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

Using the Fourier series of $\ln(\tan{x})$, \begin{align} &\int^\frac{\pi}{12}_0\ln(\tan{x})\ {\rm d}x\\ =&-2\sum^\infty_{n=0}\frac{1}{2n+1}\int^\frac{\pi}{12}_0\cos\Big{[}(4n+2)x\Big{]}\ {\rm d}x\\ =&-\sum^\infty_{n=0}\frac{\sin\Big[(2n+1)\tfrac{\pi}{6}\Big{]}}{(2n+1)^2}\\ =&\color{#E2062C}{-\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+1)^2}}\color{#6F00FF}{-\sum^\infty_{n=0}\frac{1}{(12n+3)^2}}-\color{#E2062C}{\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+5)^2}}\\ &\color{#E2062C}{+\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+7)^2}}\color{#6F00FF}{+\sum^\infty_{n=0}\frac{1}{(12n+9)^2}}\color{#E2062C}{+\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+11)^2}}\\ =&\color{#6F00FF}{-\frac{1}{9}\underbrace{\sum^\infty_{n=0}\left[\frac{1}{(4n+1)^2}-\frac{1}{(4n+3)^2}\right]}_{G}}\color{#E2062C}{-\frac{1}{2}G-\frac{1}{2}\underbrace{\sum^\infty_{n=0}\left[\frac{1}{(12n+3)^2}-\frac{1}{(12n+9)^2}\right]}_{\frac{1}{9}G}}\\ =&\left(-\frac{1}{9}-\frac{1}{2}-\frac{1}{18}\right)G=\large{-\frac{2}{3}G} \end{align}

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Things could be made clearer if we explicitly write out the terms of the sums. For the red sums, \begin{align} &-\frac{1}{2}\left(\frac{1}{1^2}+\frac{1}{5^2}-\frac{1}{7^2}-\frac{1}{11^2}+\cdots\right)\\ =&-\frac{1}{2}\left(\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}+\frac{1}{9^2}-\frac{1}{11^2}+\cdots\right)-\frac{1}{2}\left(\frac{1}{3^2}-\frac{1}{9^2}+\frac{1}{15^2}-\cdots\right)\\ =&-\frac{1}{2}G-\frac{1}{2}\cdot\frac{1}{9}\left(\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{5^2}-\cdots\right)=-\frac{5}{9}G \end{align}

First: $~\displaystyle 2\int_0^{\tfrac{\pi}{12}} \log(\tan(3x))dx=\int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx\qquad(1)$

Proof:

Let $I=\displaystyle \int_0^{\tfrac{\pi}{12}} \log(\tan(3x))dx$

$\tan(3x)=\tan(x)\tan\big(\dfrac{\pi}{3}+x\big)\tan\big(\dfrac{\pi}{3}-x\big)$

$\displaystyle I= \int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx+\int_0^{\tfrac{\pi}{12}} \log\Big(\tan\Big (\dfrac{\pi}{3}+x\Big)\Big)dx+\int_0^{\tfrac{\pi}{12}} \log\Big(\tan\Big (\dfrac{\pi}{3}-x\Big)\Big)dx$

$\displaystyle I=\int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx+\int_{\tfrac{\pi}{3}}^{\tfrac{5\pi}{12}} \log(\tan(x))dx+\int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{3}} \log(\tan(x))dx$

$\displaystyle I=\int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx+\int_{\tfrac{\pi}{4}}^{\tfrac{5\pi}{12}} \log(\tan(x))dx$

$\displaystyle I=\int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx-\int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{12}} \log\Big(\tan\Big (\dfrac{\pi}{2}-x\Big)\Big)dx$

$\tan\Big (\dfrac{\pi}{2}-x\Big)=\dfrac{1}{\tan(x)}$

So: $~\displaystyle I=\int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx+\int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{12}}\log(\tan(x))dx$

$\displaystyle I=2\int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx-\int_0^{\tfrac{\pi}{4}} \log(\tan(x))dx$

$\displaystyle I=2\int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx-3\int_0^{\tfrac{\pi}{12}} \log(\tan(3x))dx$

$\displaystyle I=2\int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx-3I$

$\displaystyle 2I=\int_0^{\tfrac{\pi}{12}}\log(\tan(x))dx$

Now perform change of variable $u=3x$ in the left member of $(1)$:

$\displaystyle 2\int_0^{\tfrac{\pi}{12}} \log(\tan(3x))dx=\dfrac{2}{3} \int_0^{\tfrac{\pi}{4}} \log(\tan(x))dx$

Since $~\displaystyle G=-\int_0^{\tfrac{\pi}{4}} \log(\tan(x))dx~$ then $~\displaystyle \int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx=-\dfrac{2}{3}G$.

$($Proof found in: Representations of Catalan's constant, David Bradley, $2001)$.

$\qquad\qquad\qquad\qquad$ Hello, there! Cleo just asked me to post this:

$$\int_0^\tfrac\pi{12}\ln(\tan x)~dx=-\dfrac23\cdot\text{Catalan}$$