Analytic continuation of Dirichlet beta funcion

Question

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.

Answer 1

The contour integral representation is correct, and it's very similar to the representation $$\zeta(s) = \frac{\Gamma(1-s)}{2 \pi i } \int_{\gamma} \frac{z^{s-1}}{e^{\color{red}{-}z}-1} \, \mathrm dz. $$

Using the principal branch of $z^{s}$ and assuming that $\Re(s) >0$, integrate the function $$f(z) = \frac{z^{s-1}}{\cosh (z)} $$ counterclockwise on a contour $(\mathcal{C})$ that consists of a ray just below the branch cut on the negative real axis, a small semicircle about the origin, and a ray just above the branch cut on the negative real axis.

We get $$\begin{align} \int_{\mathcal{C}} \frac{z^{s-1}}{\cosh(z)} \, \mathrm dz &= \int_{\infty}^{\epsilon} \frac{(te^{-i \pi})^{s-1}}{\cosh(te^{- i \pi})} \, e^{- i \pi} \, \mathrm dt + \int_{|r|= \epsilon} f(z) \, \mathrm dz + \int_{\epsilon}^{\infty} \frac{(te^{i \pi})^{s-1}}{\cosh (te^{ i \pi})} \, e^{i \pi }\mathrm \, \mathrm dt \\ &=-e^{- i \pi s}\int^{\infty}_{\epsilon} \frac{t^{s-1}}{\cosh(t)} \, \mathrm dt + \int_{|r|= \epsilon} f(z) \, \mathrm dz + e^{i \pi s} \int_{\epsilon}^{\infty} \frac{t^{s-1}}{\cosh (t)} \, \mathrm dt. \end{align}$$

Since $\Re(z) >0$, the second integral on the right vanishes as $\epsilon \to 0$, and we get

$$ \begin{align} \lim_{\epsilon \to 0}\int_{\mathcal{C}} \frac{z^{s-1}}{\cosh(z)} \, \mathrm dz &= \left(-e^{- i \pi s} + e^{i \pi s} \right) \int_{0}^{\infty} \frac{t^{s-1}}{\cosh (t)} \, \mathrm dt \\ &= 2i \sin \left(\pi s \right) \, 2 \, \Gamma(s) \, \beta(s) \\ &= \frac{4 \pi i \, \beta(s)}{\Gamma(1-s)} . \end{align}$$

However, since the magnitude of $\frac{1}{\cosh (z)}$ decays exponentially to zero as $\Re(z) \to - \infty$, the integral $$\int_{\gamma} \frac{z^{s-1}}{\cosh (z)} \, \mathrm dz$$ has the same value as long as it doesn't curve around any of the poles of $\frac{1}{\cosh(z)}$ on the imaginary axis.

(See this related question that I asked a long time ago.)

Therefore, we indeed have $$\beta(s) = \frac{\Gamma(1-s)}{4 \pi i} \int_{\mathcal{\gamma}} \frac{z^{s-1}}{\cosh(z)} \, \mathrm dz$$ for all $s \in \mathbb{C}. $

The function $\Gamma(1-s)$ has simple poles at the positive integers, but the limit of the expression on the right side is finite because the integral vanishes when $s$ is a positive integer.

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To derive the functional equation, let's close the contour with a square with vertices at $z= \pm 2N \pi \pm i 2N \pi$, where $N$ is some large positive integer.

The function $\frac{1}{\cosh(z)}$ has simple poles at the imaginary half- integers.

But on the square, $\frac{1}{\cosh (z)}$ is uniformly bounded.

So by the estimation lemma, the integral on the square vanishes as $N \to \infty$ through the positive integers if $\operatorname{Re}(s) <0.$

(A more subtle argument shows that it actually vanishes for $\Re(s) <1$. It has to do with the fact that the magnitude of $\operatorname{sech}(z)$ is going to $0$ exponentially fast as $\Re(z) \to \pm \infty$.)

Therefore, for $\Re(s) <0$, we have $$ \begin{align} \beta(s) &= - \frac{\Gamma(1-s)}{4 \pi i } \, 2 \pi i \left( \sum_{k=0}^{\infty} \operatorname{Res} \left[ f(z), \frac{(2k+1) \pi e^{-i \pi/2}}{2} \right] + \sum_{k=0}^{\infty} \operatorname{Res} \left[ f(z), \frac{(2k+1) \pi e^{i \pi/2}}{2} \right] \right) \\ &= - \frac{\Gamma(1-s)}{2} \left( \sum_{k=0}^{\infty} \frac{\left(\tfrac{(2k+1) \pi e^{-i \pi/2}}{2} \right)^{s-1}}{\sinh \left(\tfrac{(2k+1)\pi e^{- i \pi/2}}{2} \right)} + \sum_{k=0}^{\infty} \frac{\left(\tfrac{(2k+1) \pi e^{i \pi/2}}{2} \right)^{s-1}}{\sinh \left(\tfrac{(2k+1)\pi e^{i \pi/2}}{2} \right)}\right) \\ &= -\frac{\Gamma(1-s)}{2} \left(\frac{\pi}{2} \right)^{s-1} \, \frac{1}{i} \left(-e^{-i \pi(s-1)/2} +e^{i \pi(s-1)/2} \right)\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^{1-s}} \\ &= -\Gamma(1-s) \left(\frac{\pi}{2} \right)^{s-1} \sin \left(\frac{\pi(s-1)}{2} \right) \beta(1-s). \end{align}$$

Replacing $s$ with $1-s$, we have

$$ \begin{align} \beta(1-s) &= - \Gamma(s) \left(\frac{\pi}{2} \right)^{-s} \sin \left(- \frac{\pi s}{2} \right) \beta(s) \\ &= \left(\frac{\pi}{2} \right)^{-s}\sin \left( \frac{\pi s}{2} \right) \Gamma(s) \beta(s). \end{align} $$

Answer 2

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

$$\frac{1}{\cosh(z)}=2 \sum\limits_{n=0}^{\infty} \frac{(-1)^n}{e^{\,(2 n+1) z}}\tag{3}.$$

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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

$$\beta(2 k)=\underset{s\to 2 k}{\text{lim}}\left(\frac{\beta(1-s)}{\left(\frac{2}{\pi}\right)^s\, \sin\left(\frac{\pi s}{2}\right)\, \Gamma(s)}\right)\tag{5}.$$

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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

$$\beta(-K)=\frac{1}{2^{K+1}} \sum\limits_{n=0}^K \frac{(-1)^n}{(2 n+1)^{-K}} \sum\limits_{k=0}^{K-n} \binom{K+1}{K-n-k},\quad K\in\mathbb{Z}\land K\ge0\tag{7}.$$