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Some years ago, someone had shown me the formula (\ref{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 (\ref{1}).

Two days ago, reading the book of Lewin "Polylogarithms and Associated Functions" I was able to find out formula (\ref{2}).

\begin{align} \frac{1}{3}C & =\int_0^1 \frac{1}{x}\arctan\left(\frac{x(1-x)}{2-x}\right)dx\tag{1}\label{1} \\[5mm] \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\nonumber \\[2mm] & -\int_0^1 \dfrac{1}{x}\arctan\left(\dfrac{x(1-x)}{3+\sqrt{5}-x}\right)dx\tag{2}\label{2} \end{align}

$C$ being the Catalan's constant.

I have a proof for both of these formulae.

My approach relies on the following identity:

For all real $x>1$,

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

The question is: What other formulas similar to (\ref{1}) and (\ref{2}) are known?

I mean formulae like this: $\displaystyle r\text{C}=\sum_{k=1}^N\int_0^1\frac{\arctan\left(R_k(x)\right)}{x}dx$ where $C$ is the Catalan constant, $r$ a rational number, $R_k$ are rational functions on algebraic numbers

(Sorry i don't know why the formulae 1+2 are not displaying fine)

FDP
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    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. – Jack D'Aurizio Jan 11 '16 at 18:38
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    Please give an outline of the proof that you have, so we don't waste our time developing and typing it out. – Rory Daulton Jan 11 '16 at 18:57
  • @Rory: my proofs are basic ones, no use of contour integration, change of variable and integration by parts are used only – FDP Jan 11 '16 at 19:12
  • Ask Wikipedia . –  Jan 11 '16 at 20:17

5 Answers5

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

Community
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FDP
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There is indeed a family of such formulas, i.e.

$$\frac{nC}{2n+1}=\int_0^1 \frac{dx}{x}\sum_{k=0}^{n-1}(-1)^{k}\tan^{-1}\frac{x(1-x)}{\frac12\csc^2 \theta_k-x}, \>\>\>\theta_k=\frac{\pi(n-k)}{2(2n+1)}$$ which produces

\begin{align} \frac{1}{3}C & =\int_0^1 \tan^{-1}\frac{x(1-x)}{2-x}\ \frac{dx}x\\ \dfrac{2}{5}C & = \int_0^1 \bigg(\tan^{-1}\dfrac{x(1-x)}{\frac1{\sqrt{5}}+1-x}- \tan^{-1}\dfrac{x(1-x)}{3+\sqrt{5}-x}\bigg) \frac{dx}x\\ \frac{3}{7}C & =\int_0^1 \bigg(\tan^{-1}\frac{x(1-x)}{\frac12\csc^2 \frac{3\pi}{14}-x}-\tan^{-1}\frac{x(1-x)}{\frac12\csc^2 \frac{\pi}{7}-x}\\ &\hspace{15mm}+\tan^{-1}\frac{x(1-x)}{\frac12\csc^2 \frac{\pi}{14}-x}\bigg) \frac{dx}x\\ \frac{4}{9}C & =\int_0^1 \bigg(\tan^{-1}\frac{x(1-x)}{\frac12\csc^2 \frac{2\pi}{9}-x}-\tan^{-1}\frac{x(1-x)}{\frac12\csc^2 \frac{\pi}{6}-x}\\ &\hspace{15mm}+\tan^{-1}\frac{x(1-x)}{\frac12\csc^2 \frac{\pi}{9}-x}-\tan^{-1}\frac{x(1-x)}{\frac12\csc^2 \frac{\pi}{18}-x}\bigg) \frac{dx}x\\ \frac{5}{11}C & =\cdots \> \cdots \end{align}

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

Vladimir Reshetnikov
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  • a comment isn't an answer. See also: http://mathworld.wolfram.com/InverseTangentIntegral.html – FDP Jan 14 '16 at 19:23
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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: $\int_0^{\pi/12} \ln(\tan x)\,dx$

FDP
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Note $$\begin{eqnarray} I&=&\int_0^1 \dfrac{1}{x}\arctan\left(\dfrac{\sqrt{5}x(1-x)}{1+\sqrt{5}-\sqrt{5}x}\right)dx -\int_0^1 \dfrac{1}{x}\arctan\left(\dfrac{x(1-x)}{3+\sqrt{5}-x}\right)dx\\ &=&\int_0^1\dfrac{1}{x}\arctan\left(\dfrac{x(1-x)}{\frac{1+\sqrt{5}}{\sqrt{5}}-x}\right)dx -\int_0^1 \dfrac{1}{x}\arctan\left(\dfrac{x(1-x)}{3+\sqrt{5}-x}\right)dx. \end{eqnarray}$$ Note from @FDP's answer (I think there is a typo in his answer), $$ J(\alpha)=\int_0^1\dfrac{1}{x}\arctan\left(\dfrac{x(1-x)}{\tfrac{1+\alpha^2}{(1-\alpha)^2}-x}\right)dx=\int_1^{\alpha} \dfrac{\log x}{1+x^2}dx.$$ From $$ \tfrac{1+\alpha^2}{(1-\alpha)^2}=\frac{1+\sqrt{5}}{\sqrt{5}},\tfrac{1+\alpha^2}{(1-\alpha)^2}=3+\sqrt{5} $$ one has $$ \alpha_1=1+\sqrt{5}-\sqrt{5+2 \sqrt{5}}=\tan\bigg(\frac{\pi}{20}\bigg),\alpha_2=\left(\sqrt{5}-2\right) \left(3+\sqrt{5}-\sqrt{5+2 \sqrt{5}}\right)=\tan\bigg(\frac{3\pi}{20}\bigg). $$ So $$ I=J(\alpha_1)-J(\alpha_2)=-\int_{\alpha_1}^{\alpha_2}\dfrac{\log x}{1+x^2}dx=\frac25C $$ from https://math.stackexchange.com/a/3887068/686284.

xpaul
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