'Tan'ned Up

By the property of logarithm we write the integral as $\int_{0}^{\frac{\pi}{4}}\left(\ln(\tan\left(\frac{x}{3}\right)-2\ln\left(\tan x\right)\right)dx$ . It is easy to derive that or well know result that $2\int_0^{\frac{\pi}{4}}\ln\tan(x)dx=-2\sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1}=-2G$ where $G$ is Catalan's constant and we are left to evaluate the integral $\int_0^{\frac{\pi}{4}} \ln\tan\left(\frac{x}{3}\right)dx=-2\sum_{n=0}^{\infty}\frac{1}{2n+1}\int_0^{\frac{\pi}{4}}\cos\left(\frac{2x}{3}(2n+1)\right)dx = -3\sum_{n=0}^{\infty}\frac{1 }{(2n+1)^2}\sin\left(\frac{\pi}{6}(2n+1)\right)$ Now we evaluate the latter series $\frac{L}{3}=-\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{(12n+1)^2}+\frac{1}{(12n+5)^2}-\frac{1}{(12n+7)^2}-\frac{1}{(12n+11)^2}\right)-\sum_{n=0}^{\infty}\left(\frac{1}{(12n+3)^2}-\frac{1}{(12n+9)^2}\right)\\=-\frac{1}{9}\sum_{n=0}^{\infty}\left(\frac{1}{(4n+1)^2}-\frac{1}{(4n+3)^2}\right)-\frac{G}{2}-\sum_{n=0}^{\infty}\frac{1}{18}\left(\frac{1}{(4n+1)^2}-\frac{1}{(4n+3)^2}\right)=-\frac{G}{9}-\frac{G}{2}-\frac{G}{18}=-\frac{2}{3}G$ so we have $\int_0^{\frac{\pi}{4}} \ln\tan\left(\frac{x}{3}\right)=-2G$ giving us $-2G+2G=0$ .

Here $\sum_{n=0}^{\infty}\left(\frac{1}{(4n+1)^2}-\frac{1}{(4n+3)^2}\right)=\frac{1}{4}\left(\psi^1\left(\frac{1}{4}\right)-\psi^1\left(\frac{3}{4}\right)\right)=\frac{1}{16}\left(\pi^2+8G-\pi^2+8G\right)=G$ Or $\begin{aligned}\sum_{n=0}^{\infty}\dfrac{1}{(4n+3)^2}&= \sum_{n=1}^{\infty}\dfrac{1}{n^2}-\left(\sum_{n=1}^{\infty}\dfrac{1}{(2n)^2}+\sum_{n=1}^{\infty}\dfrac{1}{(4n-3)^2}\right)\\&= \dfrac{\pi^2}{8}- \left(\sum_{n=0}^{\infty}\dfrac{(-1)^n}{(2n-1)^2}+\sum_{n=0}^{\infty}\dfrac{1}{(4n+3)^2}\right)\\&\implies \sum_{n=0}^{\infty}\dfrac{1}{(4n+3)^2} =\dfrac{\pi^2}{16}-\dfrac{G}{2} \end{aligned}$ and similarly $\displaystyle \sum_{n\geq 0}\frac{1}{(4n+1)^2}=\frac{\pi^2}{16}+\frac{G}{2}$ .