Formulae for Catalan's constant. Some years ago, someone had shown me the formula (1).
I have searched for its origin and for a proof.
I wasn't able to get true origin of this formula but I was able to find out an elementary proof for it.
Since then, I'm interested in different approaches to find more formulae as (1).
What other formulas similar to ($1$) are known?
Two days ago, reading the book of Lewin "Polylogarithms and Associated Functions" I was able to find out formula (2).
$\displaystyle \dfrac{1}{3}C=\int_0^1 \dfrac{1}{x}\arctan\left(\dfrac{x(1-x)}{2-x}\right)dx\tag1$ 
$\displaystyle \dfrac{2}{5}C=\int_0^1 \dfrac{1}{x}\arctan\left(\dfrac{\sqrt{5}x(1-x)}{1+\sqrt{5}-\sqrt{5}x}\right)dx-\int_0^1 \dfrac{1}{x}\arctan\left(\dfrac{x(1-x)}{3+\sqrt{5}-x}\right)dx\tag2$ 
$C$ being the Catalan's constant.
I have a proof for both of these formulae.
My approach relies on the following identity:
For all real $x>1$,
$\displaystyle \int_0^1 \dfrac{1}{t} \arctan \left (\dfrac{t(1-t)}{\frac{x+1}{2}-t}\right) dt=\int_1^{\frac{\sqrt{x}+1}{\sqrt{x}-1}}\dfrac{\log(t)}{1+t^2}dt$
 A: For all $x\in [0,1]$ and $\alpha>1$,
$\displaystyle \arctan\left(\dfrac{x(1-x)}{\tfrac{1+\alpha^2}{(1-\alpha)^2}-x}\right)=\arctan\left(\dfrac{x}{\tfrac{1+\alpha^2}{\alpha(\alpha-1)}+\tfrac{1}{\alpha}x}\right)+\arctan\left(\dfrac{x}{\tfrac{1+\alpha^2}{1-\alpha}+\alpha x}\right)$
For all $\alpha>1$,
$\displaystyle J(\alpha)=\int_0^1\dfrac{1}{x}\arctan\left(\dfrac{x(1-x)}{\tfrac{1+\alpha^2}{(1+\alpha)^2}-x}\right)dx=\int_0^{\tfrac{\alpha-1}{\alpha+1}} \dfrac{\arctan x}{x\left(1-\tfrac{1}{\alpha}x\right)}dx-\int_0^{\tfrac{\alpha-1}{\alpha+1}} \dfrac{\arctan x}{x(1+\alpha x)}dx$
For $x \in ]0,1]$,
$\dfrac{1}{x\left(1-\tfrac{1}{\alpha}x\right)}-\dfrac{1}{x\left(1+\alpha x\right)}=\dfrac{1}{\alpha-x}+\dfrac{\alpha}{1+\alpha x}$
Thus, one obtains,
$\displaystyle J(\alpha)=\int_0^{\tfrac{\alpha-1}{\alpha+1}}\dfrac{\arctan x}{\alpha-x}dx+\int_0^{\tfrac{\alpha-1}{\alpha+1}}\dfrac{\alpha \arctan x}{1+\alpha x}dx$
$\displaystyle J(\alpha)=\Big[-\log(\alpha-x)\arctan x\Big]_0^{\tfrac{\alpha-1}{\alpha+1}}+\int_0^{\tfrac{\alpha-1}{\alpha+1}}\dfrac{\log(\alpha-x)}{1+x^2}dx+\Big[\log(1+\alpha x)\arctan x\Big]_0^{\tfrac{\alpha-1}{\alpha+1}}-\int_0^{\tfrac{\alpha-1}{\alpha+1}}\dfrac{\log(1+\alpha x)}{1+x^2}dx$
$\displaystyle J(\alpha)=\int_0^{\tfrac{\alpha-1}{\alpha+1}}\dfrac{\log\left(\tfrac{\alpha-x}{1+\alpha x}\right)}{1+x^2}dx$
Using change of variable $y=\dfrac{\alpha-x}{1+\alpha x}$ ,
$\displaystyle J(\alpha)=\int_1^{\alpha} \dfrac{\log x}{1+x^2}dx$
If $\alpha=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}$, one obtains,
For all $x>1$,
$\displaystyle \int_0^1 \dfrac{1}{t} \arctan \left (\dfrac{t(1-t)}{\tfrac{x+1}{2}-t}\right) dt=\int_1^{\tfrac{\sqrt{x}+1}{\sqrt{x}-1}}\dfrac{\log(t)}{1+t^2}dt$
when $x=3$, one obtains,
$\displaystyle \int_0^1 \dfrac{1}{t} \arctan \left (\dfrac{t(1-t)}{2-t}\right) dt=\int_1^{\tfrac{\sqrt{3}+1}{\sqrt{3}-1}}\dfrac{\log(t)}{1+t^2}dt=\int_1^{2+\sqrt{3}}\dfrac{\log(t)}{1+t^2}dt$
It's well known that:
$\displaystyle \int_1^{2+\sqrt{3}}\dfrac{\log(t)}{1+t^2}dt=\int_1^{2-\sqrt{3}}\dfrac{\log(t)}{1+t^2}dt=\dfrac{C}{3}$
(recall that, $\tan\left(\dfrac{\pi}{12}\right)=2-\sqrt{3}$ and see Integral: $\int_0^{\pi/12} \ln(\tan x)\,dx$ )
A: There is a large multitude of different representations of the Catalan constant. See the following links for some of them:


