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)

" 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)

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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.