I am trying to evaluate the infinite series
$$\sum_{n=1}^\infty (-1)^{n-1} \frac{\overline{H}_n}{n \binom{2n}{n}}$$
where $\overline{H}_n = 1 – \frac{1}{2}+ \frac{1}{3} – \ldots+ \frac{(-1)^{n-1}}{n}$ is $n$-th skew harmonic number. My attempt was to use the beta function:
\begin{align*}
\sum_{n=1}^\infty (-1)^{n-1} \frac{\overline{H}_n}{n \binom{2n}{n}}
&= \sum_{n=0}^\infty (-1)^n \frac{\overline{H}_{n+1}}{(n+1) \binom{2n+2}{n+1}}\\
&= \frac{1}{2} \sum_{n=0}^\infty (-1)^n \frac{\overline{H}_{n+1} (n!)^2}{(2n+1)!}\\
&= \frac{1}{2} \sum_{n=0}^\infty (-1)^n \overline{H}_{n+1} \frac{\Gamma^2(n+1) }{\Gamma(2n+2)}\\
&= \frac{1}{2} \sum_{n=0}^\infty (-1)^n \overline{H}_{n+1} B(n+1, n+1)\\
&= \frac{1}{2} \sum_{n=0}^\infty (-1)^n \overline{H}_{n+1} \int_0^1 t^n(1-t)^n\,dt\\
&= \frac{1}{2}\int_0^1 \sum_{n=0}^\infty (-1)^n \overline{H}_{n+1} t^n(1-t)^n\,dt.
\end{align*}
Since
$$\sum_{n=1}^\infty \overline{H}_n x^n = \frac{\log(1+x)}{1-x},$$
we get that
\begin{align*}
\frac{1}{2}\int_0^1 \sum_{n=0}^\infty (-1)^n \overline{H}_{n+1} t^n(1-t)^n\,dt
&= -\frac{1}{2}\int_0^1\frac{\log(1-t(1-t))}{t(1-t)(1+t(1-t))} \,dt\\
&= \frac{1}{2}\underbrace{\int_0^1 \frac{\log(1-t(1-t))}{t-1}\,dt}_{= \frac{\pi^2}{18}} – \frac{1}{2}\underbrace{\int_0^1 \frac{\log(1-t(1-t))}{t}\,dt}_{= \frac{\pi^2}{18}}\\
&- \frac{1}{2}\int_0^1 \frac{\log(1-t(1-t))}{t^2 -t -1}\\
&= \frac{\pi^2}{18}-\frac{1}{2}\int_0^1 \frac{\log(t^3 + 1) – \log(t+1)}{t^2 – t -1}\,dt\\
&= \frac{\pi^2}{18} + \frac{1}{2}\underbrace{\int_0^1 \frac{\log(t+1)}{t^2 – t -1}\,dt}_{= \frac{\pi^2}{5\sqrt{5}}+\frac{1}{\sqrt{5}}\left(\mathrm{Li}_2(-2 – \sqrt{5}) – \mathrm{Li}_2(-2 + \sqrt{5})\right)} – \frac{1}{2}\int_0^1 \frac{\log(t^3+1)}{t^2 – t -1}\,dt\\
&= \frac{\pi^2}{18} + \frac{\pi^2}{10\sqrt{5}} + \frac{\mathrm{Li}_2(-2 – \sqrt{5}) – \mathrm{Li}_2(-2 + \sqrt{5})}{2\sqrt{5}} – \frac{1}{2}\int_0^1 \frac{\log(t^3+1)}{t^2 – t -1}\,dt.
