오늘의 계산 29 - Alternating Euler Sum involving Binomial (2n, n)

Problem. Prove the following identities. \begin{equation}\label{eqn:wts} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^3 \binom{2n}{n}} = \frac{2}{5} \zeta(3), \qquad \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^3 2^n \binom{2n}{n}} = \frac{1}{4} \zeta(3) - \frac{1}{6}\log^3 2 \end{equation}

We divide the proof into several lemmas.

Lemma 1. For $|x| < 1$. \begin{equation}\label{eqn:wts_lem_1} \int_{0}^{x} \frac{\mathrm{arcsinh}^2 t}{t} \, dt = \frac{1}{4} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^3 \binom{2n}{n}} (2x)^{2n}. \end{equation}

Proof. Let $A_n$ and $B_n$ be the sequence of rational numbers defined as follows: \begin{equation*} A_0 = 1, \quad A_{n} = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)}, \qquad B_n = \frac{2 \cdot 4 \cdot 6 \cdots (2n)}{3 \cdot 5 \cdot 7 \cdots (2n+1)} \end{equation*} Then we claim that \begin{equation*} A_{n+1} = \sum_{k=0}^{n} \frac{A_k A_{n-k}}{2k+2}, \quad B_{n} = \sum_{k=0}^{n} \frac{A_k A_{n-k}}{2k+1}. \end{equation*} The first equality is easily deduced, since \begin{equation*} \frac{1}{\sqrt{1-x}} = \sum_{n=0}^{\infty} A_n x^n \quad \text{and} \quad \sqrt{1-x} = 1 - x \sum_{n=0}^{\infty} \frac{A_n}{2n+2} x^n. \end{equation*} The second equality can be shown by induction. Clearly, it holds for $n = 0$. Now assume it holds for $n$. Then \begin{align*} & \sum_{k=0}^{n+1} \frac{A_k A_{n+1-k}}{2k+1} \\ & = \frac{A_{n+1}}{2n+3} + \sum_{k=0}^{n} \frac{A_k A_{n-k}}{2k+1} \left( 1 - \frac{1}{2n+2-2k} \right) \\ & = \frac{A_{n+1}}{2n+3} + B_n - \frac{1}{2n+3} \sum_{k=0}^{n} A_k A_{n-k} \left( \frac{1}{2k+1} + \frac{1}{2n+2-2k} \right) \\ & = \frac{A_{n+1}}{2n+3} + \frac{2n+2}{2n+3} B_n - \frac{1}{2n+3} \sum_{k'=0}^{n} \frac{A_{k'} A_{n-k'}}{2k'+2} \qquad (k' = n-k) \\ & = B_{n+1} \end{align*} as desired. But since \begin{equation*} B_{n-1} = \frac{2^{2n-1}}{n \binom{2n}{n}}, \end{equation*} we have \begin{equation*} \frac{\mathrm{arcsinh} \, x}{\sqrt{1+x^2}} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \binom{2n}{n}} (2x)^{2n-1}, \end{equation*} which leads to \eqref{eqn:wts_lem_1}.

Lemma 2. Let $\phi = \frac{1+\sqrt{5}}{2}$ be the golden ratio. Then \begin{eqnarray*} \mathrm{Li}_2 \big( \tfrac{1}{\phi^2} \big) & = & \frac{2}{5} \zeta(2) - \log^2 \phi \\ \mathrm{Li}_3 \big( \tfrac{1}{\phi^2} \big) & = & \frac{4}{5} \zeta(3) - \frac{4}{5} \zeta (2) \log \phi + \frac{2}{3} \log^3 \phi \\ \mathrm{Li}_2 \left( \tfrac{1}{2} \right) & = & \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2 \\ \mathrm{Li}_3 \left( \tfrac{1}{2} \right) & = & \frac{7}{8} \zeta(3) - \frac{1}{2} \zeta (2) \log 2 + \frac{1}{6} \log^3 2 \end{eqnarray*}

Proof. These are consequences of the following identities:

