Mostrar que $\int_0^1 \int_0^1 {x\ln x\over (1-xy)\ln(xy)} \, dx \, dy=1-\gamma.$

Question

$$\int_{0}^{1}\int_{0}^{1}{x\ln x\over (1-xy)\ln(xy)} \, dx \, dy=1-\gamma.\tag1$$

Let $u=xy

$$\int_0^1 {1\over y^2}\int_0^y {u\ln u -u\ln y \over (1-u)\ln(u)}dudy\tag2$$

$$\int_0^y {u\over 1-u} \, du-\ln y\int_0^y {u\over \ln u} \, du\tag3$$

$$-y-\ln(1-y)-\ln y \int_0^y {u\over \ln u} \, du\tag4$$

En cuanto a esta integral

Definimos $n=1$

$$f(n,u)=\int_0^y {u^n\over \ln u} \, du\tag5$$

Podemos eliminar $\ln u$ mediante diferenciación

$${df\over dn}=\int_0^y u^n \, du = {y^{n+1}\over n+1}\tag6$$

¿Cómo puedo avanzar al siguiente paso?

Answer 1

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

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\begin{align} \color{#f00}{\large ?} & = \int_{0}^{1}{1 - x + \ln\pars{x} \over \pars{1 - x}\ln\pars{x}}\,\dd x = -\int_{0}^{1}{\pars{x - 1}/\ln\pars{x} - 1 \over 1 - x}\,\dd x = -\int_{0}^{1}{1 \over 1 - x}\int_{0}^{1}\pars{x^{t} - 1}\,\dd t\,\dd x \\[3mm] & = \int_{0}^{1}\int_{0}^{1}{1 - x^{t} \over 1 - x}\,\dd x\,\dd t = \int_{0}^{1}\bracks{\Psi\pars{t + 1} + \gamma}\,\dd t = \ln\pars{\Gamma\pars{2}} - \ln\pars{\Gamma\pars{1}} + \gamma = \color{#f00}{\gamma} \end{align}

Note that $$ \int_{0}^{1}\pars{x^{t} - 1}\,\dd t = {1 \over \ln\pars{x}}\int_{0}^{1}x^{t}\ln\pars{x}\,\dd t - 1 = {1 \over \ln\pars{x}}\int_{0}^{1}\partiald{x^{t}}{t}\,\dd t - 1 = {x - 1 \over \ln\pars{x}} - 1 $$

Answer 2

Puede demostrarse primero que: $$\iint_{(0,1)^2}\frac{x^{k+1}y^k \log(x)}{\log(xy)}\,dx\,dy = -\frac{1}{k+2}+\log(k+2)-\log(k+1)\tag{1}$$ a través de la diferenciación bajo el signo integral u otras técnicas, luego sumando el RHS de $(1)$ sobre $k\geq 0$ obtenemos: $$ I = \iint_{(0,1)^2}\frac{x\log x}{(1-xy)\log(xy)}\,dx\,dy = 1-\sum_{k\geq 1}\left(\frac{1}{k}-\log\left(1+\frac{1}{k}\right)\right)=1-\gamma\tag{2}$$ como queríamos.