From Basel problem to multiple zeta values (Q6649376)

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scientific article; zbMATH DE number 7954776
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From Basel problem to multiple zeta values
scientific article; zbMATH DE number 7954776

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    From Basel problem to multiple zeta values (English)
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    5 December 2024
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    Let \(\zeta\) denote the Riemann zeta function.The aim of the paper under review is to show the following integral evaluations\N\begin{align*}\N\int_0^1 \frac{\arcsin x \arccos x}{x} \, dx &= \frac{7}{8} \zeta(3), \\\N\frac{2}{\pi} \int_0^1 \frac{\arcsin^2 x \arccos x}{x} \, dx &= \frac{1}{8} \zeta(3),\N\end{align*}\Nand discuss their applications to central binomial series, multiple zeta values, and some other sums shown as Theorems 2.5, 2.6, 2.7, 2.11, 2.16, 2.23. The main idea is to evaluate integrals involving powers of arcsine function.\N\NTheorem 2.6.\NLet \(G\) denote the Catalan constant\N\[\NG = 1 - \frac{1}{3^2} + \frac{1}{5^2} - \frac{1}{7^2} + \cdots \approx 0.915966 \dots\N\]\NThen\N\begin{align*}\N\int_0^1 \frac{\mathrm{arcsinh} \, x}{\sqrt{1-x^2}} \, dx &= G, \\\N\int_0^1 \frac{\mathrm{arcsinh} \, x\arccos x}{x} \, dx &= \frac{\pi^3}{32}.\N\end{align*}\N\NTheorem 2.7.\N\[\N\int_0^1 \frac{\arctan x \operatorname{arcctg} x}{x} \, dx = \frac{7}{8} \zeta(3).\N\]\N\NFor positive integers \(i_1, \dots, i_k\) such that \(i_1 \geq 2\), define the multiple zeta value and multiple \(t\)-value by\N\begin{align*}\N\zeta(i_1, i_2, \dots, i_k) &= \sum_{n_1 > n_2 > \cdots > n_k > 0} \frac{1}{n_1^{i_1} n_2^{i_2} \cdots n_k^{i_k}}, \\\Nt(i_1, i_2, \dots, i_k) &= \sum_{\substack{n_1 > n_2 > \cdots > n_k > 0 \\ n_j \text{ odd}}} \frac{1}{n_1^{i_1} n_2^{i_2} \cdots n_k^{i_k}}.\N\end{align*}\NLet \(\{m\}^n\) denote the sequence \((m, m, \dots, m)\) of length \(n\).\N\NTheorem 2.16.\NFor \(n \geq 0\),\N\begin{align*}\N\zeta(3, \{2\}^n) &= \frac{2}{\pi} \int_0^1 \frac{2^{2n+3}\arcsin^{2n+2} x \arccos x}{ (2n+2)! x} \, dx, \\\Nt(3, \{2\}^n) &= \int_0^1 \frac{\arcsin^{2n+1} x \arccos x}{(2n+1)! x} \, dx.\N\end{align*}\N\NThe author finishes the article with the following conjecture.\N\NConjecture 2.26.\N\[\N\sum_{k=0}^\infty \frac{\binom{2k}{k}}{2^{2k} (2k+1)^4} = \frac{1}{48} \left( 6\pi \zeta(3) + 4\pi \log^3 2 + \pi^3 \log 2 \right).\N\]\N\NRecently, quite similar series were evaluated by \textit{J. Ablinger} in [Exp. Math. 26, No. 1, 62--71 (2017; Zbl 1365.05009)] by method of iterated integrals, integration by parts, and generating functions.The author expects that the conjecture can be proved with some similar idea.
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    Apéry constant
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    arcsine function
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    Basel problem
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    Maclaurin series
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    multiple zeta values
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    Riemann zeta function
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    Wallis integral
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