How to find ${\large\int}_1^\infty\frac{1-x+\ln x}{x \left(1+x^2\right) \ln^2 x} \mathrm dx$?

Question

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.

Answer 1

$$I=\frac{\ln2}6+\frac{\ln\pi}2-6\ln A+\frac2\pi G,$$ where $A$ is the Glaisher–Kinkelin constant and $G$ is the Catalan constant.

Answer 2

Here is another Feynman's way to evaluate the integral. Set $x=\frac1t$, then $$ \int_1^\infty\frac{1-x+\ln x}{x \left(1+x^2\right) \ln^2 x}\ dx=\int_0^1\frac{t-1-t\ln t}{(1+t^2)\ln^2t}\ dt. $$ Now consider $$ \mathcal{I}(\alpha)=\int_0^1t^\alpha\cdot\frac{t-1-t\ln t}{(1+t^2)\ln^2t}\ dt. $$ Hence \begin{align} \frac{d^2\mathcal{I}}{d\alpha^2}&=\int_0^1\frac{t^\alpha}{1+t^2}(t-1-t\ln t)\ dt\\ &=\int_0^1\sum_{n=0}^\infty(-1)^nt^{2n+\alpha}(t-1-t\ln t)\ dt\\ &=\sum_{n=0}^\infty(-1)^n\int_0^1\left(t^{2n+\alpha+1}-t^{2n+\alpha}-t^{2n+\alpha+1}\ln t\right)\ dt\tag1\\ &=\sum_{n=0}^\infty(-1)^n\left[\frac{1}{2n+\alpha+2}-\frac{1}{2n+\alpha+1}+\frac{1}{(2n+\alpha+2)^2}\right]\tag2\\ &=-\frac14\left[\psi\left(\frac{\alpha+2}{4}\right)-\psi\left(\frac{\alpha+4}{4}\right)\right]+\frac14\left[\psi\left(\frac{\alpha+1}{4}\right)-\psi\left(\frac{\alpha+3}{4}\right)\right]\\ &\quad+\frac1{16}\left[\psi_1\left(\frac{\alpha+2}{4}\right)-\psi_1\left(\frac{\alpha+4}{4}\right)\right]\tag3\\ \frac{d\mathcal{I}}{d\alpha}&=-\ln\Gamma\left(\frac{\alpha+2}{4}\right)+\ln\Gamma\left(\frac{\alpha+4}{4}\right)+\ln\Gamma\left(\frac{\alpha+1}{4}\right)-\ln\Gamma\left(\frac{\alpha+3}{4}\right)\\ &\quad+\frac14\left[\psi\left(\frac{\alpha+2}{4}\right)-\psi\left(\frac{\alpha+4}{4}\right)\right]\tag4\\ &=\ln\Gamma\left[\frac{(\alpha+1)(\alpha+4)}{(\alpha+2)(\alpha+3)}\right]+\frac14\left[\psi\left(\frac{\alpha+2}{4}\right)-\psi\left(\frac{\alpha+4}{4}\right)\right]\tag5\\ \mathcal{I}(\alpha)&=\int\ln\Gamma\left[\frac{(\alpha+1)(\alpha+4)}{(\alpha+2)(\alpha+3)}\right]\ d\alpha+\ln\Gamma\left(\frac{\alpha+2}{4}\right)-\ln\Gamma\left(\frac{\alpha+4}{4}\right). \end{align} Since $0<t<1$, then as $\alpha\to\infty$, implying $\mathcal{I}(\alpha)\to0$ and $\mathcal{I}'(\alpha)\to0$. Thus

$$ \mathcal{I}(0)=\int_1^\infty\frac{1-x+\ln x}{x \left(1+x^2\right) \ln^2 x}\ dx=\color{blue}{\frac{\ln2}6+\frac{\ln\pi}2-6\ln A+\frac2\pi G}, $$

where $A$ is the Glaisher–Kinkelin constant and $G$ is the Catalan constant.

