Tuesday 24 September 2019

complex analysis - Evaluating $zetaleft(frac{1}{2}right)$ as an integral $ zetaleft(frac{1}{2}right) = frac{1}{2} int_0^infty frac{[x]-x}{x^{3/2}} , dx$




I am reading the second chapter of Titchmarsh's book on the Riemann Zeta Function. I would have written:



$$ \zeta\left(\frac{1}{2}\right) = 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots = \infty $$



If you think about it for a moment. This doesn't decay nearly fast enough, and so the sequence diverges. Then I had to look up the actual definition of $\zeta(s)$ in the region $s = \sigma + it$ and $0 < \sigma < 1$. We have:



$$ \zeta(s) = \left\{ \begin{array}{cl}
\sum \frac{1}{n^s} & \mathrm{Re}(s) > 1 \\ \\
s \int_0^\infty \frac{[x]-x}{x^{s+1}} dx & 1 > \text{Re}(s) > 0

\end{array} \right. $$



Then if I evaluate at $s = \frac{3}{2} = \frac{3}{2} + 0i$ we use the second formula:



$$ \zeta\left(\frac{1}{2}\right) = \frac{1}{2} \int_0^\infty \frac{[x]-x}{x^{3/2}} \, dx = \sum_{n=0}^\infty \left[ 2 \sqrt{n} - 2 \sqrt{n+1} - \frac{1}{\sqrt{n+1}} \right] \stackrel{?}{<} 0$$



Is this thing negative? Is $\zeta(\frac{1}{2}) < 0$. The book overs several "analytic continuations" and I'm only looking at this first one, to make sure I understand.



Could someone help me evaluate the integral? I didn't use any fancy changes of variables. The main step is:




$$ \int_0^\infty f(x) \,dx = \sum \int_{n}^{n+1} f(x) \,dx = \int_0^1 f(x) \,dx + \int_1^2 f(x) \,dx + \dots $$


Answer



$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}

\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$




You can take advantage of the
following identity:





\begin{align}
\zeta\pars{1 \over 2} & =
\sum_{k = 1}^{N}{1 \over \root{k}} - 2\root{N} - {1 \over 2}\int_{N}^{\infty}{x - \left\lfloor x\right\rfloor \over x^{3/2}}\,\dd x\,,\qquad N = 1,2,3,\ldots
\end{align}




Note that
$\ds{0 < \verts{{1 \over 2}\int_{N}^{\infty}{x - \left\lfloor x\right\rfloor \over x^{3/2}}\,\dd x} < {1 \over \root{N}}}$ such that you don't need to evaluate the integral. Namely,





$$
\bbx{\zeta\pars{1 \over 2} =
\lim_{N \to \infty}\pars{\sum_{k = 1}^{N}{1 \over \root{k}} - 2\root{N}}}
$$


No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...