In terms of Hurwitz zeta function, Dirichlet beta function is given by
$$\beta(s)=\frac1{4^s}\left(\zeta(s,\frac14)+\zeta(s,\frac34)\right).$$
Following the links, by means of analytic continuation of Hurwitz zeta function, I found that
$$\beta(s)=\frac{\Gamma(1-s)}{4\pi i}\int_\gamma\frac{z^{s-1}}{\cosh z} dz$$
where $\gamma$ is the contour
And by using the functional equation of Dirichlet beta function, I found $\beta(2k+1)$, as shown by $(7)$, here.
I don't know much about these topics. I am following the links and formulas. Is my analytic continuation correct? Can it be useful for computation of $\beta(2k)$ which are related to Euler's zigzag numbers?
I would appreciate any help/reference in these topics. Thanks in advance.
Best Answer
The Dirichlet beta function
$$\beta(s)=4^{-s} \left(\zeta\left(s,\frac{1}{4}\right)-\zeta\left(s,\frac{3}{4}\right)\right)=\sum\limits_{n=0}^{\infty } \frac{(-1)^n}{(2 n+1)^s}\,,\quad\Re(s)>0\tag{1}$$
can be derived from the term-wise Mellin transform
$$\beta(s)=\frac{1}{2\, \Gamma(s)} \mathcal{M}_z\left[\frac{1}{\cosh (z)}\right](s)=\frac{1}{2\, \Gamma(s)} \int\limits_0^{\infty} \frac{z^{s-1}}{\cosh (z)} \, dz\tag{2}$$
of the series representation
$$\frac{1}{\cosh(z)}=2 \sum\limits_{n=0}^{\infty} \frac{(-1)^n}{e^{\,(2 n+1) z}}\tag{3}.$$
The value of $\beta(s)$ at positive even integers can be derived from the functional equation
$$\beta(1-s)=\left(\frac{2}{\pi}\right)^s\, \sin\left(\frac{\pi s}{2}\right)\, \Gamma(s)\, \beta(s)\tag{4}$$
via the formula
$$\beta(2 k)=\underset{s\to 2 k}{\text{lim}}\left(\frac{\beta(1-s)}{\left(\frac{2}{\pi}\right)^s\, \sin\left(\frac{\pi s}{2}\right)\, \Gamma(s)}\right)\tag{5}.$$
The globally convergent series
$$\beta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^{K+1}} \sum\limits_{n=0}^K \frac{(-1)^n}{(2 n+1)^s} \sum\limits_{k=0}^{K-n} \binom{K+1}{K-n-k}\right)\tag{6}$$
has the nice property that it evaluates exactly correct at a non-positive integer $n$ for all integer values of $K\ge -n$, so the Dirichlet beta function can be evaluated at non-positive integers most simply by the formula
$$\beta(-K)=\frac{1}{2^{K+1}} \sum\limits_{n=0}^K \frac{(-1)^n}{(2 n+1)^{-K}} \sum\limits_{k=0}^{K-n} \binom{K+1}{K-n-k},\quad K\in\mathbb{Z}\land K\ge0\tag{7}.$$