Computing $\int_0^{1/2}\frac{\sinh^{-1}(u)}{u} \,du=\frac{\pi^2}{20}$, $\zeta(2)=\frac53 \sum_{n=0}^\infty \frac{(-1)^n\binom{2n}{n}}{2^{4n}(2n+1)^2}$

Question

Edit: I followed @J.G. suggestion and spoted my mistake, so I fixed it and and left the correct answer below in case someone is interested in it.

I am trying to compute the following integral $$\int_0^{1/2}\frac{\sinh^{-1}(u)}{u} \,du=\frac{\pi^2}{20}$$ As I will show below, instead, I got an extra term in the solution, i.e. $\int_0^{1/2}\frac{\sinh^{-1}(u)}{u} \,du=\ln(\phi)\ln(\phi^2-1)+\frac{\pi^2}{20}$.

I appreciate if you can help me find my mistake and how to fix it.

The background for this integral is the following: I came across the following infinite series

\begin{align*} \zeta(2)=\frac53 \sum_{n=0}^\infty \frac{(-1)^n\binom{2n}{n}}{2^{4n}(2n+1)^2} \end{align*}

To compute it I started from the $\arcsin(x)$ expansion, divided both sides by $x$

\begin{align*} \frac{\arcsin(x)}{x} &=\sum_{n=0}^\infty \frac{\binom{2n}{n}}{2^{2n}}\frac{x^{2n}}{2n+1} \end{align*}

Then set $x \to ix$ to obtain

\begin{align*} \frac{\sinh^{-1}(x)}{x} &=\sum_{n=0}^\infty \frac{\binom{2n}{n}}{2^{2n}}\frac{(-1)^nx^{2n}}{2n+1} \tag{1} \end{align*}

Now, if we integrate $(1)$ from $0$ to $x$ we get

\begin{align*} \sum_{n=0}^\infty \frac{\binom{2n}{n}}{2^{2n}}\frac{(-1)^n x^{2n}}{(2n+1)^2}&=\frac{1}{x}\int_0^x\frac{\sinh^{-1}(u)}{u} \,du \tag{2} \end{align*}

letting $x=\frac12$ in $(2)$ we obtain

\begin{align*} &\sum_{n=0}^\infty \frac{\binom{2n}{n}}{2^{4n}}\frac{(-1)^n }{(2n+1)^2}\\ &=2\int_0^{1/2}\frac{\sinh^{-1}(u)}{u} \,du\\ &=2\int_0^{\ln(\phi)}\frac{x}{\sinh(x)}\cosh(x)\,dx &(u=\sinh(x))\\ &=2\int_0^{\ln(\phi)}x\coth(x)\,dx\\ &=2\left(x\ln\left(\sinh(x)\right)\Big|_0^{\ln(\phi)}-\int_0^{\ln(\phi)}\ln\left(\sinh(x)\right)\,dx\right) &(\text{i.b.p.})\\ &=2\left(\ln(\phi)\ln\left(\frac{e^{\ln(\phi)}-e^{-\ln(\phi)}}{2}\right)-\int_0^{\ln(\phi)}\left(-\ln(2)+x-\sum_{n=1}^\infty\frac{e^{-2nx}}{n} \right)\,dx\right)\\ &=2\left(-\ln(2)\ln(\phi)+\ln(\phi)\ln(\phi^2-1)-\ln^2(\phi)+\ln(2)\ln(\phi)-\frac{\ln^2(\phi)}{2}-\frac12\sum_{n=1}^\infty\frac{e^{-2nx}}{n^2}\Big|_0^{\ln(\phi)}\right)\\ &=2\left(-\ln(2)\ln(\phi)+\ln^2(\phi)-\ln^2(\phi)+\ln(2)\ln(\phi)-\frac{\ln^2(\phi)}{2}+\frac{\zeta(2)}{2}-\frac12\operatorname{Li}_2(\phi^{-2})\right)\\ &=2\left(-\frac{\ln^2(\phi)}{2}+\frac{\zeta(2)}{2}-\frac12\left(\frac{\pi^2}{15}-\ln^2(\phi) \right)\right)\\ &=\frac{\pi^2}{10} \end{align*}

where I used the fact that $ \sinh^{-1}(x)=\ln\left(x+\sqrt{1+x^2} \right)$ and therefore $\sinh^{-1}\left(\frac12\right)=\ln\left(\frac{1+\sqrt{5}}{2}\right)=\ln(\phi)$, where $\phi$ is the golden ratio. I also relied on the expansion $\ln\left( \sinh(x)\right)=-\ln 2+x-\sum_{n=1}^{\infty}\frac{e^{-2nx}}{n}$ and that $\operatorname{Li}_2\left(\phi^{-2})\right)=\frac{\pi^2}{15}-\ln^2(\phi)$.

The expansion for $\ln\left( \sinh(x)\right)$ was derived as following:

\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}

Answer 1

Letting $x=\sinh t$ in $$I=\int_{0}^{1/2} \frac{\sinh^{-1}x}{x} dx$$ $\frac12=\frac{e^t-e^{-t}}{2} \implies t=\ln \phi, \,\phi=\frac{1+\sqrt{5}}{2},$ we get $$I=\int_{0}^{\ln \phi} t \coth t~ dt=t\ln \sin t\large{|}_{0}^{\ln \phi}-\int_{0}^{\ln \phi} \ln\sin t dt$$ $$\implies I=\ln \phi \ln (\sin \ln \phi)- \int_{0}^{\ln \phi} \ln\left[\frac{e^t-e^{-t}}{2}\right]dt $$ $$\implies I=-\ln \phi \ln 2-(t^2/2-t\ln 2)\large{|}_{0}^{\ln {\phi}}-\int_{0}^{\ln \phi} \ln[1-e^{-2t}] dt$$ $$\implies I=-\frac{1}{2}ln^2 \phi +\int_{0}^{\ln \phi} \sum_{k=0}^{\infty}\frac{e^{-2kt}}{k} dt$$ $$\implies I= -\frac{1}{2}ln^2 \phi -\sum_{0}^{\infty} \frac{e^{-2kt}}{2k^2}\large{|}_{0}^{\ln \phi}$$ $$\implies I=-\frac{1}{2}\ln^2\phi-\frac{1}{2} Li_2(1/\phi^2)+\frac{1}{2}\zeta(2)$$ Using $Li_2(1/\phi^2)=[\frac{\pi^2}{15}-\ln^2 \phi]$ see https://en.wikipedia.org/wiki/Polylogarithm#Relationship_to_other_functions We finally get $$I=\frac{-\pi^2}{30}+\frac{\pi^2}{12}=\frac{\pi^2}{20}.$$