $$ $$\stackrel{IBP}{=}-\int_{0}^{1}\frac{y^{\gamma-1}}{1+y}dy=\sum_{k\geq0}\left(-1\right)^{k+1}\frac{1}{\gamma+k}=\color{red}{-\Phi\left(-1,1,\gamma\right)} $$ donde $\gamma>0$ y $\Phi\left(x,y,z\right)$ es el Lerch Trascendente . Obviamente, para algunos valores especiales de $\gamma$ tenemos bonitas formas cerradas. Por ejemplo, para $\gamma=1$ tenemos $$I\left(1\right)=\sum_{k\geq0}\left(-1\right)^{k+1}\frac{1}{1+k}=-\log\left(2\right).$$

Answer 2

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\color{#f00}{\int_{0}^{1}\int_{0}^{1}{x^{\alpha - 1}\,\,y^{\beta - 1} \over \pars{1 + xy}\ln\pars{xy}}\,\dd x\,\dd y} = -\int_{0}^{1}\int_{0}^{1}{x^{\alpha - 1}\,\,y^{\beta - 1} \over 1 + xy}\,\,\, \overbrace{\int_{0}^{\infty}\pars{xy}^{z}\,\dd z} ^{\ds{-\,{1 \over \ln\pars{xy}}}}\ \,\dd x\,\dd y \\[5mm] = &\ -\int_{0}^{\infty}\int_{0}^{1}\int_{0}^{1} {x^{\alpha -1 + z}\,\,\,y^{\beta - 1 + z} \over 1 + xy} \,\dd x\,\dd y\,\dd z = -\int_{0}^{\infty}\int_{0}^{1}\int_{0}^{1} {\pars{xy}^{\alpha -1 + z}\,\,\,y^{\beta - \alpha - 1} \over 1 + xy} \,\dd\pars{xy}\,\dd y\,\dd z \\[5mm] = &\ -\int_{0}^{\infty}\int_{0}^{1}\int_{0}^{y} {x^{\alpha -1 + z}\,\,\,y^{\beta - \alpha - 1} \over 1 + x} \,\dd x\,\dd y\,\dd z = -\int_{0}^{\infty}\int_{0}^{1}{x^{\alpha -1 + z}\,\,\, \over 1 + x}\int_{x}^{1} y^{\beta - \alpha - 1}\,\,\,\dd y\,\dd x\,\dd z \\[5mm] = &\ {1 \over \beta - \alpha}\int_{0}^{\infty}\int_{0}^{1} {x^{\beta + z - 1}\,\,\, -\,\,\, x^{\alpha + z - 1}\,\,\, \over 1 + x} \,\dd x\,\dd z \\[5mm] = &\ {1 \over \beta - \alpha}\int_{0}^{\infty}\int_{0}^{1} {x^{\beta + z - 1}\,\,\, -\,\,\, x^{\alpha + z - 1}\,\,\, -\,\,\, x^{\beta + z}\,\,\, +\,\,\, x^{\alpha + z} \over 1 - x^{2}} \,\dd x\,\dd z \\[5mm] \stackrel{x^{2}\ \mapsto\ x}{=}\ &\ {1 \over 2\pars{\beta - \alpha}} \int_{0}^{\infty}\int_{0}^{1} {x^{\beta/2 + z/2 - 1}\,\,\, -\,\,\, x^{\alpha/2 + z/2 - 1}\,\,\, -\,\,\, x^{\beta/2 + z/2 - 1/2}\,\,\, +\,\,\, x^{\alpha/2 + z/2 - 1/2} \over 1 - x} \,\,\,\,\dd x\,\dd z \end{align}

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Con la Función Digamma identidad $\pars{~\gamma\ \mbox{is the Euler-Mascheroni Constant}~}$ $$\left.\Psi\pars{\xi} = -\gamma + \int_{0}^{1}{1 - t^{\xi - 1} \over 1 - t}\,\dd t\, \right\vert_{\ \Re\pars{\xi}\ >\ 0\,\,\,\,\,} $$ la integración se reduce a: \begin{align} &\color{#f00}{\int_{0}^{1}\int_{0}^{1}{x^{\alpha - 1}\,\,y^{\beta - 1} \over \pars{1 + xy}\ln\pars{xy}}\,\dd x\,\dd y} \\[5mm] = &\ {1 \over 2\pars{\beta - \alpha}}\int_{0}^{\infty}\bracks{% \Psi\pars{z + \alpha \over 2} + \Psi\pars{z + \beta + 1 \over 2} - \Psi\pars{z + \beta \over 2} - \Psi\pars{z + \alpha + 1 \over 2}}\,\dd z \end{align}

