2
$\begingroup$

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:

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

|cite|improve this question
$\endgroup$
  • $\begingroup$ $\zeta(1/2)$ is negative, indeed (approximately $-1.4603545088095868128894991525152980125$). $\endgroup$ – Professor Vector Nov 27 '17 at 14:19
  • $\begingroup$ @cactus314, what do you mean by "evaluate the integral"? $\zeta(1/2)$ is probably transcendental, so there's probably no nice closed form for it. $\endgroup$ – Peter Humphries Nov 29 '17 at 13:54
  • $\begingroup$ In any case, note that for $\Re(s) > 1$, partial summation allows us to write $\zeta(s) = s \int_{1}^{\infty} \lfloor x \rfloor x^{-s} \, \frac{dx}{x}$, which is $s \int_{1}^{\infty} x^{-s + 1} \, \frac{dx}{x} - s \int_{1}^{\infty} \{x\} x^{-s} \, \frac{dx}{x}$. The first term is $\frac{s}{s - 1}$ by evaluating the integral, while the second converges absolutely for all $\Re(s) > 0$. So this gives the analytic continuation $\zeta(s) = \frac{s}{s - 1} - s \int_{1}^{\infty} \{x\} x^{-s} \, \frac{dx}{x}$ valid for $\Re(s) > 0$ with $s \neq 1$. $\endgroup$ – Peter Humphries Nov 29 '17 at 13:59
  • $\begingroup$ Note that this is equal to $s \int_{0}^{\infty} \{x\} x^{-s} \, \frac{dx}{x}$ for $0 < \Re(s) < 1$. Anyway, taking $s = 1/2$ shows that $\zeta(\frac{1}{2}) = -1 - \frac{1}{2} \int_{1}^{\infty} \{x\} x^{-1/2} \, \frac{dx}{x}$; both terms are negative, so clearly $\zeta(1/2) < 0$. (In fact, it is easy to see that $\zeta(\sigma) > 1$ for $\sigma > 1$ and $\zeta(\sigma) < 0$ for $0 < \sigma < 1$ via this expression.) $\endgroup$ – Peter Humphries Nov 29 '17 at 14:01
2
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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,

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

|cite|improve this answer
$\endgroup$
2
$\begingroup$

Consider that for any $s$ such that $\text{Re}(s)>0$ we have: $$ \eta(s)=\sum_{n\geq 1}\frac{(-1)^s}{n^s} = \left(1-\frac{2}{2^s}\right)\sum_{n\geq 1}\frac{1}{n^s} = \left(1-\frac{2}{2^s}\right)\zeta(s) $$ but the LHS is conditionally convergent for any $s$ such that $\text{Re}(s)>0$.
This provides an analytic continuation: $$\forall s:\text{Re}(s)>0,\qquad \zeta(s)=\left(1-\frac{2}{2^s}\right)^{-1}\sum_{n\geq 1}\frac{(-1)^{n+1}}{\Gamma(s)}\int_{0}^{+\infty}t^{s-1} e^{-nt}\,dt $$ and by the dominated convergence theorem: $$\forall s:\text{Re}(s)>0,\qquad \zeta(s)=\frac{1}{\Gamma(s)}\left(1-\frac{2}{2^s}\right)^{-1}\int_{0}^{+\infty}\frac{t^{s-1}}{e^t+1}\,dt$$ such that: $$ \zeta\left(\tfrac{1}{2}\right)=-\frac{2(1+\sqrt{2})}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{du}{1+\exp(u^2)}\approx -1.46. $$

|cite|improve this answer
$\endgroup$
  • $\begingroup$ This is a different analytic continuation than the one I'm asking about. Certainly it gives the same answer. Here I am emphasizing resummation. $\endgroup$ – cactus314 Nov 28 '17 at 3:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.