https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Travorlzh&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T14:40:23ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=Functional_equation_for_the_zeta_function&diff=142009Functional equation for the zeta function2021-01-13T07:56:58Z<p>Travorlzh: /* Resources */</p>
<hr />
<div>The '''functional equation for Riemann zeta function''' is a result due to analytic continuation of [[Riemann zeta function]]:<br />
<br />
<cmath><br />
\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)\Gamma(1-s)\zeta(1-s)<br />
</cmath><br />
<br />
== Proof ==<br />
<br />
=== Preparation ===<br />
<br />
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<br />
<br />
<cmath><br />
B_1(x)\triangleq\{x\}-\frac12=-\sum_{n=1}^\infty{\sin(2\pi nx)\over\pi n}<br />
</cmath><br />
<br />
and the Laplace transform identity that<br />
<br />
<cmath><br />
{\Gamma(z)\over r^z}=\int_0^\infty\tau^{z-1}e^{-z\tau}\mathrm d\tau<br />
</cmath><br />
<br />
where <math>-\pi/2\le\arg z\le\pi/2</math><br />
<br />
=== A formula for <math>\zeta(s)</math> in <math>-1<\sigma<0</math> ===<br />
<br />
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<br />
<br />
<cmath><br />
\zeta(s)=\frac12+{s\over s-1}-s\int_1^\infty{B_1(x)\over x^{s+1}}\mathrm dx<br />
</cmath><br />
<br />
in which we can extend the ROC of the latter integral to <math>\sigma>-1</math> via [[integration by parts]]:<br />
<br />
<cmath><br />
\begin{align*}<br />
\int_1^\infty{B_1(x)\over x^{s+1}}\mathrm dx<br />
&=\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 \\<br />
&={B_2\over2}+{s+1\over2}\int_1^\infty{B_2(x)\over x^{s+2}}\mathrm dx<br />
\end{align*}<br />
</cmath><br />
<br />
When <math>-1<\sigma<0</math> there is<br />
<br />
<cmath><br />
-s\int_0^1{B_1(x)\over x^{s+1}}\mathrm dx=\frac12+{1\over s-1}<br />
</cmath><br />
<br />
As a result, we obtain a formula for <math>\zeta(s)</math> for <math>-1<\sigma<0</math>:<br />
<br />
<cmath><br />
\zeta(s)=-s\int_0^\infty{B_1(x)\over x^{s+1}}\mathrm dx<br />
</cmath><br />
<br />
=== Expansion of <math>B_1(x)</math> into [[Fourier series]] ===<br />
<br />
In order to go deeper, let's plug<br />
<br />
<cmath><br />
B_1(x)\triangleq\{x\}-\frac12=-\sum_{n=1}^\infty{\sin(2\pi nx)\over\pi n}<br />
</cmath><br />
<br />
into the previously obtained formula, so that<br />
<br />
<cmath><br />
\begin{align*}<br />
\zeta(s)<br />
&=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} \\<br />
&=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}} \\<br />
&=\sum_{n=1}^\infty{2s\over(2\pi n)^{1-s}}\int_0^\infty{\sin y\over y^{s+1}}\mathrm dy \\<br />
&=2^s\pi^{s-1}\zeta(1-s)\int_0^\infty{\sin y\over y^{s+1}}\mathrm dy<br />
\end{align*}<br />
</cmath><br />
<br />
Therefore, the remaining step is to handle the integral<br />
<br />
=== Evaluation of <math>\int_0^\infty{\sin y\over y^{s+1}}\mathrm dy</math> ===<br />
<br />
By [[Euler's formula]], we have<br />
<br />
<cmath><br />
\sin\theta={e^{i\theta}-e^{-i\theta}\over2i}<br />
</cmath><br />
<br />
As a result, we only need to calculate<br />
<br />
<cmath><br />
\int_0^\infty{e^{\pm iy}\over y^{s+1}}\mathrm dy<br />
</cmath><br />
<br />
if we want to take down the remaining integral. According to Laplace transform identities, we can see that<br />
<br />
<cmath><br />
\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}<br />
</cmath><br />
<br />
Thus we deduce<br />
<br />
<cmath><br />
\int_0^\infty{\sin y\over y^{s+1}}\mathrm dy=-\Gamma(-s)\sin\left(\pi s\over2\right)<br />
</cmath><br />
<br />
wherein the RHS serves to be a meromorphic continuation of the LHS integral.<br />
<br />
=== Proof of the functional equation ===<br />
<br />
With everything ready, we can put everything together and obtain<br />
<br />
<cmath><br />
\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)(-s)\Gamma(-s)\zeta(1-s)<br />
</cmath><br />
<br />
and by <math>\Gamma(z+1)=z\Gamma(z)</math>, this identity becomes the functional equation:<br />
<br />
<cmath><br />
\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)\Gamma(1-s)\zeta(1-s)<br />
</cmath><br />
<br />
== Resources ==<br />
<br />
* Titchmarsh, E. C., ''The Theory of the Riemann Zeta-Function.'' Oxford Univ. Press, London and New York, 1951.</div>Travorlzhhttps://artofproblemsolving.com/wiki/index.php?title=Functional_equation_for_the_zeta_function&diff=142008Functional equation for the zeta function2021-01-13T07:56:40Z<p>Travorlzh: /* Resources */</p>
