analytic continuation of Riemann zeta (using integral)

The Riemann zeta functionDlmfDlmfMathworldPlanetmath can be analytically continued to the whole complex planeMathworldPlanetmath minus the point 1 by means of an integralDlmfPlanetmath representation. Remember that the zeta functon is defined by the series


When s>1, this series converges; furthermore, this convergence is uniform on compact subsets of this half-plane, hence the series converges to an analytic functionMathworldPlanetmath on this half plane. However, the series diverges when we have s<1, so this series cannot be used to define the zeta functionMathworldPlanetmath in the whole complex plane, which is why we must make an analytic continuation.

To make this continuation, we start by changing the variable in an integration:


This provides us with an integral representation of our summand. Substituting this into the series, we find that


We note that


because the series converges, hence it is possible to interchange integration and summation and subsequently sum a geometric series.


As it stands, the integral representation we have is not of much use for analytically continuing the zeta function because the integral diverges when s<1 on account of the fact that the integrand behaves like x-s when x is close to zero. However, it is possible to make use of the theorem of Cauchy to move the path of integration away from zero.

Given a real number r>0, define the contour Cr on the Riemann surfaceDlmfPlanetmath of zs-1 as follows: Cr passes from + to r along a lift of the real axis, then continues along the circle of radius r clockwise, and finally goes from r to +.

We now examine the integral over such a contour by breaking it into three pieces.


We may estimate the third integral in absolute valueMathworldPlanetmathPlanetmathPlanetmath like so:


The expression x/(ex-1) represents an analytic function of x, and hence a bounded function of x in a neighborhood of 0. When s>1, it happens that limx0|x|xs-2=0, so


The third integral differs from the first integral by a phase, so they may be combined by pulling out this common factor. When s>0, we may take the limit as r approaches 0 after doing so to obtain the following:


Since, aside from the branch pointMathworldPlanetmath at 0, the only singularities of our integrand occur at multiplesMathworldPlanetmathPlanetmath of 2πi, it follows from Cauchy’s theorem that


whenever 0<a<2π and 0<b<2π, which trivially implies that


for any r between 0 and 2π. Therefore,


when z>1. This integral converges for all complex s because the exponential grows more rapidly than the power. Furthermore, this integral defines an analytic function of s, so we have an analytic continuation of the zeta function to the whole complex plane minus the point 1.

Title analytic continuation of Riemann zeta (using integral)
Canonical name AnalyticContinuationOfRiemannZetausingIntegral
Date of creation 2013-03-22 16:53:59
Last modified on 2013-03-22 16:53:59
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 21
Author rspuzio (6075)
Entry type Example
Classification msc 30B40
Classification msc 30A99
Related topic EstimatingTheoremOfContourIntegral
Related topic PeriodicityOfExponentialFunction
Related topic AnalyticContinuationOfRiemannZeta