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Difference between revisions of "Functional equation for the zeta function"
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</cmath>
 
</cmath>
  
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in which we can extend the ROC of the latter integral to <math>\sigma>-1</math> via integration by parts:
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in which we can extend the ROC of the latter integral to <math>\sigma>-1</math> via [[integration by parts]]:
  
 
<cmath>
 
<cmath>
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</cmath>
 
</cmath>
  
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=== Expansion of <math>B_1(x)</math> into Fourier series ===
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=== Expansion of <math>B_1(x)</math> into [[Fourier series]] ===
  
 
In order to go deeper, let's plug
 
In order to go deeper, let's plug
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<cmath>
 
<cmath>
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\begin
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\begin{align*}
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\zeta(s)=s\int_0^\infty\sum_{n=1}^\infty{\sin(2\pi nx)\over n\pi}{\mathrm dx\over x^{s+1}}
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\zeta(s)
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&=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}
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</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>
 
</cmath>
 +
 +
== Resources ==
 +
 +
* Titchmarsh, E. C., ''The Theory of the Riemann Zeta-Function.'' Oxford Univ. Press, London and New York, 1951.

Latest revision as of 02:56, 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{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*}

Therefore, the remaining step is to handle the integral

Evaluation of $\int_0^\infty{\sin y\over y^{s+1}}\mathrm dy$

By Euler's formula, we have

\[\sin\theta={e^{i\theta}-e^{-i\theta}\over2i}\]

As a result, we only need to calculate

\[\int_0^\infty{e^{\pm iy}\over y^{s+1}}\mathrm dy\]

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

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

Thus we deduce

\[\int_0^\infty{\sin y\over y^{s+1}}\mathrm dy=-\Gamma(-s)\sin\left(\pi s\over2\right)\]

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

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

and by $\Gamma(z+1)=z\Gamma(z)$, this identity becomes the functional equation:

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

Resources

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