View source for Functional equation for the zeta function

The '''functional equation for Riemann zeta function''' is a result due to analytic continuation of [[Riemann zeta function]]:

<cmath>
\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)\Gamma(1-s)\zeta(1-s)
</cmath>

== Proof ==

=== Preparation ===

There are multiple proofs for the functional equation for Riemann zeta function, and this page presents a light-weighted approach which merely relies on the Fourier series for the first periodic [[Bernoulli polynomial]] that

<cmath>
B_1(x)\triangleq\{x\}-\frac12=-\sum_{n=1}^\infty{\sin(2\pi nx)\over\pi n}
</cmath>

and the Laplace transform identity that

<cmath>
{\Gamma(z)\over r^z}=\int_0^\infty\tau^{z-1}e^{-z\tau}\mathrm d\tau
</cmath>

where <math>-\pi/2\le\arg z\le\pi/2</math>

=== A formula for <math>\zeta(s)</math> in <math>-1<\sigma<0</math> ===

In this article, we will use the common convention that <math>s=\sigma+it</math> where <math>\sigma,t\in\mathbb R</math>. As a result, we say that the original [[Dirichlet series]] definition <math>\zeta(s)\triangleq\sum_{k=1}^\infty{1\over k^s}</math> converges only for <math>\sigma>1</math>. However, if we were to apply Euler-Maclaurin summation on this definition, we obtain

<cmath>
\zeta(s)=\frac12+{s\over s-1}-s\int_1^\infty{B_1(x)\over x^{s+1}}\mathrm dx
</cmath>

in which we can extend the ROC of the latter integral to <math>\sigma>-1</math> via [[integration by parts]]:

<cmath>
\begin{align*}
\int_1^\infty{B_1(x)\over x^{s+1}}\mathrm dx
&=\left.{B_2(x)\over2x^{s+1}}\right|_1^\infty+{s+1\over2}\int_1^\infty{B_2(x)\over x^{s+2}}\mathrm dx \\
&={B_2\over2}+{s+1\over2}\int_1^\infty{B_2(x)\over x^{s+2}}\mathrm dx
\end{align*}
</cmath>

When <math>-1<\sigma<0</math> there is

<cmath>
-s\int_0^1{B_1(x)\over x^{s+1}}\mathrm dx=\frac12+{1\over s-1}
</cmath>

As a result, we obtain a formula for <math>\zeta(s)</math> for <math>-1<\sigma<0</math>:

<cmath>
\zeta(s)=-s\int_0^\infty{B_1(x)\over x^{s+1}}\mathrm dx
</cmath>

=== Expansion of <math>B_1(x)</math> into [[Fourier series]] ===

In order to go deeper, let's plug

<cmath>
B_1(x)\triangleq\{x\}-\frac12=-\sum_{n=1}^\infty{\sin(2\pi nx)\over\pi n}
</cmath>

into the previously obtained formula, so that

<cmath>
\begin{align*}
\zeta(s)
&=s\underbrace{\int_0^\infty\sum_{n=1}^\infty{\sin(2\pi nx)\over n\pi}{\mathrm dx\over x^{s+1}}}_{y=2\pi nx} \\
&=s\int_0^\infty\sum_{n=1}^\infty{\sin y\over n\pi}{\mathrm dy/2\pi n\over(y/2\pi n)^{s+1}} \\
&=\sum_{n=1}^\infty{2s\over(2\pi n)^{1-s}}\int_0^\infty{\sin y\over y^{s+1}}\mathrm dy \\
&=2^s\pi^{s-1}\zeta(1-s)\int_0^\infty{\sin y\over y^{s+1}}\mathrm dy
\end{align*}
</cmath>

Therefore, the remaining step is to handle the integral

=== Evaluation of <math>\int_0^\infty{\sin y\over y^{s+1}}\mathrm dy</math> ===

By [[Euler's formula]], we have

<cmath>
\sin\theta={e^{i\theta}-e^{-i\theta}\over2i}
</cmath>

As a result, we only need to calculate

<cmath>
\int_0^\infty{e^{\pm iy}\over y^{s+1}}\mathrm dy
</cmath>

if we want to take down the remaining integral. According to Laplace transform identities, we can see that

<cmath>
\int_0^\infty{e^{\pm iy}\over y^{s+1}}\mathrm dy=\int_0^\infty y^{-s-1}e^{-e^{\mp i\pi/2}y}\mathrm dy=\Gamma(-s)e^{\mp i\pi s/2}
</cmath>

Thus we deduce

<cmath>
\int_0^\infty{\sin y\over y^{s+1}}\mathrm dy=-\Gamma(-s)\sin\left(\pi s\over2\right)
</cmath>

wherein the RHS serves to be a meromorphic continuation of the LHS integral.

=== Proof of the functional equation ===

With everything ready, we can put everything together and obtain

<cmath>
\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)(-s)\Gamma(-s)\zeta(1-s)
</cmath>

and by <math>\Gamma(z+1)=z\Gamma(z)</math>, this identity becomes the functional equation:

<cmath>
\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)\Gamma(1-s)\zeta(1-s)
</cmath>

== Resources ==

* Titchmarsh, E. C., ''The Theory of the Riemann Zeta-Function.'' Oxford Univ. Press, London and New York, 1951.