Leibniz formula for \(\pi\)
First, we will derive the following Leibniz formula for \(\pi\):
\[
\frac{\pi}{4} = \frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots.
\]
【Derivation of the formula】
The zeros of the function \(\sin \left( \frac{\pi (z + 1)}{4} \right)\) are \(z = -1, 3, -5, 7, -9, \cdots\), it holds that
\[
c \cdot \sin \left( \frac{\pi (z + 1)}{4} \right) = \left( 1 + \frac{z}{1} \right)\left( 1-\frac{z}{3} \right)\left( 1 + \frac{z}{5} \right)\left( 1-\frac{z}{7} \right) \cdots
\] where \(c\) is the constant. If you substitute 0 for \(z\), you will get \(c = \sqrt 2\). Therefore, it holds that
\[
\sqrt{2} \sin \left( \frac{\pi (z + 1)}{4} \right) = \left( 1 + \frac{z}{1} \right)\left( 1-\frac{z}{3} \right)\left( 1 + \frac{z}{5} \right)\left( 1-\frac{z}{7} \right) \cdots.
\] On the other hand, it holds that
\begin{align}
\sqrt{2} \sin \left( \frac{\pi (z + 1)}{4} \right)
&= \sqrt{2} \sin \left( \frac{\pi}{4} + \frac{\pi z}{4} \right) \\
&= \sqrt{2} \left( \sin \frac{\pi}{4} \cdot \cos \frac{\pi z}{4} + \cos \frac{\pi}{4} \cdot \sin \frac{\pi z}{4} \right) \\
&= \cos \frac{\pi z}{4} + \sin \frac{\pi z}{4} \\
&= \left( 1-\frac{1}{2!} \left( \frac{\pi z}{4} \right)^2 + \frac{1}{4!} \left( \frac{\pi z}{4} \right)^4-\cdots \right)
+
\left( \frac{1}{1!} \cdot \frac{\pi z}{4}-\frac{1}{3!} \left( \frac{\pi z}{4} \right)^3 + \frac{1}{5!} \left( \frac{\pi z}{4} \right)^5-\cdots \right) \\
&= 1 + \frac{\pi}{4} z-\frac{\pi ^ 2}{2! \cdot 4 ^ 2 } z^2-\frac{\pi ^ 3}{3! \cdot 4^3} z^3 + \frac{\pi ^ 4}{4! \cdot 4^4} z^4 + \cdots.
\end{align} Therefore, the coefficient of \(z\) is
\[
\frac{\pi}{4} = \frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots.
\]
The value of \(\beta(3)\)
The Dirichlet beta function is defined as
\[
\beta(s) = \sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n + 1)^s}.
\] In this section, we will derive the following formula:
\[
\beta(3) = \frac{\pi ^ 3}{32}.
\]
【Derivation of the formula】
Let \(x_1, \cdots, x_n\) be variables, denote for \(k \geq 1\) by \(p_k(x_1, \cdots, x_n)\) or \(p_k\) the k-th power sum:
\[
p_k = p_k(x_1, \cdots, x_n) = \sum_{i = 1}^n x_i^k,
\] and for \(k \geq 0\) denote by \(e_k(x_1, \cdots, x_n)\) or \(e_k\) the elementary symmetric polynomial:
\begin{align}
e_0 &= e_0(x_1, \cdots, x_n) = 1, \\
e_1 &= e_1(x_1, \cdots, x_n) = x_1 + x_2 + \cdots + x_n, \\
e_2 &= e_2(x_1, \cdots, x_n) = \sum_{1 \leq i < j \leq n} x_i x_j, \\
\vdots \\
e_n &= e_n(x_1, \cdots, x_n) = x_1 x_2 \cdots x_n.
\end{align} Then, Newton’s identities can be stated as
\[
k e_k(x_1, \cdots, x_n) = \sum_{i = 1}^k e_{k-i}(x_1, \cdots, x_n) p_i (x_1, \cdots, x_n),
\] valid for all \(n \geq k \geq 1\). In this post, let \(x_j = \frac{(-1)^{j-1}}{2j-1}\) and \(n \rightarrow \infty\). Then, it holds that
\begin{align}
\sqrt{2} \sin \left( \frac{\pi (z + 1)}{4} \right) &= (1+x_1 z)(1+x_2 z)(1+x_3 z)(1+x_4 z) \cdots \\
&= e_0 + e_1 z + e_2 z^2 + e_3 z^3 + e_4 z^4 + \cdots.
\end{align} Therefore, \(e_0 = 1, e_1 = \frac{\pi}{4}, e_2 = -\frac{\pi^2}{2! \cdot 4^2}, e_3 = -\frac{\pi^3}{3! \cdot 4^3}, \cdots\). Furthermore, \(p_1 = e_1 = \frac{\pi}{4}\). Since Newton’s identities, it holds that
\[
2 e_2 = e_1 p_1-e_0 p_2,
\] so we can get
\[
p_2 = \frac{\pi^2}{8},
\] i.e.
\[
\frac{\pi^2}{8} = \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots.
\] Since Newton’s identities again,
\[
3 e_3 = e_2 p_1-e_1 p_2 + e_0 p_3.
\] Therefore, we can get
\[
3 \cdot \left(-\frac{\pi^3}{3! \cdot 4^3} \right) =-\frac{\pi^2}{2! \cdot 4^2} \cdot \frac{\pi}{4}-\frac{\pi}{4} \cdot \frac{\pi^2}{8} + p_3
\] i.e.
\[
\beta(3) = p_3 = \frac{\pi^3}{32}.
\] In this way, we can calculate the values of \(\beta(5), \beta(7), \beta(9), \cdots\).
References
Basel problem – Wikipedia: https://en.wikipedia.org/wiki/Basel_problem
Newton’s identities – Wikipedia: https://en.wikipedia.org/wiki/Newton%27s_identities
Dirichlet beta function – Wikipedia: https://en.wikipedia.org/wiki/Dirichlet_beta_function
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