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 $x>1$,

$\displaystyle \int_0^1 \dfrac{1}{t} \arctan \left (\dfrac{t(1-t)}{\frac{x+1}{2}-t}\right) dt=\int_1^{\frac{\sqrt{x}+1}{\sqrt{x}-1}}\dfrac{\log(t)}{1+t^2}dt$

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: $\int_0^{\pi/12} \ln(\tan x)\,dx$ )

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