Riemann Zeta Function
I have a couple question regarding the Riemann zeta function and its extension onto the complex plane. I know the function is given by the reflection formula .
My first question is how is this proved?
My second question is how does this extend to the complex plane?
The way I look at it is to plug in a value already known, which would be some and solve for . But this only solves for all leaving \ unaccounted for...
I obviously know nothing about CA, but just so you can cross-check whatever anybody else here gives you, I will transcribe a proof from "Riemann's Zeta Function" H.M. Edwards (Dover)
Originally Posted by chiph588@
" For negative real values of s, Riemann evaluated the integral as follows. Let be the domain in the -plane which consists of all other points other than those which lie within of the positve real axis or within of one of the singularities of the integrand. Let be the boundary of oriented in the usual way. Then, ignoring for the moment the fact that is not compact, Cauchy's theorem gives
Now one component of this integral is the integral with the orientation reversed, whereas the others are integrals of the circles oriented clockwise. Thus when the circles are oriented in the usual counterclockwise sense becomes
The integrals over the circles can be evaluated by setting for to find
by the Cauchy integral formula. Summing over all integers other than and using then gives
Finally using the simplification
One obtains the desired formula
In order to prove rigorously that the above holds for negative s, it suffices to modify the above argument letting be the intersection of with the disk and letting .; then the integral splits into two parts, one being an integral over the circle with the points within of the positive real axis deleted, and the other being an integral whose limit as is the left side of . The first of these two parts approaches zero because the length of the part of integration is less than , because the modulus of on the circle is for . Thus the second part, which by Cauchy's theorem is the negative of the first part, also approaches zero, which implies and hence .
This completes the proof of the functional equation in the case . However, both sides of are analytic functions of , so this suffices to prove for all values of (except for where one or more of terms in have poles)
Note two things
A) In the first integral the limits of integration are meant to represent a path of integration which begins at , moves to the left down the positive real axis, circles the origin once in the counterclockwise direction, and returns up the positive real axis to
B) I only did this because I almost never see anyone on this site give an answer regarding the RZ function. If this does not make sense I am sorry. Take this proof with a grain of salt, for I know nothing of this and am just copying it from a book. I could easily have copied the wrong thing. If something confuses you or this does not seem like the pertinent proof just disgregard the thing in its entirety. And if need be there is an alternate proof.
In the interests of referencing accuracy, the exact transcription (whch starts first line of p13) is as follows:
The substitution of for as the symbol for the gamma function has also been made.
Originally Posted by Mathstud28
The substitution of various colours of * rather than numbers to label equations has also been made.
A reference to a footnote has been omitted.
The zeta function is initially defined as . This series converges for all complex numbers s whose real part is greater than 1. To extend the domain of definition, the usual procedure (as far as I understand it—I'm not an expert in this area) is to use analytic continuation to define it for all s with real part greater than 0 (except for the pole at s=1). An intricate argument using contour integration (as in the previous comments) then establishes the functional equation for s in the strip 0<re(s)<1. The functional equation is then used to define in the remainder of the complex plane.
Originally Posted by chiph588@