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.

Shashi · Answer 1 · 2018-01-20T00:23:56.207

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}

edited Jan 20 '18 at 00:23

answered Jan 19 '18 at 23:15

Shashi

8,738

thank you very very much for your help.I have only dubt: from I(a) to I'(a), how do you jump between? What do you mean into the notation? – Alex Jan 20 '18 at 13:51
@Alex you are welcome. $I(a) $ is a function of $a$ and has an integral representation. $I'(a) $ is the derivative and the closed form is obtained via differentiating under the integral sign a.k.a. Feynman's trick. Knowing the derivative can help to determine the original function through integrating and that is what is done in this answer. – Shashi Jan 20 '18 at 14:03
Thanks again. My dubt grows up because I don't know the Feynman's trick that you used to start the exercise so I am very happy that I can learn a new concept :-) – Alex Jan 20 '18 at 14:12
@Alex I wonder how you came up with these integrals. You are also trying to avoid the residue theorem which is very useful in this case. Any reason for that? – Shashi Jan 20 '18 at 14:15
The reason is because I explain this exercise to my friends and he doesn't treat the residue theorem yet so....I can reach the goal without this way :-).Thanks again! – Alex Jan 20 '18 at 22:35

score 0 · Answer 2 · answered Jan 19 '18 at 23:15

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.

thanks to you. Your solution is fastest but it is not recommended to use the Euler's Beta functions. — Alex, Jan 20 '18 at 14:07

FDP · Answer 3 · 2018-01-20T09:52:15.267

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

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,

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

thanks a lot for your assistance. If you perform the variable switch where is the term for dy=2xdx? I don't know if I wrong my point of view. The value is OK,I verified with numerical calculation.Last question is the $\int_{0}^{1}\frac{\log(1+x^2)}{x(1+x^2)}dx = 0 $? Thanks again! — Alex, Jan 20 '18 at 14:03
@Alex: $\dfrac{1}{x}=\dfrac{1}{2}\times \dfrac{2x}{x^2}\$. $\text{}\ \displaystyle \int_{0}^{1}\frac{\log(1+x^2)}{x(1+x^2)}dx$ is not $0\$. $\displaystyle \frac{\log(1+x^2)}{x(1+x^2)}=\frac{1}{2}\times \frac{2x\ln(1+x^2)}{x^2}-\frac{1}{2}\times \frac{2x\ln(1+x^2)}{1+x^2}$ — FDP, Jan 20 '18 at 17:26
Thanks again. I verified and I put wrong parenthesis into the expression's tool when i checked the solution so the result is not correct. Now I try to do the exercise with residues theorem. — Alex, Jan 21 '18 at 20:31

Felix Marin · Answer 4 · 2018-01-25T23:16:27.890

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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 $\ds{\mrm{Li}\pars{-1} = \sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^{2}} = \sum_{n = 1}^{\infty}\bracks{{1 \over \pars{2n}^{2}} - {1 \over \pars{2n - 1}^{2}}} = \sum_{n = 1}^{\infty}\bracks{-\,{1 \over n^{2}} + 2\,{1 \over \pars{2n}^{2}}}}$

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

Thank you very very much for your precious and useful details! — Alex, Jan 27 '18 at 14:38
I try to analyze the series proof expression on your "note that" post (the 3th & the 4th before the well know final relation) . What is the tip/trick that you use to reach the final step? Thank you very much for help. — Alex, Oct 05 '23 at 16:19

Quanto · Answer 5 · 2020-09-16T16:49:02.633

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

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

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}

score 0 · Answer 6 · answered Sep 15 '20 at 04:30

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\Large\left. 1\right)}$ \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}

Lai · Answer 7 · 2023-10-04T15:31:06.937

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 $$\int_0^1 \frac{\ln \left(x^2+1\right)}{x\left(1+x^2\right)} d x = \frac{\pi^2}{24}-\frac{\ln ^2 2}{4} $$

Thank you very much. I have only dubt to your variable-switch from $x$ to $1/x$. Maybe I lose some steps....Thank you again — Alex, Oct 04 '23 at 10:34
The substitution should be $x^2\mapsto x$. Sorry for my typo. Fixed. — Lai, Oct 04 '23 at 15:32

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 $

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