Odd values for Dirichlet beta function

Question

I would like to find a proof for the generating formula for odd values of Dirichlet beta function, namely: $$\beta(2k+1)=\frac{(-1)^kE_{2k}\pi^{2k+1}}{4^{k+1}(2k)!}$$

My try was to start with the cosine infinite product $$\cos{(\pi x)}=\prod_{n\ge 1} \left(1-\frac{4x^2}{(2n-1)^2}\right)\;,$$ Logarithmate and differentiate it, in order to get: $$-\pi\tan{(\pi x)}= \sum_{n\ge 1} \frac{\frac{-8x}{(2n-1)^2}}{1-\frac{4x^2}{(2n-1)^2}}$$ Then multiplying by $x$ and expanding the denominator into a geometric series, yields: $$\pi x\tan{(\pi x)}= 2\sum_{n\ge 1} \sum_{k\ge 0} \left(\frac{2x}{(2n-1)}\right)^{2k}\frac{4x^2}{(2n-1)}=2\sum_{n\ge 0} \sum_{k\ge 0} \left(\frac{2x}{(2n+1)}\right)^{2k+1}$$ Now, already in this post it has been shown how to expand $\tan(x)$ into power series, but I have $x\tan(x)$ and I don't know how to equate the coefficients this way. Can I get some help with this?

EDIT: Related: Euler numbers grow $2\left(\frac{2}{ \pi }\right)^{2 n+1}$-times slower than the factorial? However I am still interested in how to show this using my idea.

Answer 1

Here is a proof that I enjoyed :)

Consider, for $k \in \mathbb{Z}^+$, the following integral $$I_k = \frac{1}{2\pi i}\oint_{|z| = \pi (2k+1)} \left( \frac{ze^{xz}}{e^z - 1} \right) \frac{dz}{z^{2r+2}} $$

Let us name the integrand $f$, which has simple poles at $z=0$ and $z_n = 2\pi i n$ (for $n \in \mathbb{Z}, n \neq 0$). The term inside paranthesis compels us to use the generating function of the Bernoulli polynomials, namely

$$f(z) = \frac{1}{z^{2r+2}} \left( \sum_{n=0}^{\infty} B_n(x) \frac{z^n}{n!} \right) = \sum_{n=0}^{\infty} \frac{B_n(x)}{n!} z^{n - (2r+2)}$$

The residue at $z=0$ is is the coefficient of $z^{-1}$, which occurs when $n = 2r+1$. Hence we have $$\operatorname{Res}(f, z=0) = \frac{B_{2r+1}(x)}{(2r+1)!}$$

Also, the residue at $z_n$ can be evaluated $$\operatorname{Res}(f, z=z_n)= \left( \lim_{z \to z_n} \frac{e^{xz}}{z^{2r+1}} \right) \cdot \left( \lim_{z \to z_n} \frac{z - z_n}{e^z - 1} \right) = \frac{e^{2\pi i n x}}{(2\pi i n)^{2r+1}}$$

Thus, by Cauchy's Residue Formula, the integral $I_k$ is the sum of all residues inside the contour (which includes poles from $n=-k$ to $n=k$). \begin{align} I_k &= \operatorname{Res}(f, 0) + \sum_{n=1}^{k} \left[ \text{Res}(f, z_n) + \text{Res}(f, z_{-n}) \right] \\ &= \frac{B_{2r+1}(x)}{(2r+1)!} + \sum_{n=1}^{k} \left[ \frac{e^{2\pi i n x}}{(2\pi i n)^{2r+1}} + \frac{e^{2\pi i (-n) x}}{(2\pi i (-n))^{2r+1}} \right]\\ &= \frac{B_{2r+1}(x)}{(2r+1)!} + \sum_{n=1}^{k} \frac{e^{2\pi i n x} - e^{-2\pi i n x}}{(2\pi i n)^{2r+1}} \\ &= \frac{B_{2r+1}(x)}{(2r+1)!} + \sum_{n=1}^{k} \frac{2i \sin(2\pi n x)}{(2\pi n)^{2r+1} \cdot i^{2r+1}} \\ &= \frac{B_{2r+1}(x)}{(2r+1)!} + \sum_{n=1}^{k} \frac{2 \sin(2\pi n x)}{(-1)^r (2\pi)^{2r+1} n^{2r+1}} \\ &= \frac{B_{2r+1}(x)}{(2r+1)!} + \frac{2(-1)^r}{(2\pi)^{2r+1}} \sum_{n=1}^{k} \frac{\sin(2\pi n x)}{n^{2r+1}} \end{align}

If we name the contour $C_k$, by the Estimation lemma, we have that $$|I_k| \le R_k \cdot \frac{1}{R_k^{2r+1}} \cdot \max_{z \in C_k} \left| \frac{e^{xz}}{e^z - 1} \right| = \frac{1}{R_k^{2r}} \max_{z \in C_k} \left| \frac{e^{xz}}{e^z - 1} \right| \le \frac{M}{R_k^{2r}}$$

so that $\lim_{k \to \infty} I_k = 0$. Then, taking the limit as $k \rightarrow \infty$, and rearranging $0 = \sum \operatorname{Res}$, we have

$$B_{2r+1}(x) = \frac{(-1)^{r+1} 2 (2r+1)!}{(2\pi)^{2r+1}} \sum_{m \geq 1} \frac{\sin(2\pi m x)}{m^{2r+1}}$$

This is actually the Fourier expansion of the Bernoulli functions of odd order. (In fact, the proof of $\zeta(2r)$ uses a similar fourier expansion for even order Bernoulli functions; the one above famously fails any attempt at $\zeta(2r+1)$) By setting $x=1/4$ in the formula above, we obtain

$$1 - \frac{1}{3^{2r+1}} + \frac{1}{5^{2r+1}} - \frac{1}{7^{2r+1}} + \cdots = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^{2r+1}} = \frac{(-1)^{r+1} (2\pi)^{2r+1}}{2 (2r+1)!} B_{2r+1}(1/4)$$

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The form OP gave in the original post can be obtained upon using the identity given in this post.