Difference between revisions of "Functional equation for the zeta function"
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− | The '''functional equation for Riemann zeta function''' is a result due to analytic continuation of Riemann zeta function: | + | The '''functional equation for Riemann zeta function''' is a result due to analytic continuation of [[Riemann zeta function]]: |
<cmath> | <cmath> | ||
Line 5: | Line 5: | ||
</cmath> | </cmath> | ||
− | There are multiple proofs for the functional equation for Riemann zeta function, and this page presents a | + | == 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> | <cmath> | ||
− | \ | + | \zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)\Gamma(1-s)\zeta(1-s) |
</cmath> | </cmath> | ||
+ | |||
+ | == Resources == | ||
+ | |||
+ | * Titchmarsh, E. C., ''The Theory of the Riemann Zeta-Function.'' Oxford Univ. Press, London and New York, 1951. |
Latest revision as of 03:56, 13 January 2021
The functional equation for Riemann zeta function is a result due to analytic continuation of Riemann zeta function:
Contents
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
and the Laplace transform identity that
where
A formula for
in ![$-1<\sigma<0$](//latex.artofproblemsolving.com/5/2/1/5211aa0e4af21cc5f79fc5b2c07015ef665a2445.png)
In this article, we will use the common convention that where
. As a result, we say that the original Dirichlet series definition
converges only for
. However, if we were to apply Euler-Maclaurin summation on this definition, we obtain
in which we can extend the ROC of the latter integral to via integration by parts:
When there is
As a result, we obtain a formula for for
:
Expansion of
into Fourier series
In order to go deeper, let's plug
into the previously obtained formula, so that
Therefore, the remaining step is to handle the integral
Evaluation of ![$\int_0^\infty{\sin y\over y^{s+1}}\mathrm dy$](//latex.artofproblemsolving.com/5/f/e/5fe6c8dd11c978fe67bf6ed0027bfe6865776180.png)
By Euler's formula, we have
As a result, we only need to calculate
if we want to take down the remaining integral. According to Laplace transform identities, we can see that
Thus we deduce
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
and by , this identity becomes the functional equation:
Resources
- Titchmarsh, E. C., The Theory of the Riemann Zeta-Function. Oxford Univ. Press, London and New York, 1951.