Riemann's functional equation for the Zeta function

In this blog entry we will derive the functional equation for the Riemann Zeta Function. It extends the Zeta function to the entire complex plane except for the point s=1 which is a simple pole.

We have already found an analytic continuation for the Zeta function through the Euler Maclaurin summation formula. There, we were able to extend it´s domain to the left side of the complex plane step by step increasing the order of the Euler Maclaurin formula. The functional equation enables us to extend the Zeta function to the entire complex domain at once.

We will start by first introducing the Poisson summation formula which is a key ingredient in the derivation. In the end of the post we will show one small branch of it´s applicability proving the result $\zeta^\prime(2)=\zeta(2)(\gamma+\ln(2\pi)-12 \ln A)$ , which we have extensively used computing integrals.

Poisson Summation Formula

Let f(x) be a continuous function of defined for $-\infty<x<\infty$ Let

$g(x)=\sum_{n=-\infty}^{\infty} f(x+n)$

g(x) is periodic with period T=1 Proof:

$g(x+1)=\sum_{n=-\infty}^{\infty} f(x+1+n)$

Letting $n+1 \mapsto n$ we obtain

$g(x+1)=\sum_{n=-\infty}^{\infty} f(x+n)=g(x)$

Then, g(x) can be expanded in a Fourier series. To find the Fourier series coefficients we have to compute

$\begin{aligned} a_{k} &=\int_{0}^{1} g(x) e^{-2 \pi i k x} d x \\ &=\int_{0}^{1}\left(\sum_{n=-\infty}^{\infty} f(x+n)\right) e^{-2 \pi i k x} d x \\ &=\sum_{n=-\infty}^{\infty} \int_{0}^{1} f(x+n) e^{-2 \pi i k x} d x \\ &=\sum_{n=-\infty}^{\infty} \int_{n}^{n+1} f(x) e^{-2 \pi i k(x-n)} d x \quad(x+n \mapsto x) \\ &=\sum_{n=-\infty}^{\infty} \int_{n}^{n+1} f(x) e^{-2 \pi i k x} \underbrace{e^{2 \pi i k n}}_{=1} d x \\ &=\sum_{n=-\infty}^{\infty} \int_{n}^{n+1} f(x) e^{-2 \pi i k x} d x \\ &=\int_{-\infty}^{\infty} f(x) e^{-2 \pi i k x} d x \end{aligned}$

Therefore

$\begin{aligned} \sum_{n=-\infty}^{\infty} f(x+n) &=\sum_{k=-\infty}^{\infty} a_{k} e^{2 \pi i k x} \\ &=\sum_{k=-\infty}^{\infty} e^{2 \pi i k x} \int_{-\infty}^{\infty} f(t) e^{-2 \pi i k t} d t \end{aligned}$

If we let x=0 in the above equation yields the Poisson Summation formula:

$\boxed{\sum_{n=-\infty}^{\infty} f(n)=\sum_{k=-\infty}^{\infty} \int_{-\infty}^{\infty} f(t) e^{-2 \pi i k t} d t}$ (1)

We now choose $f(t)=e^{-a t^{2}}$ we get that

$\begin{aligned} a_{k} &=\int_{-\infty}^{\infty} e^{-a t^{2}} e^{-2 \pi i k t} d t \\ &=\sqrt{\frac{\pi}{a}} e^{-\frac{\pi^{2} k^{2}}{a}} \end{aligned}$

Which we recognize as Fourier transform of a Gaussian. Letting $a=\pi x$ and t=n we obtain

$a_{k}=\frac{1}{\sqrt{x}} e^{-\frac{\pi k^{2}}{x}}$ (2)

Plugging (2) in the Poisson Summation formula (1) we obtain

$\boxed{\sum_{n=-\infty}^{\infty} e^{-\pi n^{2} x}=\frac{1}{\sqrt{x}} \sum_{n=-\infty}^{\infty} e^{-\pi n^{2} / x} }$ (3)

