Proving property of $\xi(s) $ in Riemann zeta function

Question

I am trying some exercise from textbook 'Introduction to ANT' by Tom Apostol as our course instructor didn't gave any assignments.

I am struck on following question (Ch-12 , Question 6 , page 274).

I have proved that zeroes will be symmetric about $t=0$ and they lie in strip $0\leq \sigma\leq 1$, but not able to prove that they will also be symmetric about $\sigma = \frac{1}2$ .

If $s$ is a zero then $1-s$ is also a zero and I know of using $s=1-s \implies s=\frac{1}2$ is line of symmetry but I don't know why to find such line we always put $s=1-s$ , i.e,. equation in function in LHS = Equation in function in RHS.

Can someone please tell rigorously why it's done so?

Thanks!!

Answer 1

Thinking aloud: The reason we consider $s=1-s$ for symmetry about $1/2$ can be reasoned as follows: if $0<\sigma<1/2$ then that's $1/2-\sigma$ from $1/2$ (the alleged line of symmetry), so the opposite would be $1/2+(1/2-\sigma)=1-\sigma$. Adding the imaginary part into the mix gives $s$ and $1-s$.

Proof of symmetry about $\sigma=1/2$: This might be overkill for the question since it asks about zeros only, but you could prove the more general result that $\xi(s)=\xi(1-s)$. Here's one way to do it where I'll use the slightly modified definition $\xi(s)\equiv\pi^{-s/2}\Gamma(s/2)\zeta(s).$

Consider the two lemmas:

1. $\int_0^\infty x^{s/2-1}e^{-n^2\pi x}dx=M\left[\sum_{k=0}^\infty\frac{(n^2\pi)^k}{k!}(-x)^k\right](s/2)\stackrel{RMT}{=}\frac{\Gamma(s/2)}{n^s\pi^{s/2}}.$

2. Let $a\in\mathbb{R}$ and define $\psi_a(x)=\sum_{n=a}^\infty e^{-n^2\pi x}$, then $$\psi_{-\infty}(x)=\frac{\psi_{-\infty}(1/x)}{\sqrt{x}}\quad\text{and}\quad\psi_{-\infty}(x)=2\psi_1(x)+1$$ $$\implies\psi_1(x)=\frac{1}{\sqrt{x}}\left(\psi_1(1/x)+\frac{1}{2}\right)-\frac{1}{2}.$$

Now the proof I wrote based on Titchmarsh, (1986). By lemma, summing over all positive integers $n$, then for $\sigma>1$, $$\frac{\Gamma(s/2)\zeta(s)}{\pi^{s/2}}=\sum_{n=1}^\infty\int_0^\infty x^{s/2-1}e^{-n^2\pi x}dx=\int_0^\infty x^{s/2-1}\psi_1(x)dx,$$ where we swapped integral and sum by virtue of absolute convergence. Splitting the integral over $(0,1]\cup[1,\infty)$ and substituting the second lemma gives $$\int_0^1 x^{s/2-1}\left(\frac{1}{\sqrt{x}}\left(\psi_1(1/x)+\frac{1}{2}\right)-\frac{1}{2}\right)dx+\int_1^\infty x^{s/2-1}\psi_1(x)dx.\tag{$\star$}$$ Expanding the first integral gives \begin{eqnarray*} & &\int_0^1 x^{s/2-3/2}\psi_1(1/x)dx+\frac{1}{2}\int_0^1 x^{s/2-3/2}dx-\frac{1}{2}\int_0^1 x^{s/2-1}dx\\ &=&\int_0^1 x^{s/2-3/2}\psi_1(1/x)dx+\frac{1}{s(s-1)}. \end{eqnarray*} Now using the change of variable $x=1/y$ where $dx=-1/y^2dy$, $$-\int_\infty^1 y^{-s/2+3/2}\psi_1(y )\frac{dy}{y^2}+\frac{1}{s(s-1)}\equiv\int_1^\infty x^{-s/2-1/2}\psi_1(x )dx+\frac{1}{s(s-1)}.$$ Substituting back into ($\star$) gives $$\pi^{-s/2}\Gamma(s/2)\zeta(s)=\frac{1}{s(s-1)}+\int_1^\infty \left(x^{-s/2-1/2}+x^{s/2-1}\right)\psi_1(x)dx,$$ which converges for all $s$, thereby so does the LHS by analytic continuation. Now, if we replace $s$ by $1-s$ in the RHS, then (i) $(1-s)(1-s-1)=(1-s)(-s)=s(s-1)$, (ii) $-(1-s)/2-1/2=s/2-1$ and (iii) $(1-s)/2-1=-s/2-1/2$, which leaves the RHS unchanged. Therefore, $$\pi^{-s/2}\Gamma(s/2)\zeta(s)=\pi^{-(1-s)/2}\Gamma((1-s)/2)\zeta(1-s),$$ which is the functional equation $\xi(s)=\xi(1-s)$.