1. For $z \notin (-\infty, 0) \cup (1, \infty)$, \begin{equation*} \mathrm{Li}_2 (z) + \mathrm{Li}_2 (1-z) = \zeta(2) - \log z \log (1-z). \end{equation*}
2. For $z \notin (-\infty, 0]$, \begin{equation*} \mathrm{Li}_2 (1-z) + \mathrm{Li}_2 \big(1-\tfrac{1}{z}\big) = - \frac{1}{2} \log^2 z \end{equation*}
3. For $|z| < 1$, \begin{equation*} \mathrm{Li}_s (z) + \mathrm{Li}_s (-z) = 2^{1-s} \mathrm{Li}_s (z^2) \end{equation*}
4. For $z \notin (-\infty, 0] \cup (1, \infty)$, \begin{equation*} \mathrm{Li}_3 (z) + \mathrm{Li}_3 (1-z) + \mathrm{Li}_3 \big(1 - \tfrac{1}{z} \big) = \zeta(3) + \zeta(2) \log z + \frac{1}{6} \log^3 z - \frac{1}{2} \log^2 z \log (1-z) \end{equation*}
For example, plugging $z = \frac{1}{\phi}$ to these equalities and using the relation $\phi^2 - \phi - 1 = 0$ gives the linear system of equations \begin{align*} \mathrm{Li}_2 \big( \tfrac{1}{\phi} \big) + \mathrm{Li}_2 \big( \tfrac{1}{\phi^2} \big) & = \zeta(2) - 2 \log^2 \phi \\ \mathrm{Li}_2 \big( \tfrac{1}{\phi^2} \big) + \mathrm{Li}_2 \big( -\tfrac{1}{\phi} \big) & = -\frac{1}{2} \log^2 \phi \\ \mathrm{Li}_2 \big( \tfrac{1}{\phi} \big) + \mathrm{Li}_2 \big( -\tfrac{1}{\phi} \big) - \frac{1}{2} \mathrm{Li}_2 \big( \tfrac{1}{\phi^2} \big) & = 0 \end{align*} and \begin{align*} \mathrm{Li}_3 \big( \tfrac{1}{\phi} \big) + \mathrm{Li}_3 \big( \tfrac{1}{\phi^2} \big) + \mathrm{Li}_3 \big( -\tfrac{1}{\phi} \big) & = \zeta(3) - \zeta(2) \log \phi - \frac{5}{6} \log^3 \phi \\ \mathrm{Li}_3 \big( \tfrac{1}{\phi} \big) + \mathrm{Li}_3 \big( -\tfrac{1}{\phi} \big) & = \frac{1}{4} \mathrm{Li}_3 \big( \tfrac{1}{\phi^2} \big), \end{align*} which give rise to two first equalities. Similarly, plugging $z = \frac{1}{2}$ gives two other equalities.

Solution of the Problem. Using the substitution $t = \frac{u^2-1}{2u}$ to the integral in Lemma 1, we have \begin{equation*} \int_{0}^{x} \frac{\mathrm{arcsinh}^2 t}{t} \, dt = - \int_{1}^{x+\sqrt{x^2+1}} \left( \frac{2u}{1-u^2} + \frac{1}{u} \right) \log^2 u \, du. \end{equation*} Now let's consider the function $F(s)$, defined for $s > 0$ by \begin{equation*} F(s) = - \int_{1}^{s} \left( \frac{2u}{1-u^2} + \frac{1}{u} \right) \log^2 u \, du. \end{equation*} (Here, $F(s)$ is used to denote a new function and is not intended to refer to the elliptic integral.) It is easy to show that $F(s) = F(1/s)$, thus we may assume $0 < s < 1$. Then expanding the rational function part of the integrand and integrating term by term immediately yields \begin{equation*} F(s) = -\frac{1}{3} \log^3 s + \log (1-s^2) \log^2 s + \mathrm{Li}_2 (s^2) \log s + \frac{1}{2} \zeta(3) - \frac{1}{2} \mathrm{Li}_3 (s^2). \end{equation*} Then \begin{equation*} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^3 \binom{2n}{n}} = 4 \int_{0}^{\frac{1}{2}} \frac{\mathrm{arcsinh}^2 t}{t} \, dt = 4 F(\phi) = 4 F \left( \frac{1}{\phi} \right) \end{equation*} and Lemma 2 yields the first identity. Similarly, \begin{equation*} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^3 2^n \binom{2n}{n}} = 4 \int_{0}^{\frac{1}{2\sqrt{2}}} \frac{\mathrm{arcsinh}^2 t}{t} \, dt = 4 F(\sqrt{2}) = 4 F \left( \frac{1}{\sqrt{2}} \right) \end{equation*} and again Lemma 2 yields the second identity, completing the proof of \eqref{eqn:wts}.