(3) is the Transformatio formula of the Jacobi Theta function $\theta_3(x)=\sum_{n=-\infty}^{\infty} e^{-\pi n^{2} x}$

Note that $\theta_3(x)$ is an even function and we can rewrite it as

$\begin{aligned} \sum_{n=-\infty}^{\infty} e^{-\pi n^{2} x}&=\sum_{n=-\infty}^{-1} e^{-\pi n^{2} x}+1+\sum_{n=1}^{\infty} e^{-\pi n^{2} x}\\ &=1+2\sum_{n=1}^{\infty} e^{-\pi n^{2} x} \end{aligned}$

Similarly

$\begin{aligned} \frac{1}{\sqrt{x}} \sum_{n=-\infty}^{\infty} e^{-\pi n^{2} / x} &=\frac{1}{\sqrt{x}} \sum_{n=-\infty}^{-1} e^{-\pi n^{2} / x}+\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x}} \sum_{n=1}^{\infty} e^{-\pi n^{2} / x}\\ &=\frac{1}{\sqrt{x}}+\frac{2}{\sqrt{x}} \sum_{n=1}^{\infty} e^{-\pi n^{2} / x} \end{aligned}$

Therefore we conclude that

$\begin{aligned} 1+2\sum_{n=1}^{\infty} e^{-\pi n^{2} x}&=\frac{1}{\sqrt{x}}+\frac{2}{\sqrt{x}} \sum_{n=1}^{\infty} e^{-\pi n^{2} / x}\\ 2\sum_{n=1}^{\infty} e^{-\pi n^{2} x}&=-1+\frac{1}{\sqrt{x}}+\frac{2}{\sqrt{x}} \sum_{n=1}^{\infty} e^{-\pi n^{2} / x}\\ \end{aligned}$

$\boxed{\sum_{n=1}^{\infty} e^{-\pi n^{2} x}=-\frac12+\frac{1}{2\sqrt{x}}+\frac{1}{\sqrt{x}} \sum_{n=1}^{\infty} e^{-\pi n^{2} / x}}$ (4)

Functional Equation for the Riemann Zeta function

Let

$g(x)=\sum_{n=1}^{\infty} e^{-\pi n^{2} x}$

From (4) we know that

$g(x)=-\frac12+\frac{1}{2\sqrt{x}}+\frac{1}{\sqrt{x}} \,g\left(\frac{1}{x}\right)$ (5)

And consider it´s Mellin transform, i.e.

$\begin{aligned} \int_0^\infty g(x)x^{s-1}\,dx&=\int_0^\infty x^{s-1}\,\sum_{n=1}^{\infty} e^{-\pi n^{2} x}\,dx\\ &=\sum_{n=1}^{\infty}\,\int_0^\infty e^{-\pi n^{2} x} x^{s-1} \,dx\\ &=\frac{1}{\pi^s}\sum_{n=1}^{\infty}\frac{1}{n^{2s}}\,\int_0^\infty e^{- x} x^{s-1} \,dx\\ &=\frac{\Gamma(s)}{\pi^s}\sum_{n=1}^{\infty}\frac{1}{n^{2s}}\\ \end{aligned}$

If we set $s=\frac s2$ we obtain

$\int_0^\infty x^{\frac s2-1}\,g(x)\,dx=\frac{\Gamma\left(\frac s2\right)}{\pi^{\frac s2}}\sum_{n=1}^{\infty}\frac{1}{n^{s}}=\frac{\Gamma\left(\frac s2\right)\zeta(s)}{\pi^{\frac s2}}$

