Evaluate $\int_{0}^{\infty}\frac{\log(1+x^2)}{x(1+x^2)}dx $, $\int_{0}^{1}\frac{\log(1+x^2)}{x(1+x^2)}dx $

Question

thanks for opportunity to write on this forum.

I have a problem with logarithmic integral.

1) $$\int_{0}^{+\infty}\frac{\log(1+x^2)}{x(1+x^2)}dx $$

2) $$\int_{0}^{1}\frac{\log(1+x^2)}{x(1+x^2)}dx $$

I try to calculate without a residues theory that it help to reach the goal. I think that it is possible using series sum's and their integration but I have no idea how to start :-(

Can anyone help me? I thank you to all again for precious assistance A.

Answer 1

By enforcing the substitution $\frac{1}{1+x^2}$ and exploiting Euler's Beta function we have $$ \int_{0}^{+\infty}\frac{dx}{x^a (1+x^2)^{b}}=\frac{\Gamma\left(\frac{1-a}{2}\right)\,\Gamma\left(\frac{a-1+2b}{2}\right)}{2\,\Gamma(b)} $$ for any $a,b$ such that $a\in(0,1)$ and $a+2b>1$. By differentiating both sides with respect to $b$, then considering the limit as $a\to 1^-, b\to 1^-$, we get $$ \int_{0}^{+\infty}\frac{\log(1+x^2)}{x(1+x^2)}\,dx = \frac{1}{2}\,\psi'(1)=\color{red}{\frac{\pi^2}{12}}.$$ I leave to you to adapt this technique (Beta function + Feynman's trick) to the second case, proving $$ \int_{0}^{1}\frac{\log(1+x^2)}{x(1+x^2)}\,dx = \frac{1}{2}\,\psi'(1)=\color{red}{\frac{\pi^2}{24}-\frac{\log^2(2)}{4}}.$$ The reflection formula for the dilogarithm is deeply involved.

Answer 2

I only do the first one. The second one is left for you! $$I(a)=\int^\infty_0\frac{\log(1+ax^2)}{x(1+x^2)}\,dx$$ so that $$I'(a)=\int^\infty_0 \frac{x}{(1+x^2)(1+ax^2)}\,dx$$ This integral is very easy. You can do it by partial fraction decomposition. The result is $$I'(a)=\frac{\ln(a)}{2(a-1)}$$ Now we need to integrate this thing. Since $I(0)=0$ we have: \begin{align} I(a)=\int^a_0 \frac{\ln(t)}{2(t-1)}\,dt \end{align} We are interested in $I(1)$: \begin{align} I(1)=-\frac{1}{2}\int^1_0 \frac{\ln(t)}{1-t}\,dt=-\frac{1}{2}\int^1_0 \frac{\ln(t)}{1-t}\,dt \end{align} Substitute $u=1-t$ to get: \begin{align} I(1)&=-\frac{1}{2} \int^1_0 \frac{\ln(1-u)}{u}\,du\\ &= -\frac{1}{2} \int^1_0- \sum_{k=1}^\infty \frac{u^{k-1}}{k}\,du\\ &=\frac{1}{2}\sum_{k=1}^\infty \frac{1}{k^2}\\ &= \frac{1}{2}\frac{\pi^2}{6} \end{align} Surely, one needs to justify interchanging of summation and integral. Finally we arrive at:

\begin{align} \int^\infty_0 \frac{\log(1+x^2)}{x(1+x^2)}\,dx= \frac{\pi^2}{12} \end{align}

Answer 3

$\displaystyle J=\int_{0}^{+\infty}\frac{\log(1+x^2)}{x(1+x^2)}dx$

Perform the change of variable $y=x^2$,

$\begin{align}J&=\frac{1}{2}\int_{0}^{+\infty}\frac{\log(1+x)}{x(1+x)}dx\\ &=\frac{1}{2}\int_{0}^{1}\frac{\log(1+x)}{x(1+x)}dx+\frac{1}{2}\int_{1}^{+\infty}\frac{\log(1+x)}{x(1+x)}dx\\ \end{align}$

