Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Floor and ceiling functions
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
===Riemann zeta function (ΞΆ)=== The fractional part function also shows up in integral representations of the [[Riemann zeta function]]. It is straightforward to prove (using integration by parts)<ref>Titchmarsh, p. 13</ref> that if <math>\varphi(x)</math> is any function with a continuous derivative in the closed interval [''a'', ''b''], :<math>\sum_{a<n\le b}\varphi(n) = \int_a^b\varphi(x) \, dx + \int_a^b\left(\{x\}-\tfrac12\right)\varphi'(x) \, dx + \left(\{a\}-\tfrac12\right)\varphi(a) - \left(\{b\}-\tfrac12\right)\varphi(b). </math> Letting <math>\varphi(n) = n^{-s}</math> for [[real part]] of ''s'' greater than 1 and letting ''a'' and ''b'' be integers, and letting ''b'' approach infinity gives :<math>\zeta(s) = s\int_1^\infty\frac{\frac12-\{x\}}{x^{s+1}}\,dx + \frac{1}{s-1} + \frac 1 2.</math> This formula is valid for all ''s'' with real part greater than −1, (except ''s'' = 1, where there is a pole) and combined with the Fourier expansion for {''x''} can be used to extend the zeta function to the entire complex plane and to prove its functional equation.<ref>Titchmarsh, pp.14β15</ref> For ''s'' = ''Ο'' + ''it'' in the critical strip 0 < ''Ο'' < 1, :<math>\zeta(s)=s\int_{-\infty}^\infty e^{-\sigma\omega}(\lfloor e^\omega\rfloor - e^\omega)e^{-it\omega}\,d\omega.</math> In 1947 [[Balthasar van der Pol|van der Pol]] used this representation to construct an analogue computer for finding roots of the zeta function.<ref>Crandall & Pomerance, p. 391</ref>
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)