Integral of the day XX

Nice integral representation of the Apéry's constant $\zeta(3)$

Today we show the proof of the following integral representation of the Aperys constant posted by @sounansya_29 \[ \frac{32}{7}\int_{0}^{\frac{\pi}{4} } x\operatorname{arctanh}(\tan x) dx = \zeta(3)\] The proof relies on some properties of the $\operatorname{arctanh}$ and $arctan$ functions and Fourier series.

Proof

Recall the definition in terms of logarithms of $\operatorname{arctanh}(z)$ \[\operatorname{arctanh}(z) = \frac{1}{2}\ln\left(\frac{1+z}{1-z}\right)\] Therefore \begin{align*} I = \int_{0}^{\frac{\pi}{4} } x\operatorname{arctanh}(\tan x) dx =&\frac{1}{2}\int_{0}^{\frac{\pi}{4} } x\ln\left(\frac{1+\tan x}{1-\tan x}\right) dx \\ =& \frac{1}{2}\int_{1}^{\infty} \frac{\arctan\left(\frac{w-1}{w+1}\right)\ln(w)}{1+w^2} dw \quad \left( w \mapsto \frac{1+\tan x}{1-\tan x}\right) \end{align*} Recall the following addition formula for $\arctan(x)$: \[ \arctan(w) - \frac{\pi}{4} = \arctan\left(\frac{w-1}{w+1}\right)\] Hence \begin{align*} I = \int_{0}^{\frac{\pi}{4} } x\operatorname{arctanh}(\tan x) dx =& \frac{1}{2}\int_{1}^{\infty} \frac{\arctan\left(\frac{w-1}{w+1}\right)\ln(w)}{1+w^2} dw\\ =& \frac{1}{2}\int_{1}^{\infty} \frac{\left(\arctan\left(w\right) - \frac{\pi}{4}\right)\ln(w)}{1+w^2} dw\\ =& \frac{1}{2}\underbrace{\int_{1}^{\infty} \frac{\arctan(w)\ln(w)}{1+w^2}dw}_{J} - \frac{\pi}{8}\underbrace{\int_{1}^{\infty} \frac{\ln(w)}{1+w^2} dw}_{K}\\ \end{align*} \begin{align*} K = \int_{1}^{\infty} \frac{\ln(w)}{1+w^2} dw = & -\int_{0}^{1} \frac{\ln(s)}{1+s^2} ds \quad \left(s \mapsto \frac{1}{w} \right)\\ =& -\int_{0}^{1} \sum_{n=0}^{\infty} (-1)^ns^{2n} \ln(s) ds\\ =& -\sum_{n=0}^{\infty} (-1)^n\int_{0}^{1} s^{2n} \left(\frac{d}{dt}\Big|_{t=0+} s^t \right) ds\\ =& \frac{d}{dt}\Big|_{t=0+} -\sum_{n=0}^{\infty} (-1)^n\int_{0}^{1} s^{2n+t} ds\\ =& \frac{d}{dt}\Big|_{t=0+} -\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+t+1} ds\\ =& -\sum_{n=0}^{\infty} \left[ \frac{d}{dt}\Big|_{t=0+} \frac{(-1)^n}{2n+t+1} \right] ds\\ =& \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2} \\ =& \beta(2) \end{align*} \begin{align*} J = \int_{1}^{\infty} \frac{\arctan(w)\ln(w)}{1+w^2}dw =& \int_{0}^{1} \frac{\arctan\left(\frac{1}{s}\right)\ln\left(\frac{1}{s}\right)}{1+s^2} ds \quad \left( w \mapsto \frac{1}{s}\right)\\ =& -\int_{0}^{1} \frac{\left(-\arctan(s) + \frac{\pi}{2}\right)\ln(s)}{1+s^2} \quad \left(\arctan\left(\frac{1}{b}\right) = -\arctan(b)-\frac{\operatorname{sign}(b)\pi}{2}\right)\\ =& \int_{0}^{1} \frac{\arctan(s)\ln(s)}{1+s^2} ds - \frac{\pi}{2} \int_{0}^{1} \frac{\ln(s)}{1+s^2} ds \\ =& \int_{0}^{1} \frac{\arctan(s)\ln(s)}{1+s^2} ds +\frac{\pi}{2}\beta(2)\\ =& \int_{0}^{\frac{\pi}{4}} t\ln(\tan(t)) dt +\frac{\pi}{2}\beta(2)\\ =& -\int_{0}^{\frac{\pi}{4}} t\ln(\cot(t)) dt +\frac{\pi}{2}\beta(2)\\ \end{align*} Recall the following Fourier series \[ \sum_{k=1}^{\infty} \frac{\cos\left((2k-1)x\right)}{2k-1} = \frac{1}{2}\ln\left(\cot \frac{x}{2}\right)\] Hence \begin{align*} J = -\int_{0}^{\frac{\pi}{4}} t\ln(\cot(t)) dt +\frac{\pi}{2}\beta(2) =& -2\int_{0}^{\frac{\pi}{4}} t\sum_{k=1}^{\infty} \frac{\cos\left((2k-1)2t\right)}{2k-1} dt +\frac{\pi}{2}\beta(2) \\ =& -\sum_{k=1}^{\infty} \frac{2}{2k-1}\int_{0}^{\frac{\pi}{4}} t\cos\left((2k-1)2t\right) dt +\frac{\pi}{2}\beta(2)\\ =& -\frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{(2k-1)^3}\int_{0}^{\frac{\pi(2k-1)}{2}} y \cos\left(y\right) dy +\frac{\pi}{2}\beta(2)\\ \stackrel{IBP}{=}& -\frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{(2k-1)^3} \left[\sin(y)y\Big|_{0}^{\frac{\pi(2k-1)}{2}} - \int_{0}^{\frac{\pi}{8(2k-1)}}\sin(y)dy \right] +\frac{\pi}{2}\beta(2)\\ =& -\frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{(2k-1)^3} \left[\sin\left(\frac{\pi(2k-1)}{2}\right)\frac{\pi(2k-1)}{2} + \cos\left(\frac{\pi(2k-1)}{2}\right) -1\right] +\frac{\pi}{2}\beta(2)\\ =& -\frac{\pi}{4}\sum_{k=1}^{\infty} \frac{\sin\left(\frac{\pi(2k-1)}{2}\right)}{(2k-1)^2} - \frac{1}{2}\sum_{k=1}^{\infty} \frac{\cos\left(\frac{\pi(2k-1)}{2}\right)}{(2k-1)^3}+\frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{(2k-1)^3}+\frac{\pi}{2}\beta(2)\\ =& \frac{\pi}{4}\sum_{k=1}^{\infty} \frac{\cos(\pi k)}{(2k-1)^2} - \frac{1}{2}\sum_{k=1}^{\infty} \frac{\sin(\pi k)}{(2k-1)^3}+\frac{1}{2}\lambda(3)+\frac{\pi}{2}\beta(2) \end{align*} Note that \[\left(\sin(\pi k)\right)_{k\in \mathbb{N}} = (0,0,0,0,...) \] \[\left(\cos(\pi k)\right)_{k\in \mathbb{N}} = (-1,1,-1,1,-1,..) \] Therefore \[ J = \frac{\pi}{4}\sum_{k=1}^{\infty} \frac{(-1)^{k}}{(2k-1)^2} + \frac{1}{2}\lambda(3)+\frac{\pi}{2}\beta(2) = \frac{\pi}{4}\beta(2) + \frac{1}{2}\lambda(3)\] From the identity \[ \lambda(v) = (1-2^{-v}) \zeta(v) \] we have \[ J = \int_{1}^{\infty} \frac{\arctan(w)\ln(w)}{1+w^2}dw = \frac{7}{16}\zeta(3) + \frac{\pi}{4}\beta(2) \] Hence \[ I = \int_{0}^{\frac{\pi}{4} } x\operatorname{arctanh}(\tan x) dx = \frac{1}{2}J - \frac{\pi}{8}K = \frac{7}{32}\zeta(3) - \frac{\pi}{8}\beta(2) + \frac{\pi}{8}\beta(2) \] Therefore, we can conclude \[ \boxed{ \frac{32}{7}\int_{0}^{\frac{\pi}{4} } x\operatorname{arctanh}(\tan x) dx = \zeta(3)} \]