TITLE: How to find ${\large\int}_1^\infty\frac{1-x+\ln x}{x \left(1+x^2\right) \ln^2 x} \mathrm dx$?
QUESTION [40 upvotes]: Please help me to find a closed form for this integral:
$$
I=\int_1^\infty\frac{1-x+\ln x}{x \left(1+x^2\right) \ln^2 x} \mathrm dx
$$
Routine textbook methods for this complicated integral fail.

REPLY [7 votes]: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

In this answer, I'll show a $\ds{\ \ul{quite\ short\ path} }$ to an integration in terms of $\ds{\ \ul{just\ a\ pair}\ }$ of Digamma's functions.

Sooner or later, I'll need the following integral:

\begin{align}
&\begin{array}{|c|}\hline\\
\ds{\quad\left.\int_{0}^{1}{x^{t} \over 1 + x^{2}}\,\dd x\,
\right\vert_{\ \Re\pars{t}\ >\ -1}\quad}
\\ \\ \hline
\end{array} =
\int_{0}^{1}{x^{t} - x^{t + 2} \over 1 - x^{4}}\,\dd x
\\[3mm] = &\
{1 \over 4}\int_{0}^{1}{x^{\pars{t - 3}/4} - x^{\pars{t - 1}/4} \over 1 - x}\,\dd x =
{1 \over 4}\bracks{\int_{0}^{1}{1 - x^{\pars{t - 1}/4} \over 1 - x}\,\dd x
-
\int_{0}^{1}{1 - x^{\pars{t - 3}/4} \over 1 - x}\,\dd x}
\\[3mm] = &\
\begin{array}{|c|}\hline \\
\ds{\quad{1 \over 4}\bracks{%
\Psi\pars{t + 3 \over 4} - \Psi\pars{t + 1 \over 4}}\quad}
\\  \\ \hline
\end{array}\tag{1}
\\&
\end{align}
  where $\ds{\Psi}$ is the Digamma Function and we used the standard integral representation of it. Namely,
  $\ds{\left.\vphantom{\huge A}\Psi\pars{z}\,\right\vert_{\ \Re\pars{z}\ >\ 0}\
=\ -\gamma + \int_{0}^{1}{1 - t^{z - 1} \over 1 - t}\,\dd t.\ }$ $\ds{\gamma}$ is the Euler-Mascheroni Constant.


\begin{align}
\color{#f00}{I} &  \equiv
\int_{1}^{\infty}{1 - x + \ln\pars{x} \over x \pars{1 + x^{2}}\ln^{2}\pars{x}}
\,\dd x\ \stackrel{x\ \mapsto\ 1/x}{=}\
\int_{0}^{1}{x - 1 - x\ln\pars{x} \over \pars{1 + x^{2}}\ln^{2}\pars{x}}\,\dd x
\\[3mm] & =
-\int_{0}^{1}\underbrace{%
\bracks{{x \over \ln\pars{x}} + {1 - x \over \ln^{2}\pars{x}}}}
_{\ds{\vphantom{\Large A}\color{#f00}{?}}}\
\,{\dd x \over 1 + x^{2}}\tag{2}
\end{align}
\begin{equation}
\mbox{Note that}\quad\
\overbrace{{x \over \ln\pars{x}} + {1 - x \over \ln^{2}\pars{x}}}
^{\ds{\color{#f00}{\vphantom{\Large A}?}}}\ =\
\int_{0}^{1}t\,x^{t}\,\dd t\tag{3}
\end{equation}
This identity can be straightforward derived by performing two successive integration by parts with the identity $\ds{x^{t} = {1 \over \ln\pars{x}}\,\partiald{x^{t}}{t}}$

