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$\mathrm G$ 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 $(3)$.

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 $K=\frac23\mathrm G$.

clathratus
  • 17,161

3 Answers3

2

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 $$\eqalign{ -\frac12S_1&-S_3-\frac12S_5+\frac12S_7+S_9+\frac12S_{11}\cr &=-\frac12(S_1-S_3+S_5-S_7+S_9-S_{11})-\frac32S_3+\frac32S_9\cr &=-\frac12G-\frac32\frac19G\cr &=-\frac23G\ .\cr}$$

David
  • 82,662
  • This works for me. Just for clarification: "$\sin(12k+1)\frac\pi6$" is the same as $\sin\left[(12k+1)\frac\pi6\right]$? – clathratus Mar 28 '19 at 03:31
  • Also: How do we have $$\mathrm G=S_1-S_3+S_5-S_7+S_9-S_{11}$$ – clathratus Mar 28 '19 at 03:34
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    We're adding all sums $\sum 1/k^2$ with $k=12m+1,,12m+5,,12m+9$, that is, all sums with $k=4n+1$. And subtracting similar sums. – David Mar 28 '19 at 03:49
2

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 $\mathrm{G}$

robjohn
  • 345,667
1

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

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

Mark Viola
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  • If you hover your cursor above the accept-answer button, it says "Click to accept this answer because it solved your problem or it was most helpful in finding your solution..." The accepted answer was accepted because it was not a mere hint, but a full answer. Don't get me wrong-I appreciate your hint (as of now I am the only up-voter of it), but the accepted answer happens to be more helpful. Also: the first answer isn't always the best. – clathratus Mar 28 '19 at 22:22
  • Thank you for the note. Was the "HINT" useful? I really wanted to give you the best answer I could, and thought the hint would have sufficed to show the way forward. – Mark Viola Mar 28 '19 at 22:30
  • The hint was somewhat useful. I must admit I can't exactly see any patterns in the sets of values of $n$ which I could use... Perhaps your answer would be more helpful if you showed a little more how to proceed with this information. I really appreciate your dedication to the quality of your answers across the site. – clathratus Mar 29 '19 at 01:27