Desde $\ds{\Psi\pars{\xi}\ \stackrel{\mrm{def.}}{=}\ \totald{\ln\pars{\Gamma\pars{\xi}}}{\xi}}$ \begin{align} &\color{#f00}{\int_{0}^{1}\int_{0}^{1}{x^{\alpha - 1}\,\,y^{\beta - 1} \over \pars{1 + xy}\ln\pars{xy}}\,\dd x\,\dd y} = \left.{1 \over \beta - \alpha} \ln\pars{\Gamma\pars{z/2 + \alpha/2}\Gamma\pars{z/2 + \beta/2 + 1/2} \over \Gamma\pars{z/2 + \beta/2}\Gamma\pars{z/2 + \alpha/2 + 1/2}} \right\vert_{\ 0}^{\ \infty} \\[5mm] = &\ \color{#f00}{{1 \over \beta - \alpha}\, \ln\pars{\Gamma\pars{\bracks{\alpha + 1}/2}\Gamma\pars{\beta/2} \over \Gamma\pars{\alpha/2}\Gamma\pars{\bracks{\beta + 1}/2}}}\,;\qquad \Re\pars{\alpha} > 0\,,\quad\Re\pars{\beta} > 0 \end{align}

Cuando $\ds{\ul{\beta \to \alpha}}$ la solución se convierte en: $$ \half\bracks{\Psi\pars{\alpha \over 2} - \Psi\pars{\alpha + 1 \over 2}} $$

Answer 3

$I\left( \alpha ,\beta \right)=\int\limits_{0}^{1}{\int\limits_{0}^{1}{\frac{x^{\alpha -1}y^{\beta -1}}{\left( 1+xy \right)\ln xy}dxdy}}=-\int\limits_{0}^{1}{\int\limits_{0}^{1}{\int\limits_{0}^{+\infty }{\frac{x^{\alpha -1}y^{\beta -1}}{1+xy}\left( xy \right)^{j}djdxdy}}}=-\int\limits_{0}^{+\infty }{\int\limits_{0}^{1}{\int\limits_{0}^{1}{\frac{x^{\alpha -1}y^{\beta -1}}{1+xy}\left( xy \right)^{j}dxdydj}}}$

$=-\int\limits_{0}^{+\infty }{\int\limits_{0}^{1}{\int\limits_{0}^{1}{\sum\limits_{k=0}^{+\infty }{\left( -xy \right)^{k}\cdot }x^{\alpha +j-1}y^{\beta +j-1}dxdydj}}}=-\sum\limits_{k=0}^{+\infty }{\left( -1 \right)^{k}\int\limits_{0}^{+\infty }{\int\limits_{0}^{1}{\int\limits_{0}^{1}{x^{\alpha +k+j-1}y^{\beta +k+j-1}dxdydj}}}}$

$=-\sum\limits_{k=0}^{+\infty }{\left( -1 \right)^{k}\int\limits_{0}^{+\infty }{\left( \int\limits_{0}^{1}{x^{\alpha +k+j-1}dx} \right)\left( \int\limits_{0}^{1}{y^{\beta +k+j-1}dy} \right)dj}}=-\sum\limits_{k=0}^{+\infty }{\left( -1 \right)^{k}\int\limits_{0}^{+\infty }{\left( \frac{1}{\left( \alpha +k+j \right)\left( \beta +k+j \right)} \right)dj}}$

$=-\frac{1}{\alpha -\beta }\sum\limits_{k=0}^{+\infty }{\left( -1 \right)^{k}\int\limits_{0}^{+\infty }{\left( \frac{\alpha +k+j-\left( \beta +k+j \right)}{\left( \alpha +k+j \right)\left( \beta +k+j \right)} \right)dj}}$

$=-\frac{1}{\alpha -\beta }\sum\limits_{k=0}^{+\infty }{\left( -1 \right)^{k}\int\limits_{0}^{+\infty }{\left( \frac{1}{\beta +k+j}-\frac{1}{\alpha +k+j} \right)dj}}=\frac{1}{\alpha -\beta }\sum\limits_{k=0}^{+\infty }{\left( -1 \right)^{k}\ln \left( \frac{\beta +k}{\alpha +k} \right)}$

$=\frac{1}{\alpha -\beta }\left( \left( -1 \right)^{0}\ln \left( \frac{\beta +0}{\alpha +0} \right)+\left( -1 \right)^{1}\ln \left( \frac{\beta +1}{\alpha +1} \right)+\left( -1 \right)^{2}\ln \left( \frac{\beta +2}{\alpha +2} \right)+\left( -1 \right)^{3}\ln \left( \frac{\beta +3}{\alpha +3} \right)+... \right)$

$=\frac{1}{\alpha -\beta }\left( \ln \left( \frac{\beta +0}{\alpha +0} \right)-\ln \left( \frac{\beta +1}{\alpha +1} \right)+\ln \left( \frac{\beta +2}{\alpha +2} \right)-\ln \left( \frac{\beta +3}{\alpha +3} \right)+... \right)$