\end{align*}
We are left with the integral
$$\int_0^1 \frac{\log(t^3+1)}{t^2-t-1} \,dt.$$
I am having trouble evaluating this integral. When computed by Wolfram Mathematica, it gives a result with a lot of polylogarithms and unit imaginary numbers $i$:
\begin{align*}
\int_0^1 \frac{\log(t^3+1)}{t^2-t-1} \,dt
&= \frac{\text{Li}_2\left(\frac{1}{2}-\frac{\sqrt{5}}{2}\right)}{\sqrt{5}}-\frac{\text{Li}_2\left(2-\sqrt{5}\right)}{\sqrt{5}}+\frac{\text{Li}_2\left(3-\sqrt{5}\right)}{\sqrt{5}}-\frac{2 \text{Li}_2\left(\frac{\sqrt{5}-i \sqrt{3}}{\sqrt{5}+1}\right)}{\sqrt{5}}\\
&-\frac{2 \text{Li}_2\left(\frac{i \sqrt{3}+\sqrt{5}}{\sqrt{5}+1}\right)}{\sqrt{5}}-\frac{\text{Li}_2\left(\frac{2}{\sqrt{5}+3}\right)}{\sqrt{5}}-\frac{2 \text{Li}_2\left(\frac{\sqrt{5}-1}{\sqrt{5}-i \sqrt{3}}\right)}{\sqrt{5}}-\frac{2 \text{Li}_2\left(\frac{\sqrt{5}-1}{i \sqrt{3}+\sqrt{5}}\right)}{\sqrt{5}}-\frac{2 \pi ^2}{3 \sqrt{5}}\\
&+\frac{\log ^2(2)}{2 \sqrt{5}}-\frac{\log ^2\left(-\frac{i}{\sqrt{3}+i \sqrt{5}}\right)}{\sqrt{5}}-\frac{\log ^2\left(\frac{i}{\sqrt{3}-i \sqrt{5}}\right)}{\sqrt{5}}-\frac{3 \log ^2\left(\sqrt{5}+1\right)}{2 \sqrt{5}}\\
&-\frac{\log ^2\left(\sqrt{5}+2\right)}{2 \sqrt{5}}-\frac{2 \log \left(\frac{\sqrt{3}+i}{\sqrt{3}+i \sqrt{5}}\right) \log \left(\sqrt{5}-1\right)}{\sqrt{5}} -\frac{2 \log \left(\frac{\sqrt{3}-i}{\sqrt{3}-i \sqrt{5}}\right) \log \left(\sqrt{5}-1\right)}{\sqrt{5}}\\
&-\frac{2 \log \left(-\frac{i}{\sqrt{3}+i \sqrt{5}}\right) \log \left(\sqrt{5}+1\right)}{\sqrt{5}} -\frac{2 \log \left(\frac{i}{\sqrt{3}-i \sqrt{5}}\right) \log \left(\sqrt{5}+1\right)}{\sqrt{5}}\\
&-\frac{\log \left(\sqrt{5}-1\right) \log \left(\sqrt{5}+3\right)}{\sqrt{5}}+\frac{\log \left(\sqrt{5}+1\right) \log \left(\sqrt{5}+3\right)}{\sqrt{5}}\\
&+\frac{2 \log \left(\sqrt{5}+1\right) \log \left(\frac{\sqrt{3}-i}{\sqrt{3}+i \sqrt{5}}\right)}{\sqrt{5}}+\frac{2 \log \left(\sqrt{5}+1\right) \log \left(\frac{\sqrt{3}+i}{\sqrt{3}-i \sqrt{5}}\right)}{\sqrt{5}}+\frac{\log (64) \log \left(\sqrt{5}-1\right)}{3 \sqrt{5}}\\
&-\frac{\log (64) \log \left(\sqrt{5}+1\right)}{3 \sqrt{5}}-\frac{\log (64) \log \left(\sqrt{5}+3\right)}{6 \sqrt{5}} \approx -0.176256827758015.
\end{align*}
How would we evaluate the remaining integral? Is there a better way to evaluate the infinite series? I still hope that its closed form does not include any imaginary units $i$.