$\begin{aligned} \int_0^\infty x^{\frac s2-1}\,g(x)\,dx&=\int_0^1 x^{\frac s2-1}\,g(x)\,dx+\int_1^\infty x^{\frac s2-1}\,g(x)\,dx\\ &=\int_0^1 x^{\frac s2-1}\,\left( -\frac12+\frac{1}{2\sqrt{x}}+\frac{1}{\sqrt{x}} \,g\left(\frac{1}{x}\right)\right)\,dx+\int_1^\infty x^{\frac s2-1}\,g(x)\,dx\\ &=-\frac12\int_0^1 x^{\frac s2-1}\,dx +\frac{1}{2}\int_0^1 \frac{x^{\frac s2-1}}{\sqrt{x}}\,dx+\int_0^1 x^{\frac s2-\frac12-1} \,g\left(\frac{1}{x}\right)\,dx+\int_1^\infty x^{\frac s2-1}\,g(x)\,dx\\ &=-\frac1s +\frac{1}{2}\left(\frac{2}{s-1} \right)+\int_1^\infty x^{-\frac s2-\frac12} \,g\left(x\right)\,dx+\int_1^\infty x^{\frac s2-1}\,g(x)\,dx\\ &=-\frac1s +\frac{1}{s-1}+\int_1^\infty \left(x^{-\frac s2-\frac12}+ x^{\frac s2-1}\right)\,g(x)\,dx\\ \end{aligned}$

And we conclude that

$\boxed{\frac{\Gamma\left(\frac s2\right)\zeta(s)}{\pi^{s/2}}=-\frac1s +\frac{1}{s-1}+\int_1^\infty \left(x^{-\frac s2-\frac12}+ x^{\frac s2-1}\right)\,g(x)\,dx}$ (6)

Since $\left|g(x) \right| \to 0$ as $x \to \infty$ , the integral in (6) is an entire function of . It follows that $\Gamma\left(\frac s2\right)\zeta(s)$ is entire in the whole plane except for the points s=0 and s=1 where it has simple poles.

Now observe the following fact, if we let $s \to 1-s$ in (6) we obtain

$\boxed{\frac{\Gamma\left(\frac {1-s}{2}\right)\zeta(1-s)}{\pi^{\frac{s-1}{2}}}=-\frac1s +\frac{1}{s-1}+\int_1^\infty \left(x^{-\frac s2-\frac12}+ x^{\frac s2-1}\right)\,g(x)\,dx}$ (7)

Note that the Right Hand Side of (7) is identical to the R.H.S. of (6), we therefore conclude that the L.H.S. of (6) must be equal to the L.H.S. of (7), and we arrive at the remarkable equation

$\Gamma\left(\frac s2\right)\zeta(s)\pi^{-s/2}=\Gamma\left(\frac {1-s}{2}\right)\zeta(1-s)\pi^{-\frac{(s-1)}{2}}$ (8)

$\zeta(s)=\pi^{s-\frac12}}\frac{\Gamma\left(\frac {1-s}{2}\right)}{\Gamma\left(\frac s2\right)}\zeta(1-s)$ (9)

Now, recall Legendre Duplication Formula for the Gamma Function

$\Gamma\left(2x\right)2^{1-2x}=\frac{\Gamma\left(x\right)\Gamma\left(x+\frac12\right)}{\sqrt{\pi}}$ (10)

Letting $x=\frac{1-s}{2}$ in (10) we obtain

$\Gamma\left(1-s\right)2^{s}=\frac{\Gamma\left(\frac{1-s}{2}\right)\Gamma\left(1-\frac{s}{2}\right)}{\sqrt{\pi}}$

$\Gamma\left(\frac{1-s}{2}\right)=\frac{\Gamma\left(1-s\right)2^{s}\sqrt{\pi}}{\Gamma\left(1-\frac{s}{2}\right)}$ (11)

Plugging (11) in (9)

$\zeta(s)=(2\pi)^{s}\frac{\Gamma\left(1-s\right)}{\Gamma\left(\frac s2\right)\Gamma\left(1-\frac{s}{2}\right)}\zeta(1-s)$ (12)

