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I'm reading through Titchmarch's "The Theory of the Riemann Zeta-Function" and there's a part in the functional equation proof number 3 that I haven't figured out.

He defines a function $$\psi(x)=\sum_{n=1}^\infty e^{-n^2\pi x}$$ and next, for $x>0$ it is known that $$ \sum_{n=-\infty}^\infty e^{-n^2\pi x}=\frac1{\sqrt{x}}\sum_{n=-\infty}^\infty e^{-\frac{n^2\pi}x}, $$ or $$2\psi(x)+1=\frac1{\sqrt{x}}\left( 2\psi\left(\frac1{x}\right)+1\right).$$ Where does the second equation come from exactly?

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    $\begingroup$ it's Riemann original proof, but the most common today is fr.wikipedia.org/wiki/… $\endgroup$ – reuns Feb 13 '16 at 13:50
  • $\begingroup$ Thanks for the link. Is there an english version available? $\endgroup$ – Rasmus Erlemann Feb 13 '16 at 13:59
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It is an application of Poisson summation formula. To verify it you only have to compute the Fourier transform: $$\frac{1}{\sqrt{x}}e^{-\frac{n^2\pi}{x}} = \int_{-\infty}^\infty e^{-m^2\pi x -2\pi i m n} dm.$$

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  • $\begingroup$ Actually. I don't understand this answer. Maybe I need to brush up on my fourier transforms. $\endgroup$ – Peter Driscoll Aug 23 '17 at 3:20
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I spelled this argument out in some detail, so that I could understand it. The argument is based on https://www.youtube.com/watch?v=-GQFljOVZ7I, but elaborated to my tastes.

Define $\theta(x) = 2 \psi(x) + 1 $. Then, $$\theta(x)=(\sum_{n=-\infty}^1 e^{-n^2\pi x}) + 1 + (\sum_{n=1}^\infty e^{-n^2\pi x}) =\sum_{n=\infty}^\infty e^{-n^2\pi x}$$

The Poisson summation formula. is, $$ \sum_{n-\infty}^\infty{f(n)} = \sum_{k=-\infty}^\infty \int_{-\infty}^\infty f(y) e^{-2\pi iky} x dy$$ Substituting $ f(n) = e^{-n^2\pi x}$ and $ f(y) = e^{-y^2\pi x}$ gives,

$$ \theta(x) = \sum_{n-\infty}^\infty{e^{-n^2\pi x}} = \sum_{k=-\infty}^\infty \int_{-\infty}^\infty e^{-y^2\pi x} e^{-2\pi iky} dy = \sum_{k=-\infty}^\infty \int_{-\infty}^\infty e^{-\pi x(y^2 + 2iy \frac{k}{x})} dy$$

Complete the square by adding and subtracting a term $ i^2 \frac{k^2}{x^2} $ $$ \theta(x) = \sum_{k=-\infty}^\infty \int_{-\infty}^\infty e^{-\pi x(y^2 + 2iy \frac{k}{x} + i^2\frac{k^2}{x^2} - i^2\frac{k^2}{x^2})} dy$$

Substituting $ y^2 + 2iy \frac{k}{x} + i^2 y^2\frac{k^2}{x^2} = (y + i \frac{k}{x})^2 $ and $ i^2\frac{k^2}{x^2} = - \frac{k^2}{x^2}$ gives, $$ \theta(x) = \sum_{k=-\infty}^\infty \int_{-\infty}^\infty e^{-\pi x((y + i \frac{k}{x})^2 + \frac{k^2}{x^2})} dy = \sum_{k=-\infty}^\infty e^{-\pi\frac{k^2}{x}} \int_{-\infty}^\infty e^{-\pi x(y + i \frac{k}{x})^2} dy $$

as $ e^{-\pi\frac{k^2}{x}} $ is not a function of y, and so can be moved outside the integral.

An argument can then be made using path integrals that that follow a rectangle around the complex plain that, $$ \int_{-\infty}^\infty e^{-\pi x(y + i \frac{k}{x})^2} dy = \int_{-\infty}^\infty e^{-\pi x z^2} dz = \frac1{\sqrt{\pi x}} \int_{-\infty}^\infty e^{-z^2} dz = \frac{\sqrt{\pi}}{\sqrt{\pi x}} = \frac1{\sqrt{x}}$$

The argument uses the fact that a closed integral around no poles is zero, and also relies on terms at infinity going to zero. This is standard complex analysis, but it is not easy to see without a diagram. It really should be proved separately.

Also using, $$\theta(\frac1{x})=\sum_{k=1}^\infty e^{-\frac{k^2\pi}{x}}$$

$$ \theta(x) = \sum_{k=-\infty}^\infty e^{-\pi\frac{k^2}{x}} \int_{-\infty}^\infty e^{-\pi x(y + i \frac{k}{x})^2} dy = \frac{\theta(\frac1{x})}{\sqrt{x}} $$

Substituting back $ \theta(x) = 2 \psi(x) + 1 $ gives the equation in terms of $\psi$.

$$ 2 \psi(x) + 1 =\frac{2 \psi(\frac1{x}) + 1}{\sqrt{x}} $$


The proof uses a result of the form, $$ \int_{-\infty}^\infty e^{-(y + ib)^2} dy = \int_{-\infty}^\infty e^{-z^2} dz $$

