Proof clarification $-\int_0^{\pi/12}\log(\tan t)dt=\frac23\mathrm G$

Question

$\mathrm G$ is Catalan's constant.

I would like clarification on the following proof provided by @M.N.C.E.:

$$ \begin{align} &\int^\frac{\pi}{12}_0\ln(\tan{x})\ {\rm d}x\\ =&-2\sum^\infty_{n=0}\frac{1}{2n+1}\int^\frac{\pi}{12}_0\cos\Big{[}(4n+2)x\Big{]}\ {\rm d}x\\ =&-\sum^\infty_{n=0}\frac{\sin\Big[(2n+1)\tfrac{\pi}{6}\Big{]}}{(2n+1)^2}\tag1\\ =&\color{#E2062C}{-\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+1)^2}}\color{#6F00FF}{-\sum^\infty_{n=0}\frac{1}{(12n+3)^2}}-\color{#E2062C}{\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+5)^2}}\\ &\color{#E2062C}{+\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+7)^2}}\color{#6F00FF}{+\sum^\infty_{n=0}\frac{1}{(12n+9)^2}}\color{#E2062C}{+\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+11)^2}}\tag2\\ =&\color{#6F00FF}{-\frac{1}{9}\underbrace{\sum^\infty_{n=0}\left[\frac{1}{(4n+1)^2}-\frac{1}{(4n+3)^2}\right]}_{\mathrm G}}\color{#E2062C}{-\frac{1}{2}\mathrm G-\frac{1}{2}\underbrace{\sum^\infty_{n=0}\left[\frac{1}{(12n+3)^2}-\frac{1}{(12n+9)^2}\right]}_{\frac{1}{9}\mathrm G}}\tag3\\ =&\left(-\frac{1}{9}-\frac{1}{2}-\frac{1}{18}\right)\mathrm G=\large{-\frac{2}{3}\mathrm G} \end{align}$$

I understand how to obtain $(1)$, but I do not know how to go from $(1)$ to $(2)$ and then from $(2)$ to $(3)$.

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A log of my attempts to find an alternate proof for the identity in question.

I tried to represent $(1)$ as a hypergeometric series because the $1/(2n+1)^2$ terms are after all what give us the identity $$_3F_2\left(\frac12,\frac12,1;\frac32,\frac32;-1\right)=\mathrm G$$ So I was hoping that $$\frac{\sin\left[\frac\pi6(2n+1)\right]}{\sin\left[\frac\pi6(2n+3)\right]}=-1$$ But alas this is not the case. I see now that I would be missing an extra factor of $2/3$.

Another, more realistic attempt of mine comes by noting that $$\sin\left[\frac\pi6(2n+1)\right]=\frac{\sqrt3}2\sin\frac{\pi n}3+\frac12\cos\frac{\pi n}3$$ So, by letting $K$ be $-1\cdot$(the quantity $(1)$), we have that $$K=\frac{\sqrt3}2S+\frac12C$$ Where $$S=\sum_{n\geq0}\frac{\sin\frac{\pi n}3}{(2n+1)^2}$$ and $$C=\sum_{n\geq0}\frac{\cos\frac{\pi n}3}{(2n+1)^2}$$ Although neither of these seem to have a 'nice' $_pF_q$ representation. I have run out of ideas for proving that $K=\frac23\mathrm G$.

Answer 1

Firstly, any $n$ can be written as $6k$ or $6k+1$ or... or $6k+5$. So $$\eqalign{ \sum^\infty_{n=0}\frac{\sin\Big[(2n+1)\tfrac{\pi}{6}\Big{]}}{(2n+1)^2} &=\sum_{k=0}^\infty\frac{\sin(12k+1)\frac\pi6}{(12k+1)^2}+\hbox{five more sums}\cr &=\sum_{k=0}^\infty\frac{\sin\frac\pi6}{(12k+1)^2}+\cdots\cr &=\frac12\sum_{k=0}^\infty\frac{1}{(12k+1)^2}+\cdots\ .\cr}$$ Then if we write $$S_m=\sum_{k=0}^\infty\frac{1}{(12k+m)^2}$$ we have $$G=S_1-S_3+S_5-S_7+S_9-S_{11}$$ and your sum $(2)$ is $$\eqalign{ -\frac12S_1&-S_3-\frac12S_5+\frac12S_7+S_9+\frac12S_{11}\cr &=-\frac12(S_1-S_3+S_5-S_7+S_9-S_{11})-\frac32S_3+\frac32S_9\cr &=-\frac12G-\frac32\frac19G\cr &=-\frac23G\ .\cr}$$

