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I was trying to prove the well known result: $$ \sum_{k=1}^\infty \frac{1}{\binom{2k}kk^2}=\frac{\zeta(2)}{3} $$ and it came down to prove the following equation: $$ \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \ln\left(\tan(x)+\tan\left(\frac{\pi}{6}\right)\right)\tan(x)\space dx=\frac{\zeta(2)}{6} $$ Can this integral be used to prove the above result; in other words can it be proven without using the above mentionend sum? I tried applying the angle sum and difference identities but I didn't get something helpful, so any help is highly appreciated!

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Can it be shown directly, in other words without knowing $\zeta(2)=\frac{\pi^2}{6}$, that $$ \sum_{k=1}^\infty \frac{1}{\binom{2k}kk^2}=\frac{\zeta(2)}{3} $$

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  • $\begingroup$ Just to get this right, you want us to prove the integral and we can use the sum if we want? $\endgroup$ – user111187 Jan 20 '15 at 8:50
  • $\begingroup$ See the edits; I fixed it now. $\endgroup$ – Redundant Aunt Jan 20 '15 at 9:46
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We use the notation $c = \dfrac{1}{\sqrt 3}$ for convenience. Substitute $t = \tan x$ to get $$ I = \int_{-c}^c dt\frac{t \log(t+c)}{1+t^2}. $$ There exists an antiderivative, namely $\newcommand{\lb}{\left(}$ $\newcommand{\rb}{\right)}$ $\newcommand{\lbb}{\left[}$ $\newcommand{\rbb}{\right]}$ $\newcommand{\li}{\operatorname{Li}_2}$ $$ I = \int dt\frac{t \log(t+c)}{1+t^2} = \Re \lbb \log \lb \frac{i-t}{i+c} \rb \log (t+c)+ \li\lb \frac{c+t}{c-i} \rb\rbb. $$ Substituting the limits, we find that the antiderivative vanishes at the left endpoint and thus $$ I = \underbrace{\Re \lbb \log \lb \frac{i-c}{i+c} \rb \log (2c)\rbb}_0 + \Re \lbb \li\lb \frac{2c}{c-i} \rb\rbb \\= \Re \li\lb e^{ i \pi/3} \rb=\frac{1}{4}(\pi - \pi/3)^2-\frac{\pi^2}{12}=\frac{\pi^2}{36} = \frac {\zeta (2)}{ 6} . $$

Here I used formula 5.16 from Lewin's book (polylogs and associated functions) to calculate the real part of the polylogarithm. The antiderivative can be straightforwardly calculated by writing $$ \frac{1}{1+t^2} = \frac 1 2 \lbb \frac{1}{1 - i t} + \frac{1}{1+ i t} \rbb = \Re \lbb \frac{1}{1 - i t} \rbb. $$

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  • $\begingroup$ If you need more details, I can provide them. $\endgroup$ – user111187 Jan 20 '15 at 9:49
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I am not sure if this helps to focus on the sum itself. If it does, then start with $\displaystyle f(x)=2\arcsin(\frac{x}{2})^2$. Write a Taylor's expansion for $\displaystyle f(x)$ (this is rather painful but fun):$$2\arcsin(\frac{x}{2})^2=\sum_{n=1}^{\infty}\frac{x^{2n}}{\binom{2n}n n^2}.$$ Now observe that $$2\arcsin(\frac{1}{2})^2=\sum_{n=1}^{\infty}\frac{1}{\binom{2n}n n^2},$$or that$$\sum_{n=1}^{\infty}\frac{1}{\binom{2n}n n^2}=2 \times (\frac{\pi}{6})^2.$$

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