<hr />
<div>The '''functional equation for Riemann zeta function''' is a result due to analytic continuation of [[Riemann zeta function]]:<br />
<br />
<cmath><br />
\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)\Gamma(1-s)\zeta(1-s)<br />
</cmath><br />
<br />
== Proof ==<br />
<br />
=== Preparation ===<br />
<br />
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<br />
<br />
<cmath><br />
B_1(x)\triangleq\{x\}-\frac12=-\sum_{n=1}^\infty{\sin(2\pi nx)\over\pi n}<br />
</cmath><br />
<br />
and the Laplace transform identity that<br />
<br />
<cmath><br />
{\Gamma(z)\over r^z}=\int_0^\infty\tau^{z-1}e^{-z\tau}\mathrm d\tau<br />
</cmath><br />
<br />
where <math>-\pi/2\le\arg z\le\pi/2</math><br />
<br />
=== A formula for <math>\zeta(s)</math> in <math>-1<\sigma<0</math> ===<br />
<br />
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<br />
<br />
<cmath><br />
\zeta(s)=\frac12+{s\over s-1}-s\int_1^\infty{B_1(x)\over x^{s+1}}\mathrm dx<br />
</cmath><br />
<br />
in which we can extend the ROC of the latter integral to <math>\sigma>-1</math> via [[integration by parts]]:<br />
<br />
<cmath><br />
\begin{align*}<br />
\int_1^\infty{B_1(x)\over x^{s+1}}\mathrm dx<br />
&=\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 \\<br />
&={B_2\over2}+{s+1\over2}\int_1^\infty{B_2(x)\over x^{s+2}}\mathrm dx<br />
\end{align*}<br />
</cmath><br />
<br />
When <math>-1<\sigma<0</math> there is<br />
<br />
<cmath><br />
-s\int_0^1{B_1(x)\over x^{s+1}}\mathrm dx=\frac12+{1\over s-1}<br />
</cmath><br />
<br />
As a result, we obtain a formula for <math>\zeta(s)</math> for <math>-1<\sigma<0</math>:<br />
<br />
<cmath><br />
\zeta(s)=-s\int_0^\infty{B_1(x)\over x^{s+1}}\mathrm dx<br />
</cmath><br />
<br />
=== Expansion of <math>B_1(x)</math> into [[Fourier series]] ===<br />
<br />
In order to go deeper, let's plug<br />
<br />
<cmath><br />
B_1(x)\triangleq\{x\}-\frac12=-\sum_{n=1}^\infty{\sin(2\pi nx)\over\pi n}<br />
</cmath><br />
<br />
into the previously obtained formula, so that<br />
<br />
<cmath><br />
\begin{align*}<br />
\zeta(s)<br />
&=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} \\<br />
&=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}} \\<br />
&=\sum_{n=1}^\infty{2s\over(2\pi n)^{1-s}}\int_0^\infty{\sin y\over y^{s+1}}\mathrm dy \\<br />
&=2^s\pi^{s-1}\zeta(1-s)\int_0^\infty{\sin y\over y^{s+1}}\mathrm dy<br />
\end{align*}<br />
</cmath><br />
<br />
Therefore, the remaining step is to handle the integral<br />
<br />
=== Evaluation of <math>\int_0^\infty{\sin y\over y^{s+1}}\mathrm dy</math> ===<br />
<br />
By [[Euler's formula]], we have<br />
<br />
<cmath><br />
\sin\theta={e^{i\theta}-e^{-i\theta}\over2i}<br />
</cmath><br />
<br />
As a result, we only need to calculate<br />
<br />
<cmath><br />
\int_0^\infty{e^{\pm iy}\over y^{s+1}}\mathrm dy<br />
</cmath><br />
<br />
if we want to take down the remaining integral. According to Laplace transform identities, we can see that<br />
<br />
<cmath><br />
\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}<br />
</cmath><br />
<br />
Thus we deduce<br />
<br />
<cmath><br />
\int_0^\infty{\sin y\over y^{s+1}}\mathrm dy=-\Gamma(-s)\sin\left(\pi s\over2\right)<br />
</cmath><br />
<br />
wherein the RHS serves to be a meromorphic continuation of the LHS integral.<br />
<br />
=== Proof of the functional equation ===<br />
<br />
With everything ready, we can put everything together and obtain<br />
<br />
<cmath><br />
\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)(-s)\Gamma(-s)\zeta(1-s)<br />
</cmath><br />
<br />
and by <math>\Gamma(z+1)=z\Gamma(z)</math>, this identity becomes the functional equation:<br />
<br />
<cmath><br />
\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)\Gamma(1-s)\zeta(1-s)<br />
</cmath><br />
<br />
== Resources ==<br />
<br />
* Titchmarsh, E. C., ```The Theory of the Riemann Zeta-Function.``` Oxford Univ. Press, London and New York, 1951.</div>Travorlzhhttps://artofproblemsolving.com/wiki/index.php?title=Functional_equation_for_the_zeta_function&diff=142007Functional equation for the zeta function2021-01-13T07:56:12Z<p>Travorlzh: </p>
<hr />
<div>The '''functional equation for Riemann zeta function''' is a result due to analytic continuation of [[Riemann zeta function]]:<br />
<br />
<cmath><br />
\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)\Gamma(1-s)\zeta(1-s)<br />
</cmath><br />
<br />
== Proof ==<br />
<br />