In the latter integral perform the change of variable $y=\frac{1}{x}$,

$\begin{align}J&=\frac{1}{2}\int_{0}^1\left(\frac{\ln(1+x)}{x}-\frac{\ln(1+x)}{1+x}\right)\,dx+\frac{1}{2}\int_{0}^1\left(\frac{\ln(1+x)}{1+x}-\frac{\ln x}{1+x}\right)\,dx\\ &=\frac{1}{2}\int_{0}^1 \frac{\ln(1+x)}{x}\,dx-\frac{1}{2}\int_{0}^1 \frac{\ln x}{1+x}\,dx\\ &=\left(\frac{1}{2}\Big[\ln x\ln(1+x)\Big]_0^1-\frac{1}{2}\int_0^1 \frac{\ln x}{1+x}\,dx\right)-\frac{1}{2}\int_0^1 \frac{\ln x}{1+x}\,dx\\ &=-\int_0^1 \frac{\ln x}{1+x}\,dx\\ &=\int_0^1 \left(\frac{2x\ln x}{1-x^2}-\frac{\ln x}{1-x}\right)\,dx\\ &=\int_0^1 \frac{2x\ln x}{1-x^2}\,dx-\int_0^1 \frac{\ln x}{1-x}\,dx \end{align}$

In the latter integral perform the change of variable $y=x^2$,

$\begin{align} J&=\frac{1}{2}\int_0^1 \frac{\ln x}{1-x}\,dx-\int_0^1 \frac{\ln x}{1-x}\,dx\\ &=\boxed{-\frac{1}{2}\int_0^1 \frac{\ln x}{1-x}\,dx} \end{align}$

Using Taylor series expansion,

$\boxed{\displaystyle J=\frac{1}{2}\zeta(2)=\frac{\pi^2}{12}}$

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{\infty}{\ln\pars{1 + x^{2}} \over x\pars{1 + x^{2}}}\,\dd x & \,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\, {1 \over 2}\int_{0}^{\infty}{\ln\pars{1 + x} \over x\pars{1 + x}}\,\dd x \,\,\,\stackrel{1 + x\ \mapsto\ x}{=}\,\,\, {1 \over 2}\int_{1}^{\infty}{\ln\pars{x} \over \pars{x - 1}x}\,\dd x \\[5mm] & \stackrel{x\ \mapsto\ 1/x}{=}\,\,\, {1 \over 2}\int_{1}^{0}{\ln\pars{1/x} \over \pars{1/x - 1}/x}\,\pars{-\,{\dd x \over x^{2}}} = -\,{1 \over 2}\int_{0}^{1}{\ln\pars{x} \over 1 - x}\,\dd x \\[5mm] & \stackrel{x\ \mapsto\ 1 - x}{=}\,\,\, -\,{1 \over 2}\int_{0}^{1}{\ln\pars{1 - x} \over x}\,\dd x = {1 \over 2}\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\,\dd x \\[5mm] & = {1 \over 2}\,\mrm{Li}_{2}\pars{1} = {1 \over 2}\,\zeta\pars{2} = {1 \over 2}\,{\pi^{2} \over 6\phantom{^{2}}} \\[5mm] & \implies \bbx{\int_{0}^{\infty}{\ln\pars{1 + x^{2}} \over x\pars{1 + x^{2}}}\,\dd x = {\pi^{2} \over 12}} \approx 0.8225 \end{align}

$\ds{-----------------------------------------}$

\begin{align} \int_{0}^{1}{\ln\pars{1 + x^{2}} \over x\pars{1 + x^{2}}}\,\dd x & \,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\, {1 \over 2}\int_{0}^{1}{\ln\pars{1 + x} \over x\pars{1 + x}}\,\dd x \\[5mm] & = {1 \over 2}\int_{0}^{-1}{\ln\pars{1 - x} \over x}\,\dd x - {1 \over 2}\int_{0}^{1}{\ln\pars{1 + x} \over 1 + x}\,\dd x \\[5mm] & = \left.-\,{1 \over 2}\,\mrm{Li}_{2}'\pars{x}\right\vert_{\ 0}^{\ -1} - \left.{1 \over 4}\,\ln^{2}\pars{1 + x}\right\vert_{\ 0}^{\ 1} = -\,{1 \over 2}\,\ \overbrace{\mrm{Li}_{2}\pars{-1}}^{\ds{-\,{\pi^{2} \over 12}}} - {1 \over 4}\,\ln^{2}\pars{2} \\[5mm] & \implies \bbx{\int_{0}^{1}{\ln\pars{1 + x^{2}} \over x\pars{1 + x^{2}}}\,\dd x = {\pi^{2} \over 24} - {\ln^{2}\pars{2} \over 4}} \approx 0.2911 \end{align}

Note that $\ds{\mrm{Li}\pars{-1} = \sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^{2}} = \sum_{n = 1}^{\infty}\bracks{{1 \over \pars{2n}^{2}} - {1 \over \pars{2n - 1}^{2}}} = \sum_{n = 1}^{\infty}\bracks{-\,{1 \over n^{2}} + 2\,{1 \over \pars{2n}^{2}}}}$

