Lobachevsky Fourier expansion and a certified π/3 value

For $\mathrm{\mid }w\mathrm{\mid }<1$, integrating the uniformly convergent geometric series on compact subdisks gives $-\log (1-w)={\sum }_{n\ge 1}{w}^n\mathrm{/}n$. Taking real parts at $w=r{e}^{2iu}$, where $0<r<1$, yields $-\log \mathrm{\mid }1-r{e}^{2iu}\mathrm{\mid }={\sum }_{n\ge 1}{r}^n\cos (2nu)\mathrm{/}n$. This series is uniformly absolutely convergent in $u$ for fixed $r$, so integrating from $0$ to $\theta$ gives $-{\int }_0^{\theta }\log \mathrm{\mid }1-r{e}^{2iu}\mathrm{\mid }du=\frac{1}{2}{\sum }_{n\ge 1}{r}^n\sin (2n\theta )\mathrm{/}{n}^2$.

For $r\ge 1\mathrm{/}2$, $\mathrm{\mid }1-r{e}^{2iu}{\mathrm{\mid }}^2=(1-r{)}^2+4r{\sin }^{2}u$, so $\sqrt{2}\mathrm{\mid }\sin u\mathrm{\mid }\le \mathrm{\mid }1-r{e}^{2iu}\mathrm{\mid }\le 2$. Therefore $\mathrm{\mid }\log \mathrm{\mid }1-r{e}^{2iu}\mathrm{\mid }\mathrm{\mid }$ is bounded by a constant plus $\mathrm{\mid }\log \mathrm{\mid }\sin u\mathrm{\mid }\mathrm{\mid }$, which is integrable on bounded intervals. Dominated convergence sends the integral in step 1.1 to $\mathrm{\Lambda }(\theta )$ as $r\uparrow 1$; the series converges uniformly in $\theta$ because $\sum n^{-2}<\mathrm{\infty }$. This proves the Fourier expansion on $[0,\pi ]$ and continuity there.

The defining integrand is even, so $\mathrm{\Lambda }$ is odd. Its derivative away from the integral's singularities is $-\log \mathrm{\mid }2\sin \theta \mathrm{\mid }$, a $\pi$-periodic function; the Fourier formula gives $\mathrm{\Lambda }(0)=\mathrm{\Lambda }(\pi )=0$, so the integral of this derivative over each period is zero. Hence $\mathrm{\Lambda }(\theta +\pi )=\mathrm{\Lambda }(\theta )$ for every real $\theta$. Substitution $\theta =\pi \mathrm{/}2$ in the Fourier series gives $\mathrm{\Lambda }(\pi \mathrm{/}2)=0$.

The identity $\mathrm{\mid }2\sin (2u)\mathrm{\mid }=\mathrm{\mid }2\sin u\mathrm{\mid }\mathrm{\mid }2\sin (u+\pi \mathrm{/}2)\mathrm{\mid }$ and substitution $v=2u$ in the defining integral give $\mathrm{\Lambda }(2\theta )=2\mathrm{\Lambda }(\theta )+2\mathrm{\Lambda }(\theta +\pi \mathrm{/}2)-2\mathrm{\Lambda }(\pi \mathrm{/}2)$. Step 3.1 gives the stated duplication identity.

In the Fourier series at $\theta =\pi \mathrm{/}3$, $\sin (2n\pi \mathrm{/}3)$ cycles through $\sqrt{3}\mathrm{/}2,-\sqrt{3}\mathrm{/}2,0$, so grouping successive triples gives the stated positive series. For $a=3k+1$, $a^{-2}-(a+1{)}^{-2}=(2a+1)\mathrm{/}(a^2(a+1{)}^2)\le 2\mathrm{/}{a}^3\le 2\mathrm{/}(27k^3)$ for $k\ge 1$. For $N\ge 1$, the tail after $k=0,\dots ,N$ is at most $\frac{\sqrt{3}}{4}\frac{2}{27}{\sum }_{k>N}{k}^{-3}\le \frac{\sqrt{3}}{4}\frac{2}{27}{\int }_N^{\mathrm{\infty }}{x}^{-3}dx=\frac{\sqrt{3}}{108N^2}\le \frac{\sqrt{3}}{72N^2}$.