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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:

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

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

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,

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

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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: $$ \zeta\left(\tfrac{1}{2}\right)=-\frac{2(1+\sqrt{2})}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{du}{1+\exp(u^2)}\approx -1.46. $$

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  • $\begingroup$ 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

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