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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 $x>1$,

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

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    $\begingroup$ By using integration by parts they can be restated as integrals over $(0,1)$ of $\log(x)$ times a rational function. By using the residue theorem together with the fact that $\int_{0}^{1}\frac{\log(x)}{x-a}\,dx = \text{Li}_2\left(\frac{1}{a}\right)$ Catalan's constant should arise easily. $\endgroup$ – Jack D'Aurizio Jan 11 '16 at 18:38
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    $\begingroup$ Please give an outline of the proof that you have, so we don't waste our time developing and typing it out. $\endgroup$ – Rory Daulton Jan 11 '16 at 18:57
  • $\begingroup$ @Rory: my proofs are basic ones, no use of contour integration, change of variable and integration by parts are used only $\endgroup$ – FDP Jan 11 '16 at 19:12
  • $\begingroup$ Ask Wikipedia . $\endgroup$ – user65203 Jan 11 '16 at 20:17
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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: $\int_0^{\pi/12} \ln(\tan x)\,dx$ )

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There is a large multitude of different representations of the Catalan constant. See the following links for some of them:

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FDP,

This integral was detailed in a paper you wrote some years ago.

Anyway it’s an endless pleasure to see how you manage such complicated calculations.

I suggest we name this generalized integral « FDP integral »

With my reiterated well deserved warm congratulations.

fjaclot

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  • $\begingroup$ I still dont know where this integral comes from. I know only this integral was found out by a mathematician on an internet site that is not online anymore. $\endgroup$ – FDP Apr 3 at 17:46
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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: $\int_0^{\pi/12} \ln(\tan x)\,dx$

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