$=\frac{1}{\alpha -\beta }\left( \ln \left( \frac{\beta +0}{\alpha +0} \right)+\ln \left( \frac{\alpha +1}{\beta +1} \right)+\ln \left( \frac{\beta +2}{\alpha +2} \right)+\ln \left( \frac{\alpha +3}{\beta +3} \right)+... \right)$

$=\frac{1}{\alpha -\beta }\ln \left( \left( \frac{\beta +0}{\alpha +0} \right)\cdot \left( \frac{\alpha +1}{\beta +1} \right)\cdot \left( \frac{\beta +2}{\alpha +2} \right)\cdot \left( \frac{\alpha +3}{\beta +3} \right)\cdot ... \right)=\frac{1}{\alpha -\beta }\ln \prod\limits_{k=0}^{+\infty }{\left( \frac{\beta +2k}{\alpha +2k} \right)\left( \frac{\alpha +1+2k}{\beta +1+2k} \right)}$

$=\frac{1}{\alpha -\beta }\ln \frac{\beta }{\alpha }\cdot \frac{\alpha +1}{\beta +1}\cdot \prod\limits_{k=1}^{+\infty }{\left( \frac{1+\frac{\frac{\beta }{2}}{k}}{1+\frac{\frac{\alpha }{2}}{k}}\cdot \frac{1+\frac{\frac{\alpha +1}{2}}{k}}{1+\frac{\frac{\beta +1}{2}}{k}} \right)}$

$=\frac{1}{\alpha -\beta }\ln \frac{\beta }{\alpha }\cdot \frac{\alpha +1}{\beta +1}\cdot \prod\limits_{k=1}^{+\infty }{\left( \frac{e^{\frac{\frac{\alpha }{2}}{k}}\left( 1+\frac{\frac{\alpha }{2}}{k} \right)^{-1}\cdot e^{\frac{\frac{\beta +1}{2}}{k}}\left( 1+\frac{\frac{\beta +1}{2}}{k} \right)^{-1}}{e^{\frac{\frac{\beta }{2}}{k}}\left( 1+\frac{\frac{\beta }{2}}{k} \right)^{-1}\cdot e^{\frac{\frac{\alpha +1}{2}}{k}}\left( 1+\frac{\frac{\alpha +1}{2}}{k} \right)^{-1}} \right)}$

$=\frac{1}{\alpha -\beta }\ln \frac{\beta }{\alpha }\cdot \frac{\alpha +1}{\beta +1}\cdot \prod\limits_{k=1}^{+\infty }{\left( \frac{e^{\frac{\frac{\alpha }{2}}{k}}\left( 1+\frac{\frac{\alpha }{2}}{k} \right)^{-1}\cdot e^{\frac{\frac{\beta +1}{2}}{k}}\left( 1+\frac{\frac{\beta +1}{2}}{k} \right)^{-1}}{e^{\frac{\frac{\beta }{2}}{k}}\left( 1+\frac{\frac{\beta }{2}}{k} \right)^{-1}\cdot e^{\frac{\frac{\alpha +1}{2}}{k}}\left( 1+\frac{\frac{\alpha +1}{2}}{k} \right)^{-1}} \right)}$

$=\frac{1}{\alpha -\beta }\ln \left( \frac{\frac{e^{-\gamma \frac{\alpha }{2}}}{\frac{\alpha }{2}}\prod\limits_{k=1}^{+\infty }{\left( e^{\frac{\frac{\alpha }{2}}{k}}\left( 1+\frac{\frac{\alpha }{2}}{k} \right)^{-1} \right)}\cdot \frac{e^{-\gamma \frac{\beta +1}{2}}}{\frac{\beta +1}{2}}\prod\limits_{k=1}^{+\infty }{\left( e^{\frac{\frac{\beta +1}{2}}{k}}\left( 1+\frac{\frac{\beta +1}{2}}{k} \right)^{-1} \right)}}{\frac{e^{-\gamma \frac{\beta }{2}}}{\frac{\beta }{2}}\prod\limits_{k=1}^{+\infty }{\left( e^{\frac{\frac{\beta }{2}}{k}}\left( 1+\frac{\frac{\beta }{2}}{k} \right)^{-1} \right)}\cdot \frac{e^{-\gamma \frac{\alpha +1}{2}}}{\frac{\alpha +1}{2}}\prod\limits_{k=1}^{+\infty }{\left( e^{\frac{\frac{\alpha +1}{2}}{k}}\left( 1+\frac{\frac{\alpha +1}{2}}{k} \right)^{-1} \right)}} \right)$

$=\frac{1}{\alpha -\beta }\ln \left( \frac{\Gamma \left( \frac{\alpha }{2} \right)\cdot \Gamma \left( \frac{\beta +1}{2} \right)}{\Gamma \left( \frac{\beta }{2} \right)\cdot \Gamma \left( \frac{\alpha +1}{2} \right)} \right)$