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$$I=\int_{0}^{1}{1\over 1+x^2}\ln\left({1+x\over 1-x}\right)=K\tag1$$

$K=0.9159655...$ ;Catalan's constant.

Recall

$$\ln\left({1+x\over 1-x}\right)=2\sum_{n=0}^{\infty}{x^{2n+1}\over 2n+1}\tag2$$

Sub (2) into (1)$\rightarrow $(3)

$$I=2\sum_{n=0}^{\infty}{1\over 2n+1}\int_{0}^{1}{x^{2n+1}\over 1+x^2}dx\tag3$$

Can't go any further!


2nd approach

$$I=\int_{0}^{1}{1\over 1+x^2}\ln\left({1+x\over 1-x}\right)=K\tag1$$

Recall

$${1\over 1+x^2}=\sum_{n=0}^{\infty}(-1)^nx^{2n}\tag2$$

$$I=\sum_{n=0}^{\infty}(-1)^n\int_{0}^{1}x^{2n}\ln\left({1+x\over 1-x}\right)dx\tag3$$

Applying integration by parts to (3)

Let $$J=\int_{0}^{1}x^{2n}\ln\left({1+x\over 1-x}\right)dx\tag4$$

$u=\ln\left({1+x\over 1-x}\right)\rightarrow du={2\over 1-x^2}dx$

$dv=x^{2n}\rightarrow v={x^{2n+1}\over 2n+1}$

$$J=\left.{x^{2n+1}\over 2n+1}\ln{1+x\over 1-x}\right|_{0}^{1}-{2\over 2n+1}\int_{0}^{1}{x^{2n+1}\over 1-x^2}dx\tag5$$ I don't is correct to integrate (4) in this manner, because when evaluating the limits it is not making sense.

Any help how to prove I?

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  • $\begingroup$ Nice approaches, especially the first one. I like your approach of writing functions as infinite series. $\endgroup$ – user230452 Jun 6 '16 at 3:19
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Enforcing the substitution $t=\frac{1-x}{1+x}$, we have $x=\frac{1-t}{1+t}$, $dx=-\frac{2}{(1+t)^2}\,dt$. Therefore,

$$\begin{align} \int_0^1\frac{1}{1+x^2}\log\left(\frac{1+x}{1-x}\right)\,dx&=\int_1^0 \frac{(1+t)^2}{2(1+t^2)}\log(1/t)\left(-\frac{2}{(1+t)^2}\right)\,dt\\\\ &=-\int_0^1 \frac{1}{1+t^2}\log(t)\,dt\\\\ &=-\sum_{n=0}^\infty (-1)^n \int_0^1 t^{2n}\log(t)\,dt\\\\ &=\sum_{n=0}^\infty (-1)^n \frac{1}{(2n+1)^2}\\\\ &=G \end{align}$$

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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With the sub$\ldots$ $\ds{{1 + x \over 1 - x} \equiv t\ \iff\ x = {t - 1 \over t + 1}}$ \begin{align} &\color{#f00}{% \int_{0}^{1}{1 \over 1 + x^{2}}\ln\pars{{1 + x \over 1 - x}}\,\dd x} = \int_{1}^{\infty}{\ln\pars{t} \over 1 + t^{2}}\,\dd t = \color{#f00}{K} \end{align}

because it's a Catalan Constant integral representation.

Moreover, with $\ds{t \to {1 \over t}}$:

\begin{align} &\color{#f00}{% \int_{0}^{1}{1 \over 1 + x^{2}}\ln\pars{{1 + x \over 1 - x}}\,\dd x} = -\int_{0}^{1}{\ln\pars{t} \over 1 + t^{2}}\,\dd t = \Im\int_{0}^{1}{\ln\pars{t} \over \ic - t}\,\dd t = \Im\int_{0}^{1}{\ln\pars{t} \over 1 - t/\ic}\,{\dd t \over \ic} \\[3mm] = &\ \Im\int_{0}^{-\ic}{\ln\pars{t\ic} \over 1 - t}\,\dd t = \Im\int_{0}^{-\ic}{\ln\pars{1 - t } \over t}\,\dd t = -\Im\mathrm{Li}_{2}\pars{-\ic} = -\Im\sum_{n = 1}^{\infty}{\pars{-\ic}^{n} \over n^{2}} \\[3mm] = &\ \sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \pars{2n + 1}^{2}} = \color{#f00}{K} \end{align}

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Another approach. By setting $x=\tanh z$, then $z=\frac{u}{2}$, we get: $$\begin{eqnarray*}I=\int_{0}^{1}\frac{2\text{arctanh}(x)}{1+x^2}\,dx&=&\int_{0}^{+\infty}\frac{2z}{\cosh^2 z+\sinh^2 z}\,dz\\&=&\int_{0}^{+\infty}\frac{u\,du}{e^u+e^{-u}}\\&=&\sum_{n\geq 0}(-1)^n\int_{0}^{+\infty}u e^{-(2n+1)u}\,du\\&=&\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^2}=\color{red}{K}\end{eqnarray*}$$ as wanted.

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