[snip]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
denote the domain in the
-plane which consists of all
[snip] points other than those which lie within
of the positve real axis or within
of one of the singularities
of the integrand of
(*). 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
over 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
![\begin{aligned}\zeta(s)&=\sum_{n=1}^{\infty}\Gamma(-s)\left[(-2\pi i n)^{s-1}+(2\pi i n)^{{\color{red}s}-1}\right]\\&=\Gamma(-s)(2\pi)^{s-1}\left[i^{s-1}+(-i)^{s-1}\right]\sum_{n=1}^{\infty}\frac{1}{n^{s-1}}\end{aligned}](http://www.mathhelpforum.com/math-help/latex2/img/23b2b5ccd38e135f7142e41d83915a64-1.gif "\begin{aligned}\zeta(s)&=\sum_{n=1}^{\infty}\Gamma(-s)\left[(-2\pi i n)^{s-1}+(2\pi i n)^{{\color{red}s}-1}\right]\\&=\Gamma(-s)(2\pi)^{s-1}\left[i^{s-1}+(-i)^{s-1}\right]\sum_{n=1}^{\infty}\frac{1}{n^{s-1}}\end{aligned}")
Finally using the simplification
![\begin{aligned}i^{s-1}+(-i)^{s-1}&=\frac{1}{i}\left[e^{s\ln(i)}-e^{s\ln(-i)}\right]\\&=\frac{1}{i}\left[e^{\frac{s\pi i}{2}}-e^{\frac{-s\pi i}{2}}\right]\\&=2\sin\left(\frac{s\pi}{2}\right)\end{aligned}](http://www.mathhelpforum.com/math-help/latex2/img/7c8f055b9b961ff702672a2b06039938-1.gif "\begin{aligned}i^{s-1}+(-i)^{s-1}&=\frac{1}{i}\left[e^{s\ln(i)}-e^{s\ln(-i)}\right]\\&=\frac{1}{i}\left[e^{\frac{s\pi i}{2}}-e^{\frac{-s\pi i}{2}}\right]\\&=2\sin\left(\frac{s\pi}{2}\right)\end{aligned}")
, one obtains the desired formula
. This relationship between 
and 
is known as the functional equation of the zeta function.
In order to prove rigorously that
[the above
] holds for
s < 0, it suffices to modify the above argument
by 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
path of integration is less than
, because
the factor 
is bounded on the circle 
, and because the modulus of
on
this circle is
![\left|x^{s-1}\right|\leqslant \left[(2n+1)\pi\right]^{{\color{red}\delta} -1}](http://www.mathhelpforum.com/math-help/latex2/img/6b393ffa2d6dcdd6aa80fcdc9353b019-1.gif "\left|x^{s-1}\right|\leqslant \left[(2n+1)\pi\right]^{{\color{red}\delta} -1}")
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
[reference to footnote] one or more of terms
of 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
,
Mr F says: A path cannot begin at + infinity.
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
Mr F says: This is not the boundary of D.
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.