2
$\begingroup$

Define a function $F(s)$ by:

$$F(s)=\sum_{n=0}^{\infty}(\sqrt n+1/3)^{-s}$$

Is there a closed form expression for the analytic continuation of $F(s)$ to $F(-s)$?

$\endgroup$

1 Answer 1

2
+50
$\begingroup$

Not sure whether this matches the query for closed form, but Euler-Maclaurin summation gives an exact expression for the analytic continuation to any $s\in\mathbb{C}\setminus\{1,2\}$:

For abbreviation, let's write $f(s,x):=(\sqrt{x}+a)^{-s}$ for $x\in\mathbb{R}$, $0<a\in\mathbb{R}$. For $a=\frac{1}{3}$ we have the OP.

For avoiding the singularity of the square root at $0$, split off some terms from the sum and apply Euler-Maclaurin summation to the remaining terms, where $N$ and $K$ are some arbitrarily chosen positive integers that do not influence the value of $F(s)$:

$$\begin{aligned} F(s) = & \sum_{n=0}^{N-1}f(s,n)+ \int_N^\infty f(s,x)\mathrm{d}x + \frac{1}{2}f(s,N)\\ &-\sum_{k=1}^K\frac{B_{2k}}{(2k)!}f^{(2k)}(s,N) +\frac{1}{(2K+1)!}\int_N^\infty \bar{B}_{2K+1}(x)f^{(2K+1)}(s,x)\mathrm{d}x \end{aligned}\tag{*}\label{eq}$$

Here, $B_{2k}$ denote Bernoulli numbers, $\bar{B}_{2K+1}(x) = B_{2K+1}(x -[x])$ the periodically continued Bernoulli polynomial of degree $2K+1$ and $f^{(n)}(s,x) = \left(\frac{\operatorname{d}}{\operatorname{d}x}\right)^nf(s,x)$ the $n$th derivation of $f(s,x)$ by $x$.

For $\Re s>2$, the first integral can be explicitly calculated

$$\int_N^\infty f(s,x)\mathrm{d}x = \int_N^\infty (\sqrt{x}+a)^{-s}\mathrm{d}x = 2\frac{(\sqrt{N}+a)^{2-s}}{s-2}-2a\frac{(\sqrt{N}+a)^{1-s}}{s-1},$$

giving a meromorphic function in $s$. In the other terms of equation \eqref{eq} the functions $f(s,x)$ and $f^{(n)}(s,x)$ are all analytic in $s$ and the second integral converges normally for $\Re s > -4K$, so that it is also analytic in this range of $s$.

Hence, \begin{align} F(s) = & \sum_{n=0}^{N-1}f(s,n)+ 2\frac{(\sqrt{N}+a)^{2-s}}{s-2}-2a\frac{(\sqrt{N}+a)^{1-s}}{s-1} + \frac{1}{2}f(s,N)\\ &-\sum_{k=1}^K\frac{B_{2k}}{(2k)!}f^{(2k)}(s,N) +\frac{1}{(2K+1)!}\int_N^\infty \bar{B}_{2K+1}(x)f^{(2K+1)}(s,x)\mathrm{d}x \end{align}

is an expression for the analytic continuation of $F(s)$ in the half plane $\Re s > -4K$. Using integration by parts of the remaining integral, the value of $K$ can be made arbitrary large (that's why it can be freely chosen), so this is no principal bound fixed bound. We see that (the analytic continuation of) $F(s)$ has two simple poles, one at $s=2$ with residue $\operatorname{Res}(F,2)=2$ and one at $s=1$ with residue $\operatorname{Res}(F,1)=-2a$, and is analytic everywhere else.

$\endgroup$
4
  • $\begingroup$ What do mean by " no principal bound " ? Not an error term ? Not a practical bound ? Sorry , Im unfamiliar with that term. $\endgroup$
    – mick
    Aug 14, 2020 at 20:35
  • 1
    $\begingroup$ Maybe that term not well chosen by me. I wanted to express that although once you choose a certain $K$, the validity of the formula is restricted to $\Re s > -4K$, there is no obstacle to go to lower values of $\Re s$ by choosing higher values of $K$. $\endgroup$
    – Uwe
    Aug 14, 2020 at 20:52
  • $\begingroup$ ah ok now I get it. the range is not infinite but as high as you want. $\endgroup$
    – mick
    Aug 14, 2020 at 20:53
  • 1
    $\begingroup$ Yes, but when one really wants to use this formula to calculate function values, then there will be some practical limit at large $K$. Since the higher-degree Bernoulli polynomials assume large values, this must be compensated with high values of $N$ to keep the estimation for the last integral small. The integral cannot be calculated exactly, but can be made as small as one wishes by going to higher $N$. The numerical problem is then the increase of the number of terms in the first sum. $\endgroup$
    – Uwe
    Aug 14, 2020 at 21:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .