MHB Prove the following zeta property :

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$$ \zeta(s) = s \int^{\infty}_1 \,\frac{ [ t ] }{t^{s+1}} \, =\,\frac{s}{s-1} \, -s \int^{\infty}_1 \frac{ \{ t \} } {t^{s+1}}\,dt $$
 
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Hi Zaid,

This is just a special case of Abel summation formula,

$$\sum_{1 \le n \le x} a_n \phi(n) = A(x)\phi(x) - \int_1^x A(u)\phi'(u) \, \mathrm{d}u \,$$

Put $$a_n = 1$$, $$\phi(x) = \frac{1}{x^s}$$ and $$A(x) = \lfloor x \rfloor$$

Hope this helps,

Balarka
.
 
ZaidAlyafey said:
$$ \zeta(s) = s \int^{\infty}_1 \,\frac{ [ t ] }{t^{s+1}} \, =\,\frac{s}{s-1} \, -s \int^{\infty}_1 \frac{ \{ t \} } {t^{s+1}}\,dt $$

The first equality can be proved computing the integral...

$\displaystyle \int_{1}^{\infty} \frac{[t]}{t^{s+1}}\ dt = \sum_{n=1}^{\infty} \int_{n}^{n+1} \frac{n}{t^{s+1}}\ dt = \frac{1}{s}\ \sum_{n=1}^{\infty} n \{- \frac{1}{(n+1)^{s}} + \frac{1}{n^{s}}\} = \frac{\zeta(s)}{s}$

Kind regards

$\chi$ $\sigma$
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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