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Notes :

$\displaystyle(1)\ \ \int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}, \qquad\text{for }\ n=0,1,2,\ldots$

$\displaystyle(2)\ \ \sum_{k=0}^\infty\frac{(-1)^{k}}{(z+k)^{m+1}}=\frac1{(-2)^{m+1}m!}\left[\psi_{m}\left(\frac{z}{2}\right)-\psi_{m}\left(\frac{z+1}{2}\right)\right]$

$\displaystyle(3)\ \ \psi_{m}(z) := \frac{d^m}{dz^m} \psi(z) = \frac{d^{m+1}}{dz^{m+1}} \ln\Gamma(z)$

$\displaystyle(4)\ \ \int\ln\Gamma(z)\ dz=\frac{z(1-z)}{2}+\frac{z}{2}\log 2\pi +z\log\Gamma(z) -\log \text{G}(1+z)+C$,

$\qquad$where $\text{G}(\cdot)$ is the Barnes G-function and $C$ is a constant of integration. Also for $a,b>0$ $$ \int\ln\Gamma\left(\frac{x+a}{b}\right)\ dx=b\ \psi^{(-2)}\left(\frac{x+a}{b}\right)+C. $$

$\displaystyle(5)\ \ $As $\alpha\to\infty$, both of sides tend to $0$, hence the constant of integration is $0$.

Answer 3

Let's give a (sketch of the) proof of the following closed form evaluation.

$$ I=\int_1^\infty\frac{1-x+\ln x}{x \left(1+x^2\right) \ln^2 x} \mathrm dx=\frac{\ln2}6+\frac{\ln\pi}2-6\ln A+\frac2\pi G, $$

where $A$ is the Glaisher–Kinkelin constant and $G$ is the Catalan constant.

Observe that, by the change of variable $\displaystyle u=\frac 1x$, we have $$ I=\int_0^1\frac{u-1-u\ln u}{\left(1+u^2\right) \ln^2 u} \mathrm du=\int_0^1\frac{(1-u^2)(u-1-u\ln u)}{\left(1-u^4\right) \ln^2 u} \mathrm du. $$ We set $$ I(s):=\int_0^1u^s\frac{(1-u^2)(u-1-u\ln u)}{\left(1-u^4\right) \ln^2 u} \mathrm du, \quad s\geq0. $$

We are allowed to differentiate $I(s)$ twice to obtain $$ I''(s)=\int_0^1u^s\frac{(1-u^2)(u-1-u\ln u)}{\left(1-u^4\right)} \mathrm du. $$ Using the standard expansion $\displaystyle \frac{1}{1-u^4}=\sum_{k=0}^{\infty}u^{4k}, \, |u|<1,$ and performing the termwise integration, $$ \begin{align} I''(s)=\sum_{k=0}^{\infty}\int_0^1u^{s+4k}(1-u^2)(u-1-u\ln u) \mathrm du \end{align} $$ gives $$ \begin{align} I''(s)=\sum_{k=0}^{\infty}\left(\frac{1}{(4k+2+s)^2}-\frac{1}{(4k+4+s)^2}+\frac{1}{(4k+2+s)}-\frac{1}{(4k+1+s)}\right). \end{align} $$ Now, recall the following series representation of the digamma function $\displaystyle \psi : = \Gamma'/\Gamma$, $$ \psi(u+1) = -\gamma + \sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{u+k} \right), \quad u >-1, $$ where $\gamma$ is the Euler-Mascheroni constant and, by differentiation, giving$$ \psi'(u+1) = \sum_{k=1}^{\infty} \frac{1}{(u+k)^2}, \quad u>-1. $$ Hence $$ I''(s)=\frac{1}{16}\psi'\left(\frac{s+2}{4}\right)-\frac{1}{16}\psi'\left(\frac{s+4}{4}\right)+\frac{1}{4}\psi\left(\frac{s+4}{4}\right)+\frac{1}{4}\psi\left(\frac{s+1}{4}\right)-\frac{1}{4}\psi\left(\frac{s+3}{4}\right)-\frac{1}{4}\psi\left(\frac{s+2}{4}\right). $$ We have, as $s \rightarrow +\infty$, $I(s) \rightarrow 0$ and $I'(s) \rightarrow 0$, leading to $$ I(s)=\log \Gamma \left(\frac{s+2}{4}\right)-\log \Gamma \left(\frac{s+4}{4}\right) + 4 \left(\psi\left(-2, \frac{s+4}{4}\right) + \psi\left(-2, \frac{s+1}{4}\right) - \psi\left(-2, \frac{s+2}{4}\right) - \psi\left(-2, \frac{s+3}{4}\right)\right), $$ then, with the use of Wolfram Alpha, as $s \rightarrow 0$ we get

$$ I=I(0)=\frac{\ln2}6+\frac{\ln\pi}2-6\ln A+\frac2\pi G. $$

Answer 4

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

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.

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\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 $\ds{x^{t} = {1 \over \ln\pars{x}}\,\partiald{x^{t}}{t}}$

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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}

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