=== Preparation ===<br />
<br />
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<br />
<br />
<cmath><br />
B_1(x)\triangleq\{x\}-\frac12=-\sum_{n=1}^\infty{\sin(2\pi nx)\over\pi n}<br />
</cmath><br />
<br />
and the Laplace transform identity that<br />
<br />
<cmath><br />
{\Gamma(z)\over r^z}=\int_0^\infty\tau^{z-1}e^{-z\tau}\mathrm d\tau<br />
</cmath><br />
<br />
where <math>-\pi/2\le\arg z\le\pi/2</math><br />
<br />
=== A formula for <math>\zeta(s)</math> in <math>-1<\sigma<0</math> ===<br />
<br />
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<br />
<br />
<cmath><br />
\zeta(s)=\frac12+{s\over s-1}-s\int_1^\infty{B_1(x)\over x^{s+1}}\mathrm dx<br />
</cmath><br />
<br />
in which we can extend the ROC of the latter integral to <math>\sigma>-1</math> via [[integration by parts]]:<br />
<br />
<cmath><br />
\begin{align*}<br />
\int_1^\infty{B_1(x)\over x^{s+1}}\mathrm dx<br />
&=\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 \\<br />
&={B_2\over2}+{s+1\over2}\int_1^\infty{B_2(x)\over x^{s+2}}\mathrm dx<br />
\end{align*}<br />
</cmath><br />
<br />
When <math>-1<\sigma<0</math> there is<br />
<br />
<cmath><br />
-s\int_0^1{B_1(x)\over x^{s+1}}\mathrm dx=\frac12+{1\over s-1}<br />
</cmath><br />
<br />
As a result, we obtain a formula for <math>\zeta(s)</math> for <math>-1<\sigma<0</math>:<br />
<br />
<cmath><br />
\zeta(s)=-s\int_0^\infty{B_1(x)\over x^{s+1}}\mathrm dx<br />
</cmath><br />
<br />
=== Expansion of <math>B_1(x)</math> into [[Fourier series]] ===<br />
<br />
In order to go deeper, let's plug<br />
<br />
<cmath><br />
B_1(x)\triangleq\{x\}-\frac12=-\sum_{n=1}^\infty{\sin(2\pi nx)\over\pi n}<br />
</cmath><br />
<br />
into the previously obtained formula, so that<br />
<br />
<cmath><br />
\begin{align*}<br />
\zeta(s)<br />
&=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} \\<br />
&=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}} \\<br />
&=\sum_{n=1}^\infty{2s\over(2\pi n)^{1-s}}\int_0^\infty{\sin y\over y^{s+1}}\mathrm dy \\<br />
&=2^s\pi^{s-1}\zeta(1-s)\int_0^\infty{\sin y\over y^{s+1}}\mathrm dy<br />
\end{align*}<br />
</cmath><br />
<br />
Therefore, the remaining step is to handle the integral<br />
<br />
=== Evaluation of <math>\int_0^\infty{\sin y\over y^{s+1}}\mathrm dy</math> ===<br />
<br />
By [[Euler's formula]], we have<br />
<br />
<cmath><br />
\sin\theta={e^{i\theta}-e^{-i\theta}\over2i}<br />
</cmath><br />
<br />
As a result, we only need to calculate<br />
<br />
<cmath><br />
\int_0^\infty{e^{\pm iy}\over y^{s+1}}\mathrm dy<br />
</cmath><br />
<br />
if we want to take down the remaining integral. According to Laplace transform identities, we can see that<br />
<br />
<cmath><br />
\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}<br />
</cmath><br />
<br />
Thus we deduce<br />
<br />
<cmath><br />
\int_0^\infty{\sin y\over y^{s+1}}\mathrm dy=-\Gamma(-s)\sin\left(\pi s\over2\right)<br />
</cmath><br />
<br />
wherein the RHS serves to be a meromorphic continuation of the LHS integral.<br />
<br />
=== Proof of the functional equation ===<br />
<br />
With everything ready, we can put everything together and obtain<br />
<br />
<cmath><br />
\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)(-s)\Gamma(-s)\zeta(1-s)<br />
</cmath><br />
<br />
and by <math>\Gamma(z+1)=z\Gamma(z)</math>, this identity becomes the functional equation:<br />
<br />
<cmath><br />
\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)\Gamma(1-s)\zeta(1-s)<br />
</cmath><br />
<br />
== Resources ==<br />
<br />
* Titchmarsh, E. C., “The Theory of the Riemann Zeta-Function.” Oxford Univ.<br />
Press, London and New York, 1951.</div>Travorlzhhttps://artofproblemsolving.com/wiki/index.php?title=Functional_equation_for_the_zeta_function&diff=142006Functional equation for the zeta function2021-01-13T07:52:04Z<p>Travorlzh: </p>
<hr />
<div>The '''functional equation for Riemann zeta function''' is a result due to analytic continuation of [[Riemann zeta function]]:<br />
<br />
<cmath><br />
\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)\Gamma(1-s)\zeta(1-s)<br />
</cmath><br />
<br />
== Proof ==<br />
<br />
=== Preparation ===<br />
<br />
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<br />
<br />
<cmath><br />
B_1(x)\triangleq\{x\}-\frac12=-\sum_{n=1}^\infty{\sin(2\pi nx)\over\pi n}<br />
</cmath><br />
<br />
and the Laplace transform identity that<br />
<br />
<cmath><br />