By replacing $\pars{3}$ in $\pars{2}$:
\begin{align}
\color{#f00}{I} & =
-\int_{0}^{1}t\int_{0}^{1}{x^{t} \over 1 + x^{2}}\,\dd x\,\dd t =
-\,{1 \over 4}\int_{0}^{1}t\bracks{\Psi\pars{t + 3 \over 4} -
\Psi\pars{t + 1 \over 4}}\,\dd t\quad
\pars{~\mbox{see}\ \pars{1}~}
\\[3mm] & =
-\,{1 \over 4}\int_{0}^{1}t\,\partiald{}{t}
\ln\pars{\Gamma^{\,4}\pars{\bracks{t + 3}/4} \over
\Gamma^{\,4}\pars{\bracks{t + 1}/4}}\,\dd t
\\[3mm] & =
-\,{1 \over 4}
\ln\pars{\Gamma^{\,4}\pars{1} \over \Gamma^{\,4}\pars{1/2}} +
\int_{0}^{1}\ln\pars{\Gamma\pars{t + 3 \over 4}}\,\dd t -
\int_{0}^{1}\ln\pars{\Gamma\pars{t + 1 \over 4}}\,\dd t
\\[3mm] & =
\half\,\ln\pars{\pi} +
4\int_{-1/4}^{0}\ln\pars{\Gamma\pars{t + 1}}\,\dd t -
4\int_{-3/4}^{-1/2}\ln\pars{\Gamma\pars{t + 1}}\,\dd t
\\[8mm] & =
\half\,\ln\pars{\pi}
\\[3mm] & -
4\int_{0}^{-1/4}\ln\pars{\Gamma\pars{t + 1}}\,\dd t -
4\int_{0}^{-1/2}\ln\pars{\Gamma\pars{t + 1}}\,\dd t +
4\int_{0}^{-3/4}\ln\pars{\Gamma\pars{t + 1}}\,\dd t\tag{4}
\end{align}
I'll use the well know identity
( $\ds{\mathrm{G}}$ is the  Barnes $\mathrm{G}$-Function or/and Double Gamma Function
 )
\begin{align}
&\int_{0}^{z}\ln\pars{\Gamma\pars{t + 1}}\,\dd t
\\[3mm] = &\
\half\,z\ln\pars{2\pi} - \half\,z\pars{z + 1} + z\ln\pars{\Gamma\pars{z + 1}} -
\ln\pars{\mathrm{G}\pars{z + 1}}
\end{align}
which, for $\ds{\ul{z \in \mathbb{R}}}$, is conveniently written as
$$
\int_{0}^{z}\ln\pars{\Gamma\pars{t + 1}}\,\dd t =
\half\,z\ln\pars{2\pi} -
\ln\pars{\exp\pars{z\pars{z + 1} \over 2}
\Gamma^{\, -z}\pars{z + 1}\,\mathrm{G}\pars{z + 1}}
$$
It leads to explicit expressions for the integrals in $\ds{\pars{4}}$:
\begin{equation}
\left\lbrace\begin{array}{rcl}
\ds{\int_{0}^{-1/4}\ln\pars{\Gamma\pars{t + 1}}\,\dd t} & \ds{=} &
\ds{-\,{1 \over 8}\,\ln\pars{2\pi} -
\ln\pars{\expo{-3/32}\ \Gamma^{1/4}\pars{3 \over 4}\mathrm{G}\pars{3 \over 4}}}
\\[3mm]
\ds{\int_{0}^{-1/2}\ln\pars{\Gamma\pars{t + 1}}\,\dd t} & \ds{=} &
\ds{-\,{1 \over 4}\,\ln\pars{2\pi} - \ln\pars{\expo{-1/8}\ \Gamma^{1/2}\pars{\half}\mathrm{G}\pars{\half}}}
\\[3mm]
\ds{\int_{0}^{-3/4}\ln\pars{\Gamma\pars{t + 1}}\,\dd t} & \ds{=} &
\ds{-\,{3 \over 8}\,\ln\pars{2\pi} -
\ln\pars{\expo{-3/32}\ \Gamma^{3/4}\pars{1 \over 4}\mathrm{G}\pars{1 \over 4}}}
\end{array}\right.\tag{5}
\end{equation}
Fortunately, in terms of the Glaisher-Kinkelin Constant $\ds{A}$ and the Catalan Constant $\ds{K}$, the above $\ds{\ul{\ln\mbox{-arguments}}}$ are given in the MathWorld G-Barnes Function page as formulas $\ds{\pars{17}}$, $\ds{\pars{19}}$ and $\ds{\vphantom{\Large A}\pars{16}}$, respectively:
\begin{align}
\left\lbrace\begin{array}{rcl}
\ds{\expo{-3/32}\ \Gamma^{1/4}\pars{3 \over 4}\mathrm{G}\pars{3 \over 4}} & \ds{=} & \ds{A^{-9/8}\ \expo{K/\pars{4\pi}}}
\\[3mm]
\ds{\expo{-1/8}\ \Gamma^{1/2}\pars{\half}\mathrm{G}\pars{\half}} & \ds{=} & \ds{A^{-3/2\,\,2^{1/24}}}
\\[3mm]
\ds{\expo{-3/32}\ \Gamma^{3/4}\pars{1 \over 4}\mathrm{G}\pars{1 \over 4}} & \ds{=} & \ds{A^{-9/8}\ \expo{-K/\pars{4\pi}}}
\end{array}\right.
\end{align}
The integrals in expressions $\ds{\pars{5}}$ become:
\begin{equation}
\left\lbrace\begin{array}{rcl}
\ds{\int_{0}^{-1/4}\ln\pars{\Gamma\pars{t + 1}}\,\dd t} & \ds{=} &
\ds{-\,{1 \over 8}\,\ln\pars{2\pi} + {9 \over 8}\,\ln\pars{A} - {K \over 4\pi}
\approx -0.0228}
\\[3mm]
\ds{\int_{0}^{-1/2}\ln\pars{\Gamma\pars{t + 1}}\,\dd t} & \ds{=} &
\ds{-\,{1 \over 4}\,\ln\pars{2\pi} + {3 \over 2}\,\ln\pars{A} -
{1 \over 24}\,\ln\pars{2} \approx -0.1152}
\\[3mm]
\ds{\int_{0}^{-3/4}\ln\pars{\Gamma\pars{t + 1}}\,\dd t} & \ds{=} &
\ds{-\,{3 \over 8}\,\ln\pars{2\pi} + {9 \over 8}\,\ln\pars{A} + {K \over 4\pi}
\approx -0.3365}
\end{array}\right.\tag{6}
\end{equation}

The final result is given by inserting these expressions into result $\ds{\pars{4}}$:
$$
\begin{array}{|c|}\hline\ \\
\ds{\quad\color{#f00}{I} = \int_{1}^{\infty}{1 - x + \ln\pars{x} \over
x\pars{1 + x^{2}}\ln^{2}\pars{x}}\,\dd x =
\color{#f00}{%
{2K \over \pi} - 6\ln\pars{A} + \half\,\ln\pars{\pi} + {1 \over 6}\,\ln\pars{2}}
\approx -0.2215\quad}
\\ \ \\ \hline
\end{array}
$$