Best Answer
Let $\mathcal{S}$ denote the sum of the following infinite series:
$$\mathcal{S}:=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\overline{H}_{n}}{n\binom{2n}{n}},$$
where here $\overline{H}_{n}$ denotes the $n$-th skew-harmonic number, defined for each positive integer $n$ by the finite sum
$$\overline{H}_{n}:=\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k};~~~\small{n\in\mathbb{N}}.$$
As you point out, it can be shown that the skew-harmonic numbers have the following generating function
$$\sum_{n=1}^{\infty}z^{n}\,\overline{H}_{n}=\frac{\ln{\left(1+z\right)}}{1-z};~~~\small{|z|<1}.$$
Using this, as well as the fact that
$$n\binom{2n}{n}\operatorname{B}{\left(n,n\right)}=2;~~~\small{n\in\mathbb{N}},$$
we find
$$\begin{align} \mathcal{S} &=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\overline{H}_{n}}{n\binom{2n}{n}}\\ &=\frac12\sum_{n=1}^{\infty}(-1)^{n-1}\overline{H}_{n}\operatorname{B}{\left(n,n\right)}\\ &=\frac12\sum_{n=1}^{\infty}(-1)^{n-1}\overline{H}_{n}\int_{0}^{1}\mathrm{d}t\,t^{n-1}\left(1-t\right)^{n-1}\\ &=\frac12\int_{0}^{1}\mathrm{d}t\,\sum_{n=1}^{\infty}(-1)^{n-1}\overline{H}_{n}t^{n-1}\left(1-t\right)^{n-1}\\ &=\frac12\int_{0}^{1}\mathrm{d}t\,\sum_{n=1}^{\infty}\overline{H}_{n}\left[-t\left(1-t\right)\right]^{n-1}\\ &=\frac12\int_{0}^{1}\mathrm{d}t\,\frac{1}{\left[-t\left(1-t\right)\right]}\sum_{n=1}^{\infty}\overline{H}_{n}\left[-t\left(1-t\right)\right]^{n}\\ &=\frac12\int_{0}^{1}\mathrm{d}t\,\frac{(-1)}{t\left(1-t\right)}\cdot\frac{\ln{\left(1-t(1-t)\right)}}{1+t(1-t)}\\ &=-\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{t\left(1-t\right)\left(1+t-t^{2}\right)}\\ &=-\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{t}-\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1-t}+\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}}\\ &=-\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{t}-\frac12\int_{0}^{1}\mathrm{d}u\,\frac{\ln{\left(1-u+u^{2}\right)}}{u};~~~\small{\left[t=1-u\right]}\\ &~~~~~+\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}}\\ &=-\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{t}+\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}}.\\ \end{align}$$
Consider the following two-variable variant of the dilogarithm:
$$\operatorname{Li}_{2}{\left(r,\theta\right)}:=-\frac12\int_{0}^{r}\mathrm{d}t\,\frac{\ln{\left(1-2t\cos{\left(\theta\right)}+t^{2}\right)}}{t};~~~\small{\left(r,\theta\right)\in\mathbb{R}^{2}}.$$
It can be shown that
$$\operatorname{Li}_{2}{\left(1,\theta\right)}=\frac14\left(\pi-\theta\right)^{2}-\frac{\pi^{2}}{12};~~~\small{0<\theta<\pi}.$$
The following particular case can be used to evaluate the first integral of $\mathcal{S}$:
$$-\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{t}=2\operatorname{Li}_{2}{\left(1,\frac{\pi}{3}\right)}=\frac12\left(\pi-\frac{\pi}{3}\right)^{2}-\frac{\pi^{2}}{6}=\frac{\pi^{2}}{18}.$$
Let $\mathcal{I}$ denote the value of the remaining definite integral:
$$\mathcal{I}:=-\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}}\approx0.0776351.$$
We find that $\mathcal{I}$ can be expressed in terms of the two-variable dilogarithm as follows:
