Hjortnes series for zeta(3)

In this post We will derive the remarkable beatiful Hjortnes series used by Apery to prove the irrationality of zeta(3)

$\zeta(3)=\frac52\sum_{n=1}^\infty \,\frac{ (-1)^{n-1}}{n^3\left(\begin{array}{c}2n\\ n\end{array}\right)}$

Recall (proved here)

$\arcsin^2(x)=\frac12\sum_{n=1}^\infty\frac{ 4^{n}}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{x^{2n}}{n^2}$ (1)

Letting $x \to ix$ in (1) and using the fact that $\arcsin(ix)=i\,\operatorname{arcsinh}(x)$ we obtain

$\operatorname{arcsinh}^2(x)=-\frac12\sum_{n=1}^\infty\frac{ 4^{n}(-1)^n}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{x^{2n}}{n^2}$ (2)

Proof:

$\begin{aligned} &\left(\arcsin(ix)\right)^2=\frac12\sum_{n=1}^\infty\frac{ 4^{n}}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{(ix)^{2n}}{n^2}\\ &\left(i\operatorname{arcsinh}(x)\right)^2=\frac12\sum_{n=1}^\infty\frac{ 4^{n}(-1)^n}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{x^{2n}}{n^2}\\ &\operatorname{arcsinh}^2(x)=-\frac12\sum_{n=1}^\infty\frac{ 4^{n}(-1)^n}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{x^{2n}}{n^2} \qquad \blacksquare\\ \end{aligned}$

Dividing (2) by x and integrating from 0 to 1/2 we obtain

$\sum_{n=1}^\infty \,\frac{ (-1)^{n-1}}{n^3\left(\begin{array}{c}2n\\ n\end{array}\right)}=4\int_0^{1/2}\frac{\operatorname{arcsinh}^2(x)}{x}\,dx$ (3)

Let´s now focus on the R.H.S. of (3)

$\begin{aligned} 4\int_0^{1/2}\frac{\operatorname{arcsinh}^2(x)}{x}\,dx&=4\int_0^{\ln(\phi)}\frac{x^2\cosh(x)}{\sinh(x)}\,dx \qquad \left( \operatorname{arcsinh}(x) \to x \right)\\ &=4\int_0^{\ln(\phi)}\frac{x^2\left(e^{x}+e^{-x} \right)}{\left(e^{x}-e^{-x} \right)}\,dx \\ &=4\int_0^{\ln(\phi)}\frac{x^2\left(e^{2x}+1 \right)}{\left(e^{2x}-1 \right)}\,dx \\ &=4\int_0^{\ln(\phi)}x^2\left(\frac{e^{2x}+1+1-1 }{e^{2x}-1 }\right)\,dx \\ &=4\int_0^{\ln(\phi)}x^2\left(\frac{2 }{e^{2x}-1 }+1\right)\,dx \\ &=\frac{4\ln^3(\phi)}{3}+8\int_0^{\ln(\phi)}\frac{x^2 }{e^{2x}-1 }\,dx \\ &=\frac{4\ln^3(\phi)}{3}+\int_0^{2\ln(\phi)}\frac{x^2 }{e^{x}-1 }\,dx \qquad \left(2x \to x \right)\\ &=\frac{4\ln^3(\phi)}{3}+\int_1^{\phi^{-2}}\frac{\left(-\ln(x)\right)^2 }{\frac{1}{x}-1 }\,\frac{dx}{x} \qquad \left(x \to -\ln(x) \right)\\ &=\frac{4\ln^3(\phi)}{3}+\int_{\phi^{-2}}^1\frac{\ln^2(x) }{1-x }\,dx\\ &=\frac{4\ln^3(\phi)}{3}+\int_0^{1}\frac{\ln^2(x) }{1-x }\,dx-\int_0^{\phi^{-2}}\frac{\ln^2(x) }{1-x }\,dx\\ &=\frac{4\ln^3(\phi)}{3}+2\zeta(3)-\int_0^{\phi^{-2}}\frac{\ln^2(x) }{1-x }\,dx\\ &=\frac{4\ln^3(\phi)}{3}+2\zeta(3)-\left[2\operatorname{Li}_{3}(x)-2\ln(x)\operatorname{Li}_2(x)-\ln(1-x)\ln^2(x) \right]_0^{\phi^{-2}} \qquad \left( \text{equation (A.8)}\right)\\ &=\frac{4\ln^3(\phi)}{3}+2\zeta(3)-\left[2\operatorname{Li}_{3}(\phi^{-2})-2\ln(\phi^{-2})\operatorname{Li}_2(\phi^{-2})-\ln(1-\phi^{-2})\ln^2(\phi^{-2}) \right]\\ &=\frac{4\ln^3(\phi)}{3}+2\zeta(3)-2\left( \frac45\zeta(3)+\frac{2\ln ^{3}(\phi)}{3}-\frac{2\pi^{2} \ln (\phi)}{15}\right)-4\ln(\phi)\left(\frac{\pi^{2}}{15}-\ln ^{2} (\phi) \right)-4\ln^3(\phi)\\ &=2\zeta(3)-\frac{8}{5}\zeta(3)\\ &=\frac{2}{5}\zeta(3) \qquad \blacksquare \end{aligned}$

