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

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</cmath>
 
</cmath>
  
in which we can extend the ROC of the latter integral to <math>\sigma>-1</math> via repeated integration:
+
in which we can extend the ROC of the latter integral to <math>\sigma>-1</math> via integration by parts:
  
 
<cmath>
 
<cmath>
\int_1^\infty{B_1(x)\over x^{s+1}}\mathrm dx={B_2(x)\over2x^{s+1}}+{s+1\over2}\int_1^\infty{B_2(x)\over x^{s+2}}\mathrm dx
+
\begin{aligned}
 +
\int_1^\infty{B_1(x)\over x^{s+1}}\mathrm dx
 +
={B_2(x)\over2x^{s+1}}+{s+1\over2}\int_1^\infty{B_2(x)\over x^{s+2}}\mathrm dx
 +
\end{aligned}
 
</cmath>
 
</cmath>

Revision as of 02:19, 13 January 2021

Introduction

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

Two useful identities

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

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

From $\sigma>1$ to $\sigma>-1$

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{aligned}
\int_1^\infty{B_1(x)\over x^{s+1}}\mathrm dx
={B_2(x)\over2x^{s+1}}+{s+1\over2}\int_1^\infty{B_2(x)\over x^{s+2}}\mathrm dx
\end{aligned} (Error compiling LaTeX. Unknown error_msg)