Consider $ \int e^{-z^2} dz $ arround a rectangular path P given by $ -\infty \to \infty \to \infty+ib \to -\infty+ib \to -\infty $. From Cauchy's integral theorem this integral must be zero because it is a closed loop, and $ e^{-z^2} $ has no poles. $$ \int_{P} e^{-z^2} dz = 0 $$ so, $$ 0 = \int_{-\infty}^{\infty} e^{-y^2} dy + \lim_{Y \to \infty} \int_{0}^{b} e^{-(Y + ix)^2} dx + \int_{\infty}^{-\infty} e^{-(y+ib)^2} dy + \lim_{Y \to -\infty} \int_{b}^{0} e^{-(Y + ix)^2} dx $$

$$ 0 = \int_{-\infty}^{\infty} e^{-y^2} dy + \lim_{Y \to \infty} \int_{0}^{b} e^{-(Y + ix)^2} dx - \int_{-\infty}^{\infty} e^{-(y+ib)^2} dy - \lim_{Y \to \infty} \int_{0}^{b} e^{-(Y - ix)^2} dx $$

And I claim that, $$ \lim_{Y \to \infty} \int_{0}^{b} e^{-(Y + ix)^2} dx = \lim_{Y \to \infty} \int_{0}^{b} e^{-(Y - ix)^2} dx = 0 $$

which gives, $$ 0 = \int_{-\infty}^{\infty} e^{-y^2} dy - \int_{-\infty}^{\infty} e^{-(y+ib)^2} dy $$

To prove the claim, firstly, $$ \lim_{Y \to \infty} \int_{0}^{b} e^{-(Y \pm ix)^2} dx = \lim_{Y \to \infty} \int_{0}^{b} e^{-(Y^2 \pm 2iYx -x^2)} dx = \lim_{Y \to \infty} \int_{0}^{b} e^{-Y^2} e^{\pm iYx} e^{x^2} dx $$ Consider the complex magnitude, $$ |\lim_{Y \to \infty} \int_{0}^{b} e^{-Y^2} e^{\pm 2iYx} e^{x^2} dx| \le \lim_{Y \to \infty} \int_{0}^{b} |e^{-Y^2}| |e^{\pm 2iYx}| |e^{x^2}| dx $$

But, $ |e^{\pm 2iYx}| = 1 $ and $ \lim_{Y \to \infty} {|e^{-Y^2}|} = 0 $ and $\int_{0}^{b} |e^{x^2}| dx$ is finite giving,

$$ \lim_{Y \to \infty} \int_{0}^{b} |e^{-Y^2}| |e^{\pm iYx}| |e^{-x^2}| dx = \lim_{Y \to \infty} \int_{0}^{b} |e^{-Y^2}| |e^{-x^2}| dx = (\lim_{Y \to \infty} {|e^{-Y^2}|}) \int_{0}^{b} |e^{x^2}| dx = 0$$ So,

$$ |\lim_{Y \to \infty} \int_{0}^{b} e^{-(Y \pm ix)^2} dy| \le 0 \implies \lim_{Y \to \infty} \int_{0}^{b} e^{-(Y \pm ix)^2} dy = 0 $$

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  • $\begingroup$ Yes the core formula is $\int_{-\infty}^\infty e^{-\pi t^2 x} e^{-2i \pi n t}dt = x^{-1/2} e^{-\pi n^2/x} $. Then let $\psi_x(t) = \sum_{k=-\infty}^\infty e^{-\pi (t-k)^2 x}$ which is $C^\infty$ and $1$-periodic. Write its Fourier series $\psi_x(t) = \sum_{n=-\infty}^\infty c_n e^{2i \pi nt}$ where $c_n = \int_0^1 \psi_x(t) e^{-2i \pi n t}dt = \int_{-\infty}^\infty e^{-\pi t^2 x} e^{-2i \pi n t}dt = x^{-1/2} e^{-\pi n^2/x}$ and hence $$\sum_{k=-\infty}^\infty e^{-\pi k^2 x}=\psi_x(0) = \sum_{n=-\infty}^\infty x^{-1/2} e^{-\pi n^2/x}=x^{-1/2}\psi_{1/x}(0)$$ $\endgroup$ – reuns Aug 19 '17 at 13:52
  • $\begingroup$ Finally $$\pi^{-s/2}\Gamma(s/2)\zeta(s) = \int_0^\infty (\psi_x(0)-1) x^{s/2-1}dx \\= \int_1^\infty (\psi_x(0)-1) x^{s/2-1}dx+\int_1^\infty (\psi_{1/x}(0)-1) x^{-s/2-1}dx \\= \frac{1}{s-1}-\frac{1}{s}+\int_1^\infty (\psi_{x}(0)-1) (x^{-s/2-1}+x^{(1-s)/2-1})dx$$ $\endgroup$ – reuns Aug 19 '17 at 13:52
  • $\begingroup$ @reuns Yes I appreciate that I am taking a fairly mechanical approach. $\endgroup$ – Peter Driscoll Aug 19 '17 at 13:56
  • $\begingroup$ @PeterDriscoll, at least you leave no room for doubt! :) But, yes, it might still be convincing if you were to omit a significant fraction of the steps! :) But, yes, still, I think everyone has to go through these details at some point in their lives, to truly believe... $\endgroup$ – paul garrett Aug 21 '17 at 0:36
  • $\begingroup$ @ paul garrett It is the OCD in me ;) $\endgroup$ – Peter Driscoll Aug 22 '17 at 1:05

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