From the Reflection formula of the Gamma function we can rewrite

$\Gamma\left(\frac s2\right)\Gamma\left(1-\frac{s}{2}\right)=\frac{\pi}{\sin\left(\frac{s \pi}{2} \right)}$

And we finally obtain the more familiar form of the functional equation

$\boxed{\zeta(s)=\frac{1}{\pi}(2\pi)^{s}\sin\left(\frac{s \pi}{2} \right)\Gamma\left(1-s\right)\zeta(1-s)}$ (13)

We have previously shown that $\zeta(-1)=-\frac{1}{12}$ using the Euler Maclaurin formula . Let´s now obtain the same result by means of the functional equation. Let s=-1 in (13):

$\begin{aligned} \zeta(-1)&=\frac{1}{\pi}(2\pi)^{-1}\sin\left(-\frac{ \pi}{2} \right)\Gamma\left(2\right)\zeta(2)\\ &=-\frac{1}{2 \pi^2}\frac{\pi^2}{6}\\ &=-\frac{1}{12} \qquad \blacksquare \end{aligned}$

Tough we can use the Euler Maclaurin to extend analytically the domain of the Zeta function for the the entire plane except for the point s=1 , in some cases the functional equation is more convenient to work with. And this is the case for obtaining an expression for $\zeta^\prime(2)$ . To this end, we start by differentiating (13) w.r. to s

$\begin{aligned} \zeta^\prime(s)&=\frac{1}{\pi}\left((2\pi)^{s}\sin\left(\frac{s \pi}{2} \right)\right)^\prime\Gamma\left(1-s\right)\zeta(1-s)+\frac{1}{\pi}(2\pi)^{s}\sin\left(\frac{s \pi}{2} \right)\left(\Gamma\left(1-s\right)\zeta(1-s) \right)^\prime\\ &=\frac{(2 \pi)^{s}}{\pi}\left( \log(2 \pi) \sin \left( \frac{\pi s}{2} \right) + \frac{\pi}{2} \cos \left(\frac{\pi s}{2} \right)\right) \Gamma(1-s) \zeta(1-s)- \frac{(2 \pi)^{s}}{\pi} \sin \left(\frac{\pi s}{2} \right)\left(\Gamma^{'}(1-s) \zeta(1-s) + \Gamma(1-s) \zeta'(1-s)\right)\\ \end{aligned}$ (14)

let $s \to -1$ in (14)

$\begin{aligned} \zeta^\prime(-1)&=\frac{(2 \pi)^{-1}}{\pi}\left( \ln(2 \pi) \sin \left( -\frac{\pi }{2} \right) + \frac{\pi}{2} \cos \left(-\frac{\pi }{2} \right)\right) \Gamma(2) \zeta(2)- \frac{(2 \pi)^{-1}}{\pi} \sin \left(-\frac{\pi }{2} \right)\left(\Gamma^{'}(2) \zeta(2) + \Gamma(2) \zeta'(2)\right)\\ &=-\frac{\ln(2 \pi)}{2 \pi^2}\frac{\pi^2}{6}+\frac{1}{2 \pi^2}\frac{\pi^2}{6}\Gamma(2)\psi(2)+\frac{\zeta^\prime(2)}{2 \pi^2}\\ &=-\frac{\ln(2 \pi)}{12}+\frac{(1-\gamma)}{12}+\frac{\zeta^\prime(2)}{2 \pi^2} \end{aligned}$

Recall that $\zeta^{\prime}(-1)=\frac{1}{12}-\ln A$ , therefore

$\begin{aligned} \frac{\zeta^\prime(2)}{2 \pi^2}&=\frac{\ln(2 \pi)}{12}-\frac{1}{12}+\frac{\gamma}{12}+\frac{1}{12}-\ln A\\ &=\frac{(\gamma+\ln(2\pi)-12 \ln A)}{12} \end{aligned}$

And finally

$\boxed{\zeta^\prime(2)=\zeta(2)(\gamma+\ln(2\pi)-12 \ln A)}$ (15)

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