{\Gamma(z)\over r^z}=\int_0^\infty\tau^{z-1}e^{-z\tau}\mathrm d\tau<br />
</cmath><br />
<br />
where <math>-\pi/2\le\arg z\le\pi/2</math><br />
<br />
=== A formula for <math>\zeta(s)</math> in <math>-1<\sigma<0</math> ===<br />
<br />
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<br />
<br />
<cmath><br />
\zeta(s)=\frac12+{s\over s-1}-s\int_1^\infty{B_1(x)\over x^{s+1}}\mathrm dx<br />
</cmath><br />
<br />
in which we can extend the ROC of the latter integral to <math>\sigma>-1</math> via [[integration by parts]]:<br />
<br />
<cmath><br />
\begin{align*}<br />
\int_1^\infty{B_1(x)\over x^{s+1}}\mathrm dx<br />
&=\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 \\<br />
&={B_2\over2}+{s+1\over2}\int_1^\infty{B_2(x)\over x^{s+2}}\mathrm dx<br />
\end{align*}<br />
</cmath><br />
<br />
When <math>-1<\sigma<0</math> there is<br />
<br />
<cmath><br />
-s\int_0^1{B_1(x)\over x^{s+1}}\mathrm dx=\frac12+{1\over s-1}<br />
</cmath><br />
<br />
As a result, we obtain a formula for <math>\zeta(s)</math> for <math>-1<\sigma<0</math>:<br />
<br />
<cmath><br />
\zeta(s)=-s\int_0^\infty{B_1(x)\over x^{s+1}}\mathrm dx<br />
</cmath><br />
<br />
=== Expansion of <math>B_1(x)</math> into [[Fourier series]] ===<br />
<br />
In order to go deeper, let's plug<br />
<br />
<cmath><br />
B_1(x)\triangleq\{x\}-\frac12=-\sum_{n=1}^\infty{\sin(2\pi nx)\over\pi n}<br />
</cmath><br />
<br />
into the previously obtained formula, so that<br />
<br />
<cmath><br />
\begin{align*}<br />
\zeta(s)<br />
&=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} \\<br />
&=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}} \\<br />
&=\sum_{n=1}^\infty{2s\over(2\pi n)^{1-s}}\int_0^\infty{\sin y\over y^{s+1}}\mathrm dy \\<br />
&=2^s\pi^{s-1}\zeta(1-s)\int_0^\infty{\sin y\over y^{s+1}}\mathrm dy<br />
\end{align*}<br />
</cmath><br />
<br />
Therefore, the remaining step is to handle the integral<br />
<br />
=== Evaluation of <math>\int_0^\infty{\sin y\over y^{s+1}}\mathrm dy</math> ===<br />
<br />
By [[Euler's formula]], we have<br />
<br />
<cmath><br />
\sin\theta={e^{i\theta}-e^{-i\theta}\over2i}<br />
</cmath><br />
<br />
As a result, we only need to calculate<br />
<br />
<cmath><br />
\int_0^\infty{e^{\pm iy}\over y^{s+1}}\mathrm dy<br />
</cmath><br />
<br />
if we want to take down the remaining integral. According to Laplace transform identities, we can see that<br />
<br />
<cmath><br />
\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}<br />
</cmath><br />
<br />
Thus we deduce<br />
<br />
<cmath><br />
\int_0^\infty{\sin y\over y^{s+1}}\mathrm dy=-\Gamma(-s)\sin\left(\pi s\over2\right)<br />
</cmath><br />
<br />
wherein the RHS serves to be a meromorphic continuation of the LHS integral.<br />
<br />
=== Proof of the functional equation ===<br />
<br />
With everything ready, we can put everything together and obtain<br />
<br />
<cmath><br />
\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)(-s)\Gamma(-s)\zeta(1-s)<br />
</cmath><br />
<br />
and by <math>\Gamma(z+1)=z\Gamma(z)</math>, this identity becomes the functional equation:<br />
<br />
<cmath><br />
\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)\Gamma(1-s)\zeta(1-s)<br />
</cmath></div>Travorlzhhttps://artofproblemsolving.com/wiki/index.php?title=Functional_equation_for_the_zeta_function&diff=142003Functional equation for the zeta function2021-01-13T07:37:03Z<p>Travorlzh: </p>
<hr />
<div>The '''functional equation for Riemann zeta function''' is a result due to analytic continuation of [[Riemann zeta function]]:<br />
<br />
<cmath><br />
\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)\Gamma(1-s)\zeta(1-s)<br />
</cmath><br />
<br />
== Proof ==<br />
<br />
=== Preparation ===<br />
<br />
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<br />
<br />
<cmath><br />
B_1(x)\triangleq\{x\}-\frac12=-\sum_{n=1}^\infty{\sin(2\pi nx)\over\pi n}<br />
</cmath><br />
<br />
and the Laplace transform identity that<br />
<br />
<cmath><br />
{\Gamma(z)\over r^z}=\int_0^\infty\tau^{z-1}e^{-z\tau}\mathrm d\tau<br />
</cmath><br />
<br />
where <math>-\pi/2\le\arg z\le\pi/2</math><br />
<br />
=== A formula for <math>\zeta(s)</math> in <math>-1<\sigma<0</math> ===<br />
<br />
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<br />
<br />
<cmath><br />
\zeta(s)=\frac12+{s\over s-1}-s\int_1^\infty{B_1(x)\over x^{s+1}}\mathrm dx<br />
</cmath><br />
<br />
in which we can extend the ROC of the latter integral to <math>\sigma>-1</math> via integration by parts:<br />