Answer 2

I understand how to obtain $(1)$, but I do not know how to go from $(1)$ to $(2)$ and then from $(2)$ to $(3)$.

HINT:

In order to arrive at $(2)$, note that $\sin\left((2n+1)\frac\pi6\right)$ takes on the values

$$\sin\left((2n+1)\frac\pi6\right)=\begin{cases}\frac12 &,n=0,6,12,\dots\\\\ 1 &,n=1,7,13,\dots\\\\ 1/2&,n=2,8,14,\dots\\\\ -\frac12&,n=3,9,15,\cdots\\\\ -1&,n=4,10,16,\cdots\\\\ -\frac12&,n=5,11,17,\cdots \end{cases}$$

Answer 3

As shown in this answer, $$ \log(2\cos(x))=\cos(2x)-\frac{\cos(4x)}2+\frac{\cos(6x)}3-\dots\tag1 $$ Substituting $x\mapsto\frac\pi2-x$ gives $$ \log(2\sin(x))=-\cos(2x)-\frac{\cos(4x)}2-\frac{\cos(6x)}3-\dots\tag2 $$ Subtracting $(1)$ from $(2)$ yields $$ \log(\tan(x))=-2\left(\cos(2x)+\frac{\cos(6x)}3+\frac{\cos(10x)}5+\dots\right)\tag3 $$ Notice that $\sin\left((2k+1)\frac\pi6\right)$ has period $6$:

$$ \begin{array}{c} k\pmod6&0&1&2&3&4&5\\ \color{#C00}{\sin\left((2k+1)\frac\pi6\right)}&\color{#C00}{\frac12}&\color{#C00}{1}&\color{#C00}{\frac12}&\color{#C00}{-\frac12}&\color{#C00}{-1}&\color{#C00}{-\frac12}\\ \color{#090}{2k+1}&\color{#090}{1}&\color{#090}{3}&\color{#090}{5}&\color{#090}{7}&\color{#090}{9}&\color{#090}{11}\\ \color{#C00}{\frac{(-1)^k}2}&\color{#C00}{\frac12}&\color{#C00}{-\frac12}&\color{#C00}{\frac12}&\color{#C00}{-\frac12}&\color{#C00}{\frac12}&\color{#C00}{-\frac12}\\ \color{#C00}{\frac32(-1)^{\frac{k-1}3}}&&\color{#C00}{\frac32}&&&\color{#C00}{-\frac32}&&k\equiv1\pmod3\\ \color{#090}{3\left(2\frac{k-1}3+1\right)}&&\color{#090}{3}&&&\color{#090}{9}&&k\equiv1\pmod3 \\ \end{array}\tag4 $$

Note that the upper red row is the sum of the two lower red rows

Thus, $$ \begin{align} \int_0^{\pi/12}\log(\tan(x))\,\mathrm{d}x &=-\sum_{k=0}^\infty\frac{\sin\left((2k+1)\frac\pi6\right)}{(2k+1)^2}\tag5\\ &=-\sum_{k=0}^\infty\left(\frac{(-1)^k}2\frac1{(2k+1)^2}+\frac{(-1)^k}9\frac32\frac1{(2k+1)^2}\right)\tag6\\ &=-\frac23\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2}\tag7\\[3pt] &=-\frac23\mathrm{G}\tag8 \end{align} $$ Explanation:
$(5)$: integrate $(3)$ term by term
$(6)$: apply $(4)$; $k$ in the right term is $\frac{k-1}3$ from $(4)$
$(7)$: collect terms
$(8)$: definition of $\mathrm{G}$