$$\begin{align} \mathcal{I} &=-\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}}\\ &=-\frac12\int_{0}^{\frac12}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}}-\frac12\int_{\frac12}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}}\\ &=-\int_{0}^{\frac12}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}};~~~\small{symmetry}\\ &=-\int_{0}^{\frac12}\mathrm{d}t\,\frac{\ln{\left(\frac34+\left(\frac12-t\right)^{2}\right)}}{\frac54-\left(\frac12-t\right)^{2}}\\ &=-\int_{0}^{\frac12}\mathrm{d}u\,\frac{\ln{\left(\frac34+u^{2}\right)}}{\frac54-u^{2}};~~~\small{\left[t=\frac12-u\right]}\\ &=-\frac{2}{\sqrt{5}}\int_{0}^{\frac{1}{\sqrt{5}}}\mathrm{d}v\,\frac{\ln{\left(\frac34+\frac54v^{2}\right)}}{1-v^{2}};~~~\small{\left[u=\frac{\sqrt{5}}{2}v\right]}\\ &=-\frac{1}{\sqrt{5}}\int_{1}^{\frac{1-\frac{1}{\sqrt{5}}}{1+\frac{1}{\sqrt{5}}}}\mathrm{d}w\,\frac{(-2)}{(1+w)^{2}}\cdot\frac{(1+w)^{2}}{2w}\ln{\left(\frac{3+5\left(\frac{1-w}{1+w}\right)^{2}}{4}\right)};~~~\small{\left[v=\frac{1-w}{1+w}\right]}\\ &=-\frac{1}{\sqrt{5}}\int_{\frac{1-\frac{1}{\sqrt{5}}}{1+\frac{1}{\sqrt{5}}}}^{1}\mathrm{d}w\,\frac{1}{w}\ln{\left(\frac{3(1+w)^{2}+5(1-w)^{2}}{4(1+w)^{2}}\right)}\\ &=-\frac{1}{\sqrt{5}}\int_{\frac{\sqrt{5}-1}{\sqrt{5}+1}}^{1}\mathrm{d}w\,\frac{1}{w}\ln{\left(\frac{2-w+2w^{2}}{(1+w)^{2}}\right)}\\ &=-\frac{1}{\sqrt{5}}\int_{\phi^{-2}}^{1}\mathrm{d}w\,\frac{1}{w}\ln{\left(\frac{2-w+2w^{2}}{(1+w)^{2}}\right)};~~~\small{\phi:=\frac{1+\sqrt{5}}{2}}\\ &=-\frac{1}{\sqrt{5}}\int_{\phi^{-2}}^{1}\mathrm{d}w\,\frac{1}{w}\left[\ln{\left(2\right)}-2\ln{\left(1+w\right)}+\ln{\left(1-\frac12w+w^{2}\right)}\right]\\ &=-\frac{1}{\sqrt{5}}\int_{\phi^{-2}}^{1}\mathrm{d}w\,\frac{\ln{\left(2\right)}}{w}+\frac{1}{\sqrt{5}}\int_{\phi^{-2}}^{1}\mathrm{d}w\,\frac{2\ln{\left(1+w\right)}}{w}\\ &~~~~~-\frac{1}{\sqrt{5}}\int_{\phi^{-2}}^{1}\mathrm{d}w\,\frac{\ln{\left(1-\frac12w+w^{2}\right)}}{w}\\ &=-\frac{\ln{\left(2\right)}}{\sqrt{5}}\ln{\left(\phi^{2}\right)}-\frac{2}{\sqrt{5}}\int_{-1}^{-\phi^{-2}}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}}{x};~~~\small{\left[w=-x\right]}\\ &~~~~~-\frac{1}{\sqrt{5}}\int_{\phi^{-2}}^{1}\mathrm{d}w\,\frac{\ln{\left(1-2w\cos{\left(\theta\right)}+w^{2}\right)}}{w};~~~\small{\left[\theta:=\frac{\pi}{2}-\arcsin{\left(\frac14\right)}\right]}\\ &=-\frac{2\ln{\left(2\right)}\ln{\left(\phi\right)}}{\sqrt{5}}+\frac{2}{\sqrt{5}}\left[\operatorname{Li}_{2}{\left(-\phi^{-2}\right)}-\operatorname{Li}_{2}{\left(-1\right)}\right]\\ &~~~~~+\frac{2}{\sqrt{5}}\left[\operatorname{Li}_{2}{\left(1,\theta\right)}-\operatorname{Li}_{2}{\left(\phi^{-2},\theta\right)}\right]\\ &=\frac{2}{\sqrt{5}}\left[-\ln{\left(2\right)}\ln{\left(\phi\right)}+\frac14\left(\pi-\theta\right)^{2}+\operatorname{Li}_{2}{\left(-\phi^{-2}\right)}-\operatorname{Li}_{2}{\left(\phi^{-2},\theta\right)}\right].\\ \end{align}$$
Hence,
$$\begin{align} \mathcal{S} &=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\overline{H}_{n}}{n\binom{2n}{n}}\\ &=-\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{t}+\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}}\\ &=\frac{\pi^{2}}{18}-\mathcal{I}\\ &=\frac{\pi^{2}}{18}-\frac{2}{\sqrt{5}}\left[-\ln{\left(2\right)}\ln{\left(\phi\right)}+\frac14\left(\pi-\theta\right)^{2}+\operatorname{Li}_{2}{\left(-\phi^{-2}\right)}-\operatorname{Li}_{2}{\left(\phi^{-2},\theta\right)}\right]\\ &=\frac{\pi^{2}}{18}+\frac{2}{\sqrt{5}}\ln{\left(2\right)}\ln{\left(\phi\right)}-\frac{1}{2\sqrt{5}}\left(\pi-\theta\right)^{2}-\frac{2}{\sqrt{5}}\operatorname{Li}_{2}{\left(-\phi^{-2}\right)}+\frac{2}{\sqrt{5}}\operatorname{Li}_{2}{\left(\phi^{-2},\theta\right)}.\\ \end{align}$$