Plugging the result of the integral back in (3) we obtain the remarkable result

$\zeta(3)=\frac52\sum_{n=1}^\infty \,\frac{ (-1)^{n-1}}{n^3\left(\begin{array}{c}2n\\ n\end{array}\right)}$

Appendix 1

Recall the following relations regarding the Golden Ratio

$\phi=\frac{1+\sqrt{5}}{2}$

$\begin{gathered} \phi=\frac{1}{\phi}+1, \quad \phi^{2}=\phi+1 \\ \phi^{-2}+\phi^{-1}=1 \Rightarrow \frac{1}{\phi^{2}}=1-\frac{1}{\phi} \end{gathered}$ (A.1)

We have proven the following relation in this post

$\begin{aligned} \mathrm{Li}_{2}\left(\frac{1}{\phi}\right) &=\frac{\pi^{2}}{10}-\ln ^{2} \phi \\ \mathrm{Li}_{2}\left(-\frac{1}{\phi}\right) &=\frac{\ln ^{2} \phi}{2}-\frac{\pi^{2}}{15} \\ \mathrm{Li}_{2}\left(\frac{1}{\phi^{2}}\right) &=\frac{\pi^{2}}{15}-\ln ^{2} \phi \end{aligned}$ (A.2)

Also, recall the Trilogarithm identity proved here

$\operatorname{Li}_{3}(x)+\operatorname{Li}_{3}(1-x)+\operatorname{Li}_{3}\left(1-\frac{1}{x}\right)=\zeta(3)+\frac{\ln ^{3}(x)}{6}+\frac{\pi^{2} \ln (x)}{6}-\frac{\ln ^{2}(x) \ln (1-x)}{2}$ (A.3)

And the Polylogarithm relation proved here

$\operatorname{Li}_{n}(x)+\operatorname{Li}_{n}(-x)=2^{1-n}\operatorname{Li}_{n}(x^2)$ (A.4)

Example, letting $x=\phi^{-1}$ in (A.4) we obtain

$\operatorname{Li}_{n}(\phi^{-1})+\operatorname{Li}_{n}(-\phi^{-1})=\frac14\operatorname{Li}_{n}(\phi^{-2})$ (A.5)

Claim:

$\operatorname{Li}_{3}\left(\frac{1}{\phi^{2}}\right)=\frac45\zeta(3)+\frac{2\ln ^{3}(\phi)}{3}-\frac{2\pi^{2} \ln (\phi)}{15}$ (A.6)

Proof:

