Editing Riemann zeta function (section)

== Riemann's functional equation ==
This zeta function satisfies the [[functional equation]]
<math display="block"> \zeta(s) = 2^s \pi^{s-1}\ \sin\left( \frac{\pi s}{2} \right)\ \Gamma(1-s)\ \zeta(1-s)\ ,</math>
where {{math|Γ(''s'')}} is the [[gamma function]]. This is an equality of meromorphic functions valid on the whole [[complex plane]]. The equation relates values of the Riemann zeta function at the points {{mvar|s}} and {{math|1 − ''s''}}, in particular relating even positive integers with odd negative integers. Owing to the zeros of the sine function, the functional equation implies that {{math|''ζ''(''s'')}} has a simple zero at each even negative integer {{math|1=''s'' = −2''n''}}, known as the '''[[Triviality (mathematics)|trivial]] zeros''' of {{math|''ζ''(''s'')}}. When {{mvar|s}} is an even positive integer, the product {{nobr|{{math|sin({{sfrac| ''&pi; s'' | 2 }}) &Gamma;(1 − ''s'')}}}} on the right is non-zero because {{math|Γ(1 − ''s'')}} has a simple [[pole (complex analysis)|pole]], which cancels the simple zero of the sine factor.

{{Collapse top|title=Proof of Riemann's functional equation}}

A proof of the functional equation proceeds as follows:
We observe that if <math>\ s > 0\ ,</math> then
<math display="block"> \int_0^\infty x^{ \frac{1}{2} s - 1 } e^{-n^2\pi x}\ \operatorname{d} x\ =\ \frac{\ \Gamma\!\left( \frac{s}{2} \right)\ }{\ n^s\ \pi^{\frac{s}{2}}\ } ~.</math>

As a result, if <math>\ s > 1\ </math> then
<math display="block"> \frac{\ \Gamma\!\left(\frac{s}{2}\right)\ \zeta(s)\ }{\ \pi^{ \frac{s}{2} }\ }\ =\ \sum_{n=1}^\infty\ \int_0^\infty\ x^{{s\over 2}-1}\ e^{-n^2 \pi x}\ \operatorname{d} x\ =\ \int_0^\infty x^{{s\over 2}-1} \sum_{n=1}^\infty e^{-n^2 \pi x}\ \operatorname{d} x\ ,</math>
with the inversion of the limiting processes justified by absolute convergence (hence the stricter requirement on <math>s</math>).

For convenience, let
<math display="block"> \psi(x)\ := \ \sum_{n=1}^\infty\ e^{-n^2 \pi x} </math>

which is a special case of the [[theta function]].

Because <math>e^{-n^2 \pi x}</math> and <math>\frac1\sqrt{x} e^{\frac{-n^2 \pi}{x}}</math> are [[Fourier transform#Definition|Fourier transform pairs]],<ref name='Damm-Johnsen'>{{cite book |last=Damm-Johnsenn |first=Håvard |title=Theta functions and their applications |year=2019 |pages=5|url=https://users.ox.ac.uk/~quee4127/theta.pdf}}</ref> then, by the [[Poisson summation formula]], we have 
<math display="block"> \sum_{n=-\infty}^\infty\ e^{ - n^2 \pi\ x }\ =\ \frac{ 1 }{\ \sqrt{x\ }\ }\ \sum_{n=-\infty}^\infty\ e^{ -\frac{\ n^2 \pi\ }{ x } }\ ,</math>

so that
<math display="block">\ 2\ \psi(x) + 1\ =\ \frac{ 1 }{\ \sqrt{x\ }\ } \left(\ 2\ \psi\!\left( \frac{ 1 }{ x } \right) + 1\ \right) ~.</math>

Hence
<math display="block"> \pi^{ -\frac{s}{2} }\ \Gamma\!\left( \frac{s}{2} \right)\ \zeta(s)\ =\ \int_0^1\ x^{ \frac{s}{2} - 1 }\ \psi(x)\ \operatorname{d} x + \int_1^\infty x^{ \frac{s}{2} - 1 } \psi(x)\ \operatorname{d} x ~.</math>