<br />
<cmath><br />
\begin{align*}<br />
\int_1^\infty{B_1(x)\over x^{s+1}}\mathrm dx<br />
&=\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 \\<br />
&={B_2\over2}+{s+1\over2}\int_1^\infty{B_2(x)\over x^{s+2}}\mathrm dx<br />
\end{align*}<br />
</cmath><br />
<br />
When <math>-1<\sigma<0</math> there is<br />
<br />
<cmath><br />
-s\int_0^1{B_1(x)\over x^{s+1}}\mathrm dx=\frac12+{1\over s-1}<br />
</cmath><br />
<br />
As a result, we obtain a formula for <math>\zeta(s)</math> for <math>-1<\sigma<0</math>:<br />
<br />
<cmath><br />
\zeta(s)=-s\int_0^\infty{B_1(x)\over x^{s+1}}\mathrm dx<br />
</cmath><br />
<br />
=== Expansion of <math>B_1(x)</math> into Fourier series ===<br />
<br />
In order to go deeper, let's plug<br />
<br />
<cmath><br />
B_1(x)\triangleq\{x\}-\frac12=-\sum_{n=1}^\infty{\sin(2\pi nx)\over\pi n}<br />
</cmath><br />
<br />
into the previously obtained formula, so that<br />
<br />
<cmath><br />
\begin<br />
\zeta(s)=s\int_0^\infty\sum_{n=1}^\infty{\sin(2\pi nx)\over n\pi}{\mathrm dx\over x^{s+1}}<br />
</cmath></div>Travorlzhhttps://artofproblemsolving.com/wiki/index.php?title=Functional_equation_for_the_zeta_function&diff=142002Functional equation for the zeta function2021-01-13T07:22:54Z<p>Travorlzh: /* Two useful identities */</p>
<hr />
<div>The '''functional equation for Riemann zeta function''' is a result due to analytic continuation of [[Riemann zeta function]]:<br />
<br />
<cmath><br />
\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)\Gamma(1-s)\zeta(1-s)<br />
</cmath><br />
<br />
== Proof ==<br />
<br />
=== Preparation ===<br />
<br />
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<br />
<br />
<cmath><br />
B_1(x)\triangleq\{x\}-\frac12=-\sum_{n=1}^\infty{\sin(2\pi nx)\over\pi n}<br />
</cmath><br />
<br />
=== A formula for <math>\zeta(s)</math> in <math>-1<\sigma<0</math> ===<br />
<br />
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<br />
<br />
<cmath><br />
\zeta(s)=\frac12+{s\over s-1}-s\int_1^\infty{B_1(x)\over x^{s+1}}\mathrm dx<br />
</cmath><br />
<br />
in which we can extend the ROC of the latter integral to <math>\sigma>-1</math> via integration by parts:<br />
<br />
<cmath><br />
\begin{align*}<br />
\int_1^\infty{B_1(x)\over x^{s+1}}\mathrm dx<br />
&=\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 \\<br />
&={B_2\over2}+{s+1\over2}\int_1^\infty{B_2(x)\over x^{s+2}}\mathrm dx<br />
\end{align*}<br />
</cmath></div>Travorlzhhttps://artofproblemsolving.com/wiki/index.php?title=Functional_equation_for_the_zeta_function&diff=142001Functional equation for the zeta function2021-01-13T07:22:28Z<p>Travorlzh: </p>
<hr />
<div>The '''functional equation for Riemann zeta function''' is a result due to analytic continuation of [[Riemann zeta function]]:<br />
<br />
<cmath><br />
\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)\Gamma(1-s)\zeta(1-s)<br />
</cmath><br />
<br />
== Proof ==<br />
<br />
=== Two useful identities ===<br />
<br />
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<br />
<br />
<cmath><br />
B_1(x)\triangleq\{x\}-\frac12=-\sum_{n=1}^\infty{\sin(2\pi nx)\over\pi n}<br />
</cmath><br />
<br />
=== A formula for <math>\zeta(s)</math> in <math>-1<\sigma<0</math> ===<br />
<br />
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<br />
<br />
<cmath><br />
\zeta(s)=\frac12+{s\over s-1}-s\int_1^\infty{B_1(x)\over x^{s+1}}\mathrm dx<br />
</cmath><br />
<br />
in which we can extend the ROC of the latter integral to <math>\sigma>-1</math> via integration by parts:<br />
<br />
<cmath><br />
\begin{align*}<br />
\int_1^\infty{B_1(x)\over x^{s+1}}\mathrm dx<br />
&=\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 \\<br />
&={B_2\over2}+{s+1\over2}\int_1^\infty{B_2(x)\over x^{s+2}}\mathrm dx<br />
\end{align*}<br />
</cmath></div>Travorlzhhttps://artofproblemsolving.com/wiki/index.php?title=Functional_equation_for_the_zeta_function&diff=142000Functional equation for the zeta function2021-01-13T07:19:38Z<p>Travorlzh: </p>
<hr />
<div>== Introduction ==<br />
<br />
The '''functional equation for Riemann zeta function''' is a result due to analytic continuation of [[Riemann zeta function]]:<br />
<br />
<cmath><br />
\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)\Gamma(1-s)\zeta(1-s)<br />
</cmath><br />
<br />
== Proof ==<br />
<br />
=== Two useful identities ===<br />
<br />
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<br />
<br />
<cmath><br />