*

*http://www.cs.cmu.edu/~adamchik/articles/catalan/catalan.htm

*http://functions.wolfram.com/Constants/Catalan/07/ShowAll.html

*http://functions.wolfram.com/Constants/Catalan/06/ShowAll.html

*http://en.wikipedia.org/wiki/Catalan%27s_constant

*http://mathworld.wolfram.com/CatalansConstant.html

*Representations of Catalan's Constant

*Ten New Representations Of Catalan's Constant
A: A more natural proof.
\begin{align}
\beta&=\sqrt{3}-1\\    
J&=\int_0^1 \frac{\arctan\left(\frac{x(1-x)}{2-x}\right)}{x}dx\\
&\overset{\text{IBP}}=\left[\arctan\left(\frac{x(1-x)}{2-x}\right)\ln x\right]_0^1-\int_0^1 \frac{(x^2-4x+2)\ln x}{(x^2+\beta x+2)(x^2-(\beta+2)x+2)}dx\\
&=-\int_0^1 \frac{(x^2-4x+2)\ln x}{(x^2+\beta x+2)(x^2-(\beta+2)x+2)}dx\\
&=\int_0^1 \frac{\beta\ln x}{2(x^2-(2+\beta) x+2)}-\int_0^1 \frac{(2+\beta)\ln x}{2(x^2+\beta x+2)}\\
&=\underbrace{\int_0^1 \frac{2\ln x}{\beta\left(\left(\frac{2x-2-\beta}{\beta}\right)^2+1\right)}dx}_{y=\frac{\beta}{2+\beta-2x}}-\underbrace{\int_0^1 \frac{2\ln x}{(2+\beta)\left(\left(\frac{2x+\beta}{2+\beta}\right)^2+1\right)}dx}_{y=\frac{2x+\beta}{2+\beta}}\\
&=-\int_{\frac{\beta}{\beta+2}}^1 \frac{\ln y}{1+y^2}dy\\
&\overset{y=\tan \theta}=-\int_{\frac{\pi}{12}}^{\frac{\pi}{4}}\ln\left(\tan \theta\right)d\theta\\
&=-\int_{0}^{\frac{\pi}{4}}\ln\left(\tan \theta\right)d\theta+\int_0^{\frac{\pi}{12}}\ln\left(\tan \theta\right)d\theta\\
&=\text{G}-\frac{2}{3}\text{G}\\
&=\boxed{\dfrac{1}{3}\text{G}}
\end{align}
NB: For the latter integral see Integral: $\int_0^{\pi/12} \ln(\tan x)\,dx$
