$\newcommand{\ángulos}[1]{\left\langle\,{#1}\,\right\rangle}
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\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\mitad}{{1 \over 2}}
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La Pregunta:$\ds{\quad\int_{0}^{1}\ln\pars{x}
\bracks{{1 \over \ln\pars{x}} + {1\over 1-x}}^{2}\,\dd x =
\color{blue}{\gamma - 1}\,?}$.
\begin{align}
&\color{#f00}{\int_{0}^{1}\ln\pars{x}
\bracks{{1 \over \ln\pars{x}} + {1\over 1-x}}^{2}\,\dd x} =
\int_{0}^{1}\bracks{{1 \over \ln\pars{x}} + {2 \over 1 - x} + {\ln\pars{x} \over \pars{1 - x}^{2}}}\,\dd x
\end{align}
Al $\ds{x \lesssim 1}$ ambos $\ds{1 \over \ln\pars{x}}$ y
$\ds{\ln\pars{x} \over \pars{1 - x}^{2}}$ $\ds{\sim\,-\,{1 \over 1 - x}}$ de manera tal que la división de la integral original en tres 'piezas' conduce a la divergencia de las integrales aunque la suma de ellos converge. El mencionado comportamiento, al $\ds{x\lesssim 1}$, sugests la siguiente división:
\begin{align}
&\color{#f00}{\int_{0}^{1}\ln\pars{x}
\bracks{{1 \over \ln\pars{x}} + {1\over 1-x}}^{2}\,\dd x}
\\[3mm] = &\
\underbrace{\int_{0}^{1}\bracks{{1 \over \ln\pars{x}} + {1 \over 1 - x}}\,\dd x}
_{\ds{J_{1}}}\ +\
\underbrace{\int_{0}^{1}\bracks{{1 \over 1 - x} +
{\ln\pars{x} \over \pars{1 - x}^{2} }}\,\dd x}_{\ds{J_{2}}}\ =\
J_{1} + J_{2}\tag{1}
\end{align}
- $\ds{\large J_{1} =\, ?}$.
\begin{align}
&\int_{0}^{1}\bracks{{1 \over \ln\pars{x}} + {1 \over 1 - x}}\,\dd x =
\int_{0}^{1}\int_{0}^{\infty}\bracks{-x^{y} + \expo{-\pars{1 - x}y}}
\,\dd y\,\dd x
\\[3mm] = &\
\int_{0}^{\infty}\int_{0}^{1}\bracks{-x^{y} + \expo{-\pars{1 - x}y}}
\,\dd x\,\dd y =
\int_{0}^{\infty}\bracks{-\,{1 \over y + 1} +
\expo{-y}\,{\expo{y} - 1 \over y}}\,\dd y
\\[3mm] = &\
\lim_{\epsilon \to 0^{+}}\int_{\epsilon}^{\infty}\bracks{%
-\,{1 \over y + 1} + {1 \over y} - {\expo{-y} \over y}}\,\dd y
\\[3mm] = &\
\lim_{\epsilon \to 0^{+}}\bracks{-\ln\pars{\epsilon \over 1 + \epsilon} +
\ln\pars{\epsilon}\expo{-\epsilon} -
\int_{\epsilon}^{\infty}\ln\pars{y}\expo{-y}\,\dd y} =
-\,\lim_{\epsilon \to 0}\partiald{}{\epsilon}
\int_{0}^{\infty}y^{\epsilon}\expo{-y}\,\dd y
\\[3mm] = &\
-\Gamma\,'\pars{1} = -\Gamma\pars{1}\Psi\pars{1}
\end{align}
\begin{equation}\fbox{$\ds{\
J_{1} = \int_{0}^{1}\bracks{{1 \over \ln\pars{x}} + {1 \over 1 - x}}\,\dd x =
\color{#f00}{\gamma}\ }$}\tag{2}
\end{equation}
- $\ds{\large J_{2} =\, ?}$.
Tenga en cuenta que
\begin{align}
\int{\ln\pars{x} \over \pars{1 - x}^{2}}\,\dd x & =
\int\ln\pars{x}\,\dd\pars{1 \over 1 - x} =
{\ln\pars{x} \over 1 - x} - \int{1 \over 1 - x}\,{1 \over x}\,\dd x
\\[3mm] & =
{\ln\pars{x} \over 1 - x} - \int\pars{{1 \over 1 - x} + {1 \over x}}\,\dd x
\\[3mm] \mbox{such that}\quad &
\int\bracks{{1 \over 1 - x} +
{\ln\pars{x} \over \pars{1 - x}^{2} }}\,\dd x =
{\ln\pars{x} \over 1 - x} - \ln\pars{x}
\\[3mm] \mbox{and}\quad &
\left\lbrace\begin{array}{rcl}
\ds{\lim_{x \to 1}\bracks{{\ln\pars{x} \over 1 - x} - \ln\pars{x}}} & \ds{=} &
\ds{\color{#f00}{-1}}
\\[2mm]
\ds{\lim_{x \to 0^{+}}\bracks{{\ln\pars{x} \over 1 - x} - \ln\pars{x}}} & \ds{=} &
\ds{\color{#f00}{0}}
\end{array}\right.
\\[3 mm] \imp\quad &
\fbox{$\ds{\
J_{2} = \int_{0}^{1}\bracks{{1 \over 1 - x} +
{\ln\pars{x} \\pars{1 - x}^{2} }}\,\dd x =
\color{#f00}{-1} - \color{#f00}{0} = \color{#f00}{-1}\
}$}\etiqueta{3}
\end{align}
Con $\pars{1}$, $\pars{2}$ y $\pars{3}$:
$$
\color{#f00}{\int_{0}^{1}\ln\pars{x}
\bracks{{1 \over \ln\pars{x}} + {1\over 1-x}}^{2}\,\dd x} =
J_{1} + J_{2} =
\color{#f00}{\gamma - 1}
$$