B_1(x)\triangleq\{x\}-\frac12=-\sum_{n=1}^\infty{\sin(2\pi nx)\over\pi n}<br />
</cmath><br />
<br />
=== From <math>\sigma>1</math> to <math>\sigma>-1</math> ===<br />
<br />
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<br />
<br />
<cmath><br />
\zeta(s)=\frac12+{s\over s-1}-s\int_1^\infty{B_1(x)\over x^{s+1}}\mathrm dx<br />
</cmath><br />
<br />
in which we can extend the ROC of the latter integral to <math>\sigma>-1</math> via integration by parts:<br />
<br />
<cmath><br />
\begin{aligned}<br />
\int_1^\infty{B_1(x)\over x^{s+1}}\mathrm dx<br />
={B_2(x)\over2x^{s+1}}+{s+1\over2}\int_1^\infty{B_2(x)\over x^{s+2}}\mathrm dx<br />
\end{aligned}<br />
</cmath></div>Travorlzhhttps://artofproblemsolving.com/wiki/index.php?title=Functional_equation_for_the_zeta_function&diff=141999Functional equation for the zeta function2021-01-13T07:17:47Z<p>Travorlzh: </p>
<hr />
<div>== Introduction ==<br />
<br />
The '''functional equation for Riemann zeta function''' is a result due to analytic continuation of [[Riemann zeta function]]:<br />
<br />
<cmath><br />
\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)\Gamma(1-s)\zeta(1-s)<br />
</cmath><br />
<br />
== Proof ==<br />
<br />
=== Two useful identities ===<br />
<br />
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<br />
<br />
<cmath><br />
B_1(x)\triangleq\{x\}-\frac12=-\sum_{n=1}^\infty{\sin(2\pi nx)\over\pi n}<br />
</cmath><br />
<br />
=== From <math>\sigma>1</math> to <math>\sigma>-1</math> ===<br />
<br />
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<br />
<br />
<cmath><br />
\zeta(s)=\frac12+{s\over s-1}-s\int_1^\infty{B_1(x)\over x^{s+1}}\mathrm dx<br />
</cmath><br />
<br />
in which we can extend the ROC of the latter integral to <math>\sigma>-1</math> via repeated integration:<br />
<br />
<cmath><br />
\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<br />
</cmath></div>Travorlzhhttps://artofproblemsolving.com/wiki/index.php?title=Riemann_zeta_function&diff=141997Riemann zeta function2021-01-13T07:04:09Z<p>Travorlzh: Changed the definition of $\xi(s)$</p>
<hr />
<div>The '''Riemann zeta function''' is a function very important in<br />
[[number theory]]. In particular, the [[Riemann Hypothesis]] is a conjecture<br />
about the roots of the zeta function.<br />
<br />
The function is defined by <br />
<cmath>\zeta (s)=\sum_{n=1}^{\infty}\frac{1}{n^s}=<br />
1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\cdots</cmath><br />
when the [[real part]] <math>\Re(s)</math> is greater than 1. (When <math>\Re(s) \le<br />
1</math> the [[series]] '''does not''' converge, but it can be extended to all<br />
[[complex number]]s except <math>s = 1</math>&mdash;see<br />
[[#Extending_the_zeta_function | below]].)<br />
<br />
[[Leonhard Euler]] showed that when <math>s=2</math>, the sum is equal to<br />
<math>\frac{\pi^2}{6}</math>. Euler also found that since every number is the product<br />
of a unique combination of [[prime number]]s, the zeta function can be<br />
expressed as an infinite product: <br />
<cmath>\zeta(s) = \left(\frac{1}{(2^0)^s} + \frac{1}{(2^1)^s}+<br />
\frac{1}{(2^2)^s} + \cdots\right) \left(\frac{1}{(3^0)^s} + \frac{1}<br />
{(3^1)^s} + \frac{1}{(3^2)^s} + \cdots\right) \left(\frac{1}{(5^0)^s}<br />
+ \frac{1}{(5^1)^s} + \frac{1}{(5^2)^s} + \cdots\right) \cdots.</cmath><br />
By summing up each of these [[geometric series]] in parentheses, we arrive<br />
at the following identity (the [[Euler Product]]): <br />
<cmath>\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} = \prod_{p \text{ prime}}<br />
(1-p^{-s})^{-1}.</cmath><br />
<br />
This gives a hint of why an [[analysis | analytic]] object like the<br />
zeta function could be related to number theoretic results.<br />
<br />
<br />
== Extending the zeta function ==<br />
<br />
The most important properties of the zeta function are based on the<br />
fact that it extends to a [[meromorphic]] function on the full<br />
[[complex plane]] which is [[holomorphic]] except at <math>s=1</math>, where<br />
there is a [[simple pole]] of [[residue]] 1. Let us see how this is done.<br />
<br />
First, we wish to extend <math>\zeta(s)</math> to the strip <math>\Re(s)>0</math>. To do this,<br />
we introduce the ''alternating zeta function''<br />
<cmath>\zeta_a(s) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s} .</cmath><br />
For <math>\Re(s) > 1</math>, we have<br />
<cmath>\zeta(s) = \zeta_a(s) + \frac{2}{2^s} + \frac{2}{4^s} + \frac{2}{6^s}<br />