If we let $x=\phi^{-1}$ in (1) we obtain

$\begin{aligned} &\operatorname{Li}_{3}\left(\phi^{-1}\right)+\operatorname{Li}_{3}(1-\phi^{-1})+\operatorname{Li}_{3}\left(1-\frac{1}{\phi^{-1}}\right)=\zeta(3)-\frac{\ln ^{3}(\phi)}{6}-\frac{\pi^{2} \ln (\phi)}{6}-\frac{\ln ^{2}(\phi) \ln (1-\phi^{-1})}{2}\\ &\operatorname{Li}_{3}\left(\phi^{-1}\right)+\operatorname{Li}_{3}(\phi^{-2})+\operatorname{Li}_{3}\left(-\phi^{-1}}\right)=\zeta(3)-\frac{\ln ^{3}(\phi)}{6}-\frac{\pi^{2} \ln (\phi)}{6}-\frac{\ln ^{2}(\phi) \ln (\phi^{-2})}{2}\\ &\frac14\operatorname{Li}_{3}\left(\phi^{-2}\right)+\operatorname{Li}_{3}(\phi^{-2})=\zeta(3)-\frac{\ln ^{3}(\phi)}{6}-\frac{\pi^{2} \ln (\phi)}{6}+\ln ^{3}(\phi) \\ &\frac54\operatorname{Li}_{3}\left(\phi^{-2}\right)=\zeta(3)+\frac{5\ln ^{3}(\phi)}{6}-\frac{\pi^{2} \ln (\phi)}{6} \\ &\operatorname{Li}_{3}\left(\phi^{-2}\right)=\frac45\zeta(3)+\frac{2\ln ^{3}(\phi)}{3}-\frac{2\pi^{2} \ln (\phi)}{15} \qquad \blacksquare \\ \end{aligned}$

Appendix 2

$\begin{aligned} \operatorname{Li}_{3}(x)&=\int_0^x\frac{\operatorname{Li}_{2}(t)}{t}\,dt\\ &=-\int_0^x\frac{1}{t}\left(\int_0^t\frac{\ln(1-w)}{w}\,dw \right)\,dt \qquad \left(\operatorname{Li}_{2}(t)=-\int_0^t\frac{\ln(1-w)}{w}\,dw \right)\\ &=-\int_0^x \frac{\ln(1-w)}{w}\left(\int_w^x\frac{dt}{t} \right)\,dw \qquad \text{switched order of integration}\\ &=-\int_0^x \frac{\ln(1-w)}{w}\left(\ln(x)-\ln(w) \right)\,dw\\ &=\int_0^x \frac{\ln(1-w)\ln(w)}{w}\,dw-\ln(x)\int_0^x \frac{\ln(1-w)}{w}\,dw\\ &=\int_0^x \frac{\ln(1-w)\ln(w)}{w}\,dw+\ln(x)\operatorname{Li}_2(x) \end{aligned}$

And we get

$\int_0^x \frac{\ln(1-w)\ln(w)}{w}\,dw=\operatorname{Li}_{3}(x)-\ln(x)\operatorname{Li}_2(x)$ (A.7)

Now we focus on the integral on the L.H.S.

$\begin{aligned} \int_0^x \frac{\ln(1-w)\ln(w)}{w}\,dw&=uv-\int v \,du\\ &=\frac{\ln(1-w)\ln^2(w)}{2}\Big|_0^x+\frac12\int_0^x\frac{\ln^2(w)}{1-w}\,dw\\ &=\frac{\ln(1-x)\ln^2(x)}{2}+\frac12\int_0^x\frac{\ln^2(w)}{1-w}\,dw\\ \end{aligned}$

Plugging this result back in (A.7) we obtain

$\int_0^x\frac{\ln^2(w)}{1-w}\,dw=2\operatorname{Li}_{3}(x)-2\ln(x)\operatorname{Li}_2(x)-\ln(1-x)\ln^2(x)$ (A.8)

Example, letting x=1 in (A.8) we obtain

$\int_0^1\frac{\ln^2(w)}{1-w}\,dw=2\operatorname{Li}_{3}(1)=2\zeta(3)$

Reference

POLYLOGARITHMS, MULTIPLE ZETA VALUES, AND THE SERIES OF HJORTNAES AND COMTET Notes by Tim Jameson (December 2009)

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