# Why does zeta have infinitely many zeros in the critical strip?

I want a simple proof that $\zeta$ has infinitely many zeros in the critical strip.

The function $$\xi(s) = \frac{1}{2} s (s-1) \pi^{\tfrac{s}{2}} \Gamma(\tfrac{s}{2})\zeta(s)$$ has exactly the non-trivial zeros of $\zeta$ as its zeros ($\Gamma$ cancels all the trivial ones out). It also satisfies the functional equation $\xi(s) = \xi(1-s)$.

If we assume it has finitely many zeros, what analysis could get a contradiction?

I found an outline for a way to do it here but I can't do the details myself: http://mathoverflow.net/questions/13647/why-does-the-riemann-zeta-function-have-non-trivial-zeros/13762#13762

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The first answer of that MO post is much shorter, comparing to the answer you are quoting. Do you somehow find the first answer unsatisfactory? – Sanchez Feb 8 at 21:34

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Questions on the RH should be barred from SE imo. What real purpose does this question serve? – Jp McCarthy Feb 9 at 3:24

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@JpMcCarthy, there are (two) crucial logical differences between infinitely many zeroes in the critical strip (the subject of OPs question) and every nontrivial zero is on the critical line (the Riemann hypothesis). Are you and those who upvoted your comment aware of this? While this question is on the topic of the zeta zeroes, which are also the subject of RH, this question (in contrast) is by design amenable to objective, definitive, enlightening and accessible answers by relevant experts in the here-and-now, and therefore constitutes a legitimate query for this site. – anon Feb 9 at 16:40

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@JpMcCarthy: I don't understand your objection at all. First, the OP is not asking about RH, but about a very much simpler and more basic property of the Riemann zeta function (one which was well known to Riemann in the 19th century). Second, you ask: "What real purpose does this question serve?" Well, the OP wants a simple proof that the zeta function has infinitely many nontrivial zeros. Answering the question will fulfill this desire of the OP and may be useful to others. In other words, it serves the same purpose as most other questions asked on this site. – Pete L. Clark Feb 9 at 20:34

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While I do understand the difference you allude to anon, I must admit that I have misjudged the situation. I withdraw my comment. – Jp McCarthy Feb 12 at 9:59
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## 2 Answers

Hardy proved in 1914 that an infinity of zeros were on the critical line ('Sur les zéros de la fonction $\zeta(s)$ de Riemann' Comptes rendus hebdomadaires des séances de l'Académie des sciences. 1914).
Of course other zeros could exist elsewhere in the critical strip.

Let's exhibit the main idea starting with the Xi function defined by : $$\Xi(t):=\xi\left(\frac 12+it\right)=-\frac 12\left(t^2+\frac 14\right)\,\pi^{-\frac 14-\frac{it}2}\,\Gamma\left(\frac 14+\frac{it}2\right)\,\zeta\left(\frac 12+it\right)$$ $\Xi(t)$ is an even integral function of $t$, real for real $t$ because of the functional equation (applied to $s=\frac 12+it$) : $$\xi(s)=\frac 12s(s-1)\pi^{-\frac s2}\,\Gamma\left(\frac s2\right)\,\zeta(s)=\frac 12s(s-1)\pi^{\frac {s-1}2}\,\Gamma\left(\frac {1-s}2\right)\,\zeta(1-s)=\xi(1-s)$$ We observe that a zero of $\zeta$ on the critical line will give a real zero of $\,\Xi(t)$.

Now it can be proved (Ramanujan's $(2.16.2)$) that : $$\int_0^\infty\frac{\Xi(t)}{t^2+\frac 14}\cos(x t)\,dt=\frac{\pi}2\left(e^{\frac x2}-2e^{-\frac x2}\psi\left(e^{-2x}\right)\right)$$ where $\,\displaystyle \psi(s):=\sum_{n=1}^\infty e^{-n^2\pi s}\$ is the theta function used by Riemann

Setting $\ x:=-i\alpha\$ and after $2n$ derivations relatively to $\alpha$ we get (see Titchmarsh $10.2$) : $$\lim_{\alpha\to\frac{\pi}4}\,\int_0^\infty\frac{\Xi(t)}{t^2+\frac 14}t^{2n}\cosh(\alpha t)\,dt=\frac{(-1)^n\,\pi\,\cos\bigl(\frac{\pi}8\bigr)}{4^n}$$ Let's suppose that $\Xi(t)$ doesn't change sign for $\,t\ge T\,$ then the integral will be uniformly convergent with respect to $\alpha$ for $0\le\alpha\le\frac{\pi}4$ so that, for every $n$, we will have (at the limit) : $$\int_0^\infty\frac{\Xi(t)}{t^2+\frac 14}t^{2n}\cosh\left(\frac {\pi t}4\right)\,dt=\frac{(-1)^n\,\pi\,\cos\bigl(\frac{\pi}8\bigr)}{4^n}$$

But this is not possible since, from our hypothesis, the left-hand side has the same sign for sufficiently large values of $n$ (c.f. Titchmarsh) while the right part has alternating signs.
This proves that $\Xi(t)$ must change sign infinitely often and that $\zeta\left(\frac 12+it\right)$ has an infinity of real solutions $t$.

Probably not as simple as you hoped but a stronger result!

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This is nice, but let's be clear that it's a much stronger result than what the OP asked for. – Pete L. Clark Feb 9 at 20:35
– Raymond Manzoni Feb 10 at 10:44

It is known that $$\xi(s)=\frac12\prod_{\xi(s)=0}\left(1-\frac s\rho\right),$$ i.e. $\xi$ would turn out to be a polynomial of degree $n$, say. Then we conclude that $\ln \xi(s)\sim n\ln s$ as $\mathbb R \ni s\to \infty$, but it is known that $\ln \xi(s)\sim s\ln s$.

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I don't know how to prove any of those 3 claims. – user58512 Feb 8 at 21:36
You missed the $e^{s/\rho}$ factor in the product. – Sanchez Feb 8 at 21:37
@user58512, you may want to look up Weierstrass factorization theorem. – Sanchez Feb 8 at 21:38

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– Hagen von Eitzen Feb 9 at 13:13
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