Hjortnes series for zeta(3)

$\zeta(3)=\frac52\sum_{n=1}^\infty \,\frac{ (-1)^{n-1}}{n^3\binom{2n}{n}}$

Today we will revisit this remarkable result proving it by a different method, namely, we will show that

$\sum_{n=1}^\infty\frac{ (-1)^{n-1}}{\binom{2n}{n}n^3}=-2\int_0^{2\ln(\phi)}x\ln\left(2 \sinh\left( \frac{x}{2}\right) \right)\,dx$

And compute this integral.

First, lets find a series expansion for $\ln\left( \sinh(x)\right)$ :

$\begin{aligned} \ln\left( \sinh(x)\right)&=\ln\left( \frac{1}{2} \left( e^{x}-e^{-x} \right)\right)\\ &=-\ln 2+\ln\left( e^{x}-e^{-x} \right)\\ &=-\ln 2+\ln\left( \frac{e^{-x}}{e^{-x}} \left( e^{x}-e^{-x} \right)\right)\\ &=-\ln 2+x+\ln\left( 1-e^{-2x} \right)\\ &=-\ln 2+x-\sum_{n=1}^{\infty}\frac{e^{-2nx}}{n} \qquad \blacksquare \end{aligned}$

Letting $x \to \frac{x}{2}$ we obtain

$\ln\left(2 \sinh\left(\frac{x}{2}\right)\right)=\frac{x}{2}-\sum_{k=1}^{\infty}\frac{e^{-kx}}{k}$ (1)

Now recall (see here)

$\frac{\arcsin\left(\frac{x}{2}\right)}{\sqrt{1-\left(\frac{x}{2}\right)^2}}=\sum_{n=1}^\infty \frac{x^{2n-1}}{\binom{2n}{n}n}$ (2)

Letting $x \to ix$ we obtain

$\frac{\operatorname{arcsinh}\left(\frac{x}{2}\right)}{\sqrt{1+\left(\frac{x}{2}\right)^2}}=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^{2n-1}}{\binom{2n}{n}n}$ (3)

Claim:

$\sum_{n=1}^\infty\frac{ (-1)^{n-1}}{\binom{2n}{n}n^k}=\frac{(-2)^{k-2}}{(k-2)!}\int_0^{2\ln(\phi)}x\ln^{k-2}\left(2 \sinh\left( \frac{x}{2}\right) \right)\,dx$

Proof:

$\begin{aligned} \sum_{n=1}^\infty\frac{ (-1)^{n-1}}{\binom{2n}{n}n^k}&=\frac{(-1)^{k-1}}{(k-1)!}\sum_{n=1}^\infty\frac{ (-1)^{n-1}}{\binom{2n}{n}}\int_0^1 \ln^{k-1}(x) x^{n-1}\,dx\\ &=\frac{2(-1)^{k-1}}{(k-1)!}\sum_{n=1}^\infty\frac{ (-1)^{n-1}}{\binom{2n}{n}}\int_0^1 \ln^{k-1}\left(x^2\right) x^{2n-1}\,dx & \left(x \to x^2\right)\\ &=\frac{2(-2)^{k-1}}{(k-1)!}\sum_{n=1}^\infty\frac{ (-1)^{n-1}}{\binom{2n}{n}}\int_0^1 \ln^{k-1}\left(x\right) x^{2n-1}\,dx \\ &=\frac{2(-2)^{k-1}}{(k-1)!}\sum_{n=1}^\infty\frac{ (-1)^{n-1}}{\binom{2n}{n}}\left(\frac{x^{2n}\ln^{k-1}(x)}{2n}\Bigg|_0^1-\frac{(k-1)}{2n}\int_0^1 \ln^{k-2}\left(x\right) x^{2n-1}\,dx \right)\\ &=-\frac{(-2)^{k-1}}{(k-2)!}\sum_{n=1}^\infty\frac{ (-1)^{n-1}}{\binom{2n}{n}n}\int_0^1 \ln^{k-2}\left(x\right) x^{2n-1}\,dx \\ &=-\frac{(-2)^{k-1}}{(k-2)!}\int_0^1 \ln^{k-2}\left(x\right)\left(\sum_{n=1}^\infty\frac{(-1)^{n-1} x^{2n-1}}{\binom{2n}{n}n} \right) \,dx \\ &=-\frac{(-2)^{k-1}}{(k-2)!}\int_0^1 \ln^{k-2}\left(x\right)\left(\frac{ \operatorname{arcsinh}\left(\frac{x}{2} \right)}{\sqrt{1+\left( \frac{x}{2}\right)^2}} \right) \,dx & \left( \text{by eq. (3)}\right)\\ &=\frac{(-2)^{k-2}}{(k-2)!}\int_0^{2\ln(\phi)} \frac{x\ln^{k-2}\left(2\sinh\left(\frac{x}{2}\right)\right) \cosh\left(\frac{x}{2} \right)}{\sqrt{1+\sinh^2\left( \frac{x}{2}\right)}} \,dx & \left( \frac{x}{2} \to \sinh\left(\frac{x}{2} \right)\right)\\ &=\frac{(-2)^{k-2}}{(k-2)!}\int_0^{2\ln(\phi)}x\ln^{k-2}\left(2 \sinh\left( \frac{x}{2}\right) \right)\,dx \qquad \blacksquare \end{aligned}$