$\ds{= -\,{1 \over 2}\sum_{n = 1}^{\infty}{1 \over n^{2}} = \color{#f00}{-\,{\pi^{2} \over 12}}}$

Answer 5

To reduce the power of $x$, we put $x^2\mapsto x $ and get

$$ \begin{aligned} 2\int_0^1 \frac{\ln \left(x^2+1\right)}{x\left(1+x^2\right)} d x=& \int_0^1 \frac{\ln (x+1)}{x(1+x)} d x \\= & \int_0^1 \frac{(1+x-x) \ln (x+1)}{x(1+x)} d x \\ = & \int_0^1 \frac{\ln (x+1)}{x} d x-\int_0^1 \frac{\ln (x+1)}{x+1} d x \\ = & \int_0^1 \ln (x+1) d(\ln x)-\left[\frac{\ln ^2(x+1)}{x+1}\right]_0^1 \\ = & -\int_0^1 \frac{\ln x}{1+x} d x-\frac{\ln ^2 2}{2} \\ = & -\sum_{n=0}^{\infty} \frac{(-1)^n}{2} \int_0^1 x^n \ln x d x -\frac{\ln ^2 2}{2} \\ = & \sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)^2} -\frac{\ln ^2 2}{2}\\ = & \frac{\pi^2}{12}-\frac{\ln ^2 2}{2} \end{aligned} $$ Now we can conclude that $$\int_0^1 \frac{\ln \left(x^2+1\right)}{x\left(1+x^2\right)} d x = \frac{\pi^2}{24}-\frac{\ln ^2 2}{4} $$

Answer 6

1. Use $\int_0^1\frac {\ln t}{t-1}dt=\frac{\pi^2}{6}$

$$\int_0^\infty \frac{\ln(1+x^2)}{x(1+x^2)}dx \overset{x^2=\frac{1-t}t}=\frac12\int_0^1\frac {\ln t}{t-1}dt = \frac12\cdot\frac{\pi^2}6 =\frac{\pi^2}{12} $$

2. Use $\int_0^1\frac {\ln (1+t)}{t}dt =\frac{\pi^2}{12}$

\begin{align} \int_0^1 \frac{\ln(1+x^2)}{x(1+x^2)}dx & \overset{x^2=t} =\frac12\int_0^1 \frac {\ln (1+t)}{t}dt - \frac12\int_0^1 \frac {\ln (1+t)}{1+t}dt \\ & =\frac12 \cdot \frac{\pi^2}{12}- \frac12\cdot \frac12\ln^22 = \frac{\pi^2}{24}-\frac14\ln^22 \end{align}

Answer 7

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\Large\left. 1\right)}$ \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}{\ln\pars{1 + x^{2}} \over x\pars{1 + x^{2}}}\dd x} \,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\, {1 \over 2}\lim_{\epsilon\ \to\ 0^{+}} \int_{0}^{\infty}x^{\color{red}{\epsilon} - 1}\, {\ln\pars{1 + x} \over 1 + x}\dd x \end{align} Note that $\ds{{\ln\pars{1 + x} \over 1 + x} = -\sum_{k = 0}^{\infty}H_{k}\,\pars{-x}^{k} = \sum_{k = 0}^{\infty}\bracks{\color{red}{-H_{k}\,\Gamma\pars{k + 1}}}\,{\pars{-x}^{k} \over k!}}$. With Ramanujan-MT, I'll have \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}{\ln\pars{1 + x^{2}} \over x\pars{1 + x^{2}}}\dd x} = \lim_{\epsilon\ \to\ 0^{+}}\braces{{1 \over 2}\,\Gamma\pars{\epsilon} \bracks{-H_{-\epsilon}\,\Gamma\pars{-\epsilon + 1}}} \\[5mm] = &\ -\,{1 \over 2}\,\pi\lim_{\epsilon\ \to\ 0^{+}} {\Psi\pars{-\epsilon + 1} + \gamma \over \sin\pars{\pi\epsilon}} = -\,{1 \over 2}\,\pi\lim_{\epsilon\ \to\ 0^{+}} {\Psi\, '\pars{-\epsilon + 1}\pars{-1} \over \cos\pars{\pi\epsilon}\pi} \\[5mm] = &\ {1 \over 2}\ \underbrace{\Psi\, '\pars{1}}_{\ds{\pi^{2} \over 6}} = \bbx{\pi^{2} \over 12} \\ & \end{align}