The right side is equivalent to
<math display="block"> \int_0^1 x^{ \frac{s}{2} - 1 } \left( \frac{ 1 }{\ \sqrt{x\ }\ }\ \psi\!\left( \frac{1}{x} \right) + \frac{ 1 }{\ 2 \sqrt{x\ }\ } - \frac{ 1 }{ 2 }\ \right) \ \operatorname{d} x + \int_1^\infty x^{{s\over 2}-1} \psi(x)\ \operatorname{d} x </math>
or
<math display="block">
\frac{ 1 }{\ s - 1\ } - \frac{ 1 }{\ s\ } + \int_0^1\ x^{ \frac{s}{2} - \frac{3}{2}}\ \psi\!\left( \frac{ 1 }{\ x\ } \right)\ \operatorname{d} x + \int_1^\infty\ x^{ \frac{s}{2} - 1 }\ \psi(x)\ \operatorname{d} x
~.</math>

So
<math display="block">
\pi^{ -\frac{ s }{ 2 } }\ \Gamma\!\left( \frac{\ s\ }{ 2 } \right)\ \zeta(s)\ =\ \frac{ 1 }{\ s ( s - 1 )\ } + \int_1^\infty\ \left( x^{ -\frac{ s }{ 2 } - \frac{ 1 }{ 2 } } + x^{ \frac{ s }{ 2 } - 1 } \right)\ \psi(x)\ \operatorname{d} x
</math>

which is convergent for all {{mvar|s}}, because <math>\psi(x)\to0</math> quicker than any power of {{mvar|x}} for <math>x>1</math>, so the integral converges. As the RHS remains the same if {{mvar|s}} is replaced by {{nobr|{{math| 1 − ''s''}} .}},

<math display="block"> \frac{\ \Gamma\!\left(\ \frac{s}{2}\ \right)\ \zeta\!\left(\ s\ \right)\ }{\ \pi^{ \frac{s}{2}\ }\ }\ =\ \frac{\ \Gamma\!\left(\ \frac{1}{2} - \frac{s}{2}\ \right)\ \zeta\!\left(\ 1 - s\ \right)\ }{\ \pi^{ \frac{1}{2} - \frac{s}{2} }\ }  </math>

which is the functional equation attributed to [[Bernhard Riemann]].<ref>{{cite book |first=E.C. |last=Titchmarsh |year=1986 |title=The Theory of the Riemann Zeta Function |edition=2nd |publisher=Oxford Science Publications |place=[[Oxford]], UK |isbn=0-19-853369-1 |pages=21–22 }}</ref>

The functional equation above can be obtained using both the [[reflection formula]] and the [[Multiplication theorem#Gamma function–Legendre formula|duplication formula]].

First collect terms of <math>\pi</math>:

<math display="block">\Gamma\left(\frac{s}{2}\right)\zeta\left(s\right) = \Gamma\left(\frac{1}{2} - \frac{s}{2}\right)\zeta\left(1 - s\right)\pi^{s-\frac{1}{2}}</math>

Then multiply both sides by <math>\Gamma\left(1-\frac s2\right)</math> and use the reflection formula:

<math display="block">\Gamma\left(1-\frac s2\right)\Gamma\left(\frac{s}{2}\right)\zeta\left(s\right) = \Gamma\left(1-\frac s2\right)\Gamma\left(\frac{1}{2} - \frac{s}{2}\right)\zeta\left(1 - s\right)\pi^{s-\frac{1}{2}}</math>

<math display="block">\zeta\left(s\right) = \sin\left(\frac{\pi s}2\right)\Gamma\left(1-\frac s2\right)\Gamma\left(\frac{1}{2} - \frac{s}{2}\right)\zeta\left(1 - s\right)\pi^{s-\frac{3}{2}}</math>

Use the duplication formula with <math>z=\frac{1}{2} - \frac{s}{2}</math>

<math display="block">\zeta\left(s\right) = \sin\left(\frac{\pi s}2\right)2^{1-1+s}\sqrt{\pi}\Gamma\left(1-s\right)\zeta\left(1 - s\right)\pi^{s-\frac{3}{2}}</math>

so that

<math display="block">\zeta\left(s\right) = \sin\left(\frac{\pi s}2\right)2^s\Gamma\left(1-s\right)\zeta\left(1 - s\right)\pi^{s-1}</math>

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The functional equation was established by Riemann in his 1859 paper "[[On the Number of Primes Less Than a Given Magnitude]]" and used to construct the analytic continuation in the first place.