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Difference between revisions of "Functional equation for the zeta function"

(→‎Two useful identities)
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B_1(x)\triangleq\{x\}-\frac12=-\sum_{n=1}^\infty{\sin(2\pi nx)\over\pi n}
 
B_1(x)\triangleq\{x\}-\frac12=-\sum_{n=1}^\infty{\sin(2\pi nx)\over\pi n}
 
</cmath>
 
</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> ===
 
=== A formula for <math>\zeta(s)</math> in <math>-1<\sigma<0</math> ===
Line 31: Line 39:
 
&={B_2\over2}+{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*}
 
\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
 +
\zeta(s)=s\int_0^\infty\sum_{n=1}^\infty{\sin(2\pi nx)\over n\pi}{\mathrm dx\over x^{s+1}}
 
</cmath>
 
</cmath>

Revision as of 02:37, 13 January 2021

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

\[\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)\Gamma(1-s)\zeta(1-s)\]

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

\[B_1(x)\triangleq\{x\}-\frac12=-\sum_{n=1}^\infty{\sin(2\pi nx)\over\pi n}\]

and the Laplace transform identity that

\[{\Gamma(z)\over r^z}=\int_0^\infty\tau^{z-1}e^{-z\tau}\mathrm d\tau\]

where $-\pi/2\le\arg z\le\pi/2$

A formula for $\zeta(s)$ in $-1<\sigma<0$

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

\[\zeta(s)=\frac12+{s\over s-1}-s\int_1^\infty{B_1(x)\over x^{s+1}}\mathrm dx\]

in which we can extend the ROC of the latter integral to $\sigma>-1$ via integration by parts:

\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*}

When $-1<\sigma<0$ there is

\[-s\int_0^1{B_1(x)\over x^{s+1}}\mathrm dx=\frac12+{1\over s-1}\]

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

\[\zeta(s)=-s\int_0^\infty{B_1(x)\over x^{s+1}}\mathrm dx\]

Expansion of $B_1(x)$ into Fourier series

In order to go deeper, let's plug

\[B_1(x)\triangleq\{x\}-\frac12=-\sum_{n=1}^\infty{\sin(2\pi nx)\over\pi n}\]

into the previously obtained formula, so that 

\[\begin
\zeta(s)=s\int_0^\infty\sum_{n=1}^\infty{\sin(2\pi nx)\over n\pi}{\mathrm dx\over x^{s+1}}\] (Error compiling LaTeX. Unknown error_msg)
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