Evaluating $\zeta\left(\frac{1}{2}\right)$ as an integral $ \zeta\left(\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:
$$  \int_0^\infty f(x) \,dx = \sum \int_{n}^{n+1} f(x) \,dx  = \int_0^1 f(x) \,dx + \int_1^2 f(x) \,dx + \dots $$
 A: 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:
$$ \zeta\left(\tfrac{1}{2}\right)=-\frac{2(1+\sqrt{2})}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{du}{1+\exp(u^2)}\approx -1.46. $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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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,

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