+ \cdots = \zeta_a(s) + 2^{1-s}{\zeta(s)}, </cmath><br />
or<br />
<cmath> \zeta(s) = \frac{1}{1- 2^{1-s}} \zeta_a(s) . </cmath><br />
We may thus use the alternating zeta function to extend the zeta<br />
function.<br />
<br />
'''Proposition.''' The series <math>\zeta_a(s)</math> converges whenever<br />
<math>\Re(s) \ge 0</math>.<br />
<br />
''Proof.'' We have<br />
<cmath> \zeta_a(s) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^s}<br />
= \sum_{n=1}^{\infty} \frac{1}{(2n-1)^s} - \frac{1}{(2n)^s} .</cmath><br />
Since<br />
<cmath> \lvert d(x^{-s})/dx \rvert = \lvert s x^{-s-1} \rvert \le<br />
\left\lvert \frac{s}{(2n-1)^{s+1}} \right\rvert </cmath><br />
for <math>x \in [2n-1, 2n]</math>, it follows that<br />
<cmath> \left\lvert \frac{1}{(2n-1)^s} - \frac{1}{(2n)^s} \right\rvert<br />
\le \left\lvert \frac{s}{(2n-1)^{s+1}} \right\rvert . </cmath><br />
Since <math>\Re(s+1) > 1</math>, the series in question converges.<br />
<math>\blacksquare</math><br />
<br />
Now we can extend the zeta function.<br />
<br />
'''Theorem 1.''' The function <math>\zeta(s)</math> has a meromorphic extension<br />
to <math>\Re(s) > 0</math>, and it is holomorphic there except at <math>s=1</math>, where<br />
it has a simple pole of residue 1.<br />
<br />
''Proof.'' For <math>s \neq 1</math>, we have the extension<br />
<cmath> \zeta(s) - \frac{1}{s-1} = \frac{1}{1 - 2^{1-s}}\zeta_a(s) -<br />
\frac{1}{s-1} . </cmath><br />
For <math>s= 1</math>, we have<br />
<cmath> \lim_{s\to 1} \frac{(s-1) \zeta_a(s)}{1- 2^{1-s}} = \lim_{s\to1}<br />
\frac{\zeta_a(s)}{\log 2 \cdot 2^{1-s}} = \frac{\zeta_a(1)}{<br />
\log 2} ,</cmath><br />
by [[l'Hôpital's Rule]], so the pole at <math>s=1</math> is simple, and its<br />
residue is <math>\zeta_a(1) / \log 2</math>.<br />
<br />
Now, for all integers <math>n\geq 1</math>,<br />
<cmath> \frac{d^n(\log t)}{(dt)^n} = \frac{(-1)^{n-1}(n-1)!}{t^n} . </cmath><br />
It follows that the [[Taylor series]] expansion of <math>\log x</math><br />
about <math>x=1</math> is<br />
<cmath> \sum_{k=1}^{\infty} \frac{(-1)^{k-1}(x-1)^k}{k} . </cmath><br />
It follows that <math>\zeta_a(1) = \log 2</math>. Thus the residue of the<br />
pole is 1. <math>\blacksquare</math><br />
<br />
The next step is the<br />
[[functional equation for the zeta function|functional equation]]:<br />
Let<br />
<cmath>\xi(s)=\frac12s(s-1)\pi^{-s/2}\Gamma\left(\frac s2\right)\zeta(s).</cmath><br />
Then <math>\xi(s)=\xi(1-s)</math>. This gives us an analytic continuation<br />
of <math>\zeta(s)</math> to all of <math>\mathbb{C}</math>.<br />
<br />
== Zeroes of the Zeta Function ==<br />
<br />
Using the Euler product, it is not too difficult to show that<br />
<math>\zeta(s)</math> has no zeros for <math>\Re s > 1</math>. Indeed, suppose this<br />
is the case; let <math>x = \Re s</math>. Then<br />
<cmath> \begin{align*}<br />
\sum_{p} \bigl\lvert \log \lvert (1-p^{-s})^{-1} \rvert \bigr\rvert<br />
&= \sum_p \log \lvert p^s \rvert - \log \lvert p^s -1 \rvert<br />
= \sum_p \int\limits_{\lvert p^s - 1 \rvert}^{\lvert p^s \rvert}<br />
\frac{dt}{t} \\<br />
&\approx \sum_p \frac{1}{\lvert p^s - 1 \rvert} \\<br />
&< \sum_p 1/p^s,<br />
\end{align*} </cmath><br />
which converges. It follows that<br />
<cmath> \prod_p (1 - p^{-s})^{-1} \neq 0 . </cmath><br />
<br />
From the functional equation<br />
<cmath> \zeta(1-s) = (2\pi)^{-s} 2 \cos(\pi s/2) \Gamma(s) \zeta(s), </cmath><br />
it is evident that the zeta function has zeroes at <math>s= -2n</math>, for<br />
<math>n</math> a postive integer. These are called the trivial zeros.<br />
Since the [[gamma function]] has no zeros, it follows that these<br />
are the only zeros with real part less than 0.<br />
<br />
In 1859, Georg Friedrich Bernhard Riemann, after whom the<br />
function is named, established the functional equation and<br />
proved that <math>\zeta(s)</math> has infinitely many zeros in the strip<br />
<math>0 \le \Re(s) \le 1</math>. He conjectured that they all lie on the<br />
line <math>\Re s = 1/2</math>. This is the famous [[Riemann Hypothesis]],<br />
and to this day it remains one of the great unsolved problems<br />
of mathematics. Recently it has been proven that the function's<br />
first ten trillion zeros lie on the line<br />
<math>\Re s = 1/2</math>[http://mathworld.wolfram.com/RiemannHypothesis.html], but<br />
proof of the Riemann hypothesis still eludes us.<br />
<br />
