Welcome to our community

Be a part of something great, join today!

Prove the following zeta property :

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle \zeta(s) = s \int^{\infty}_1 \,\frac{ [ t ] }{t^{s+1}} \, =\,\frac{s}{s-1} \, -s \int^{\infty}_1 \frac{ \{ t \} } {t^{s+1}}\,dt \)
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Hi Zaid,

This is just a special case of Abel summation formula,

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

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

Hope this helps,

Balarka
.
 

chisigma

Well-known member
Feb 13, 2012
1,704
\(\displaystyle \zeta(s) = s \int^{\infty}_1 \,\frac{ [ t ] }{t^{s+1}} \, =\,\frac{s}{s-1} \, -s \int^{\infty}_1 \frac{ \{ t \} } {t^{s+1}}\,dt \)
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

$\chi$ $\sigma$