Letting k=3 we obtain

$\sum_{n=1}^\infty\frac{ (-1)^{n-1}}{\binom{2n}{n}n^3}=-2\int_0^{2\ln(\phi)}x\ln\left(2 \sinh\left( \frac{x}{2}\right) \right)\,dx$

Let´s now calculate this integral

$\begin{aligned} \sum_{n=1}^\infty\frac{ (-1)^{n-1}}{\binom{2n}{n}n^3}&=-2\int_0^{2\ln(\phi)}x\ln\left(2 \sinh\left( \frac{x}{2}\right) \right)\,dx \\ &=-2\int_0^{2\ln(\phi)}x\left(\frac{x}{2}-\sum_{k=1}^{\infty}\frac{e^{-kx}}{k} \right)\,dx & \left( \text{by eq. (1)}\right)\\ &=-\int_0^{2\ln(\phi)}x^2\,dx+2\sum_{k=1}^{\infty}\frac{1}{k}\int_0^{2\ln(\phi)} xe^{-kx}\,dx\\ &=-\frac{8}{3}\ln^3(\phi)+2\sum_{k=1}^{\infty}\frac{1}{k}\left(-\frac{2\ln(\phi)\phi^{-2k}}{k}+\frac{1}{k}\int_0^{2\ln(\phi)} e^{-kx}\,dx \right)\\ &=-\frac{8}{3}\ln^3(\phi)-4\ln(\phi)\sum_{k=1}^{\infty}\frac{(\phi^{-2})^k}{k^2}+2\sum_{k=1}^{\infty}\frac{1}{k}\left(\frac{1}{k^2}-\frac{(\phi^{-2})^k}{k^2}\right)\\ &=-\frac{8}{3}\ln^3(\phi)-4\ln(\phi)\operatorname{Li}_2(\phi^{-2})+2\zeta(3)-2\sum_{k=1}^{\infty}\frac{(\phi^{-2})^k}{k^3}\\ &=-\frac{8}{3}\ln^3(\phi)-4\ln(\phi)\operatorname{Li}_2(\phi^{-2})-2\operatorname{Li}_3(\phi^{-2})+2\zeta(3)\\ &=-\frac{8}{3}\ln^3(\phi)-4\ln(\phi)\left( \frac{\pi^{2}}{15}-\ln ^{2} \phi\right)-2\left(\frac45\zeta(3)+\frac{2\ln ^{3}(\phi)}{3}-\frac{2\pi^{2} \ln (\phi)}{15} \right)+2\zeta(3)\\ &=-\frac{8}{3}\ln^3(\phi)+4\ln^3(\phi)-\frac{4}{3}\ln^3(\phi)-\frac85\zeta(3)+2\zeta(3)\\ &=\frac25\zeta(3) \qquad \blacksquare \end{aligned}$

Where we used (see here)

$\mathrm{Li}_{2}\left(\frac{1}{\phi^{2}}\right) &=\frac{\pi^{2}}{15}-\ln ^{2} \phi$

$\operatorname{Li}_{3}\left(\frac{1}{\phi^{2}}\right)=\frac45\zeta(3)+\frac{2\ln ^{3}(\phi)}{3}-\frac{2\pi^{2} \ln (\phi)}{15}$

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Hjortnes series for zeta(3) - PART II

Claim:

Proof:

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