In 1896, Jacque Hadamard and Charles-Jean de la Vallée Poussin<br />
independently proved that <math>\zeta(s)</math> has no zeros on the line<br />
<math>\Re(s) = 1</math>. From this they proved the [[prime number theorem]].<br />
We prove this result here.<br />
<br />
We first define the phi function,<br />
<cmath> \phi(s) = \sum_{p \text{ prime}} \frac{\log p}{p^s} . </cmath><br />
<br />
'''Theorem 2.''' The function <math>\phi(s)</math> has a meromorphic<br />
continuation to <math>\Re(s) > 1/2</math> with simple poles at<br />
the poles and zeros of <math>\zeta(s)</math>, and with no other poles.<br />
The continuation is<br />
<cmath> \phi(s) = - \frac{\zeta'(s)}{\zeta(s)} - \sum_p<br />
\frac{\log p}{p^s(p^s-1)} .</cmath><br />
<br />
''Proof.'' It follows from the Euler product formula that for<br />
<math>\Re(s) > 1</math>,<br />
<cmath> \begin{align*}<br />
\frac{\zeta'(s)}{\zeta(s)} &= \sum_p \frac{d (1- p^{-s})^{-1}/ds}<br />
{(1-p^{-s})^{-1}} = -\sum_p \frac{d(1-p^{-s})/ds}{(1-p^{-s})} \\<br />
&= -\sum_p \frac{\log p \cdot p^{-s}}{1-p^{-s}} \\<br />
&= -\sum_p \frac{\log p}{p^s -1 } <br />
= - \phi(s) - \sum_p \frac{\log p}{p^s (p^s - 1)}.<br />
\end{align*} </cmath><br />
Since <math>\sum_p \frac{\log p}{p^s (p^s- 1)}</math> converges when<br />
<math>\Re s > 1/2</math>, the theorem statement follows. <math>\blacksquare</math><br />
<br />
Now we proceed to the main result.<br />
<br />
'''Theorem 3.''' The zeta function has no zeros on the<br />
line <math>\Re(s) = 1</math>.<br />
<br />
''Proof.'' We use the fact that <math>\zeta(\bar s) = \overline{<br />
\zeta(s)}</math>.<br />
<br />
Let <math>g(s) = 1/ \zeta(s)</math>. Then 1 is a zero of <math>g</math> of order 1.<br />
Thus<br />
<cmath><br />
\lim_{\epsilon \to 0} \epsilon \phi(1+\epsilon)<br />
= \lim_{\epsilon \to 0} -\frac{\epsilon(1/g(1+\epsilon))'}{1/g(1+<br />
\epsilon)}<br />
= \lim_{\epsilon \to 0} \frac{\epsilon g'(1+\epsilon)}{g(1+\epsilon)}<br />
= 1 . </cmath><br />
<br />
Suppose now that <math>1+ki</math> and <math>1+2ki</math> are zeros of <math>\zeta(s)</math> of<br />
<math>\zeta(s)</math> of order <math>m</math> and <math>n</math>, respectively. (Note that <math>m</math><br />
and <math>n</math> may be zero.)<br />
Then<br />
<cmath> \begin{align*}<br />
\lim_{\epsilon\to 0}\epsilon \phi(1+\epsilon\pm ki)<br />
&= \lim_{\epsilon\to 0} - \frac{\epsilon \zeta'(1+\epsilon \pm ki)}{<br />
\zeta(1+\epsilon \pm ki)} = -m , \\<br />
\lim_{\epsilon\to 0}\epsilon \phi(1+\epsilon\pm 2ki)<br />
&= \lim_{\epsilon\to 0} - \frac{\epsilon \zeta'(1+\epsilon \pm 2ki)}{<br />
\zeta(1+\epsilon \pm 2ki)} = -n .<br />
\end{align*} </cmath><br />
<br />
Now for real, positive <math>\epsilon</math>,<br />
<cmath> \sum_{a=0}^{4} \binom{4}{a} \phi(1+\epsilon - 2ki+4kai)<br />
= \sum_p \frac{\log p}{p^{1+\epsilon}} (p^{ki/2} + p^{-ki/2})^4<br />
\ge 0, </cmath><br />
since <math>p^{-ki/2} = \overline{p^{ki/2}}</math>. It follows that<br />
<cmath> -2n - 8m + 6 \ge 0 . </cmath><br />
Since <math>m</math> and <math>n</math> must be nonnegative integers, it follows that <math>m=0</math>.<br />
Thus <math>\zeta(1+ki) \neq 0</math>. Since <math>k</math> was arbitrary, it follows<br />
that <math>\zeta(s)</math> has no zeros on the line <math>\Re s = 1</math>.<br />
<math>\blacksquare</math><br />
<br />
== Resources ==<br />
<br />
* Koch, Helmut (trans. David Kramer), ''Number Theory: Algebraic Numbers and Functions.'' AMS 2000, ISBN 0-8218-2054-0.<br />
<br />
== See also ==<br />
<br />
* [[Riemann Hypothesis]]<br />
* [[Prime number theorem]]<br />
<br />
[[Category:Number theory]]<br />
[[Category:Analytic number theory]]<br />
[[Category:Complex analysis]]</div>Travorlzhhttps://artofproblemsolving.com/wiki/index.php?title=Functional_equation_for_the_zeta_function&diff=141996Functional equation for the zeta function2021-01-13T07:00:48Z<p>Travorlzh: Created page with "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\..."</p>
<hr />
<div>The '''functional equation for Riemann zeta function''' is a result due to analytic continuation of Riemann zeta function:<br />
<br />
<cmath><br />
\zeta(s)=2^s\pi^{s-1}\sin\left(\pi s\over2\right)\Gamma(1-s)\zeta(1-s)<br />
</cmath><br />
<br />
There are multiple proofs for the functional equation for Riemann zeta function, and this page presents a lightweighted approach which merely relies on Fourier series that<br />
<br />
<cmath><br />
\frac12-\{x\}=\sum_{n=1}^\infty{\sin(2\pi nx)\over\pi n}<br />
</cmath></div>Travorlzh