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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:

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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 '13 at 3:24
@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 '13 at 16:40
@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 '13 at 20:34
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 '13 at 9:59
the trick is considering $\Xi(s) = \xi(1/2+s^{1/2})$ which is entire and (with the help of the Stirling formula for $\Gamma(s)$) of non-integer order ($1/2$) hence it has an infinite number of zerosêta_de_Riemann#Les_zéros_non_triviaux – user1952009 Apr 10 at 18:15
up vote 13 down vote accepted

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 (using Ramanujan's $(2.16.2)$ reproduced at the end) 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's first proof $10.2$, alternative proofs follow in the book...) : $$\lim_{\alpha\to\frac{\pi}4}\,\int_0^\infty\frac{\Xi(t)}{t^2+\frac 14}t^{2n}\cosh(\alpha t)\,dt=\frac{(-1)^n\,\pi\,\cos\bigl(\frac{\pi}8\bigr)}{4^n}$$ 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! $$-$$

From Titchmarsh's book "The Theory of the Riemann Zeta-function" p. $35-36\;$ and $\;255-258$ :

p 35 p 36

p 256 p 257b p 257 p 258

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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 '13 at 20:35
@PeteL.Clark: yes you are right of course but my answer had the $\xi$ function and the proof by contradiction! :-) Let's add that the OP asked a more focused question after that. – Raymond Manzoni Feb 10 '13 at 10:44
@RaymondManzoni - Your links to Ramanujan's and Titchmarsh's proofs no longer work. Can you comment the titles of those publications or update those links because I really want to read those proofs. – Ted Jul 9 '14 at 7:11
@nick: the two links were to Titchmarsh's book "The Theory of the Riemann Zeta-function". I updated my answer and added a link to Hardy's paper. Excellent reading, – Raymond Manzoni Jul 9 '14 at 11:20
@RaymondManzoni : please look at my answer below and tell me what you think, it is the same argument that a finite number of non-trivial zeros would make everything convergent applied to $\frac{\zeta'(s)}{\zeta(s)}$ – user1952009 Apr 10 at 18:28

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 '13 at 21:36
You missed the $e^{s/\rho}$ factor in the product. – user27126 Feb 8 '13 at 21:37
@user58512, you may want to look up Weierstrass factorization theorem. – user27126 Feb 8 '13 at 21:38
@Sanchez Did I? Actually, all I used about $\xi$ is copied from Wikipedia ... – Hagen von Eitzen Feb 9 '13 at 13:13

In 1912 Harald Bohr proved that there are infinitely many zeros of $\zeta(s)$ in $0\leq \sigma \leq 1$ by contradiction. He assumes that there are no zeros in the critical strip and deduces from this that that $1/|\zeta(s)|$ is bounded, contradicting his (earlier extrinsic) result that there exist arbitrarily small values of $\zeta(s)$ for a given range of $\sigma.$

This is as I recall simpler than Hadamard's proof and far simpler than the proof of Hardy's stronger result. As far as I know there is no English version but the Danish is accessible with online translation tools. As I recall the theorem makes use of Caratheodory's theorem to establish a bound.

If I make it back to my lending library I will try to post a better synopsis but for anyone who is interested here is the cite: Et nyt for, at den Riemann'ske Zetafunktion $\zeta(s) = \zeta(\sigma + it)$ har uendelig mange Nulpunkter idenfor Parallelstrimlen $0\leq \sigma\leq 1.$ B9, Harald Bohr, Collected Math. Works, Vol. I.

Otherwise there is Hadamard's theorem which is given in Titchmarsh's Theory of the Riemann Zeta Function at p. 30. A better synopsis of this approach than I could give is found in a brief note by Paul Garrett but I think Hagen von Eitzen's answer is also a good distillation of this.

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why can not use a representation for $ \int_{0}^{\infty} \frac{ \xi(a+it)}{a^{2}+t^{2}}cos(xt)=g(x) $ for some g(x) and then prove this integral does not change of sign her a is different from 1/2

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(I think this is an elementary proof)

everybody here knows that the zeros of $\zeta(s)$ are useful mainly for the Riemann explicit formula :

$$\psi(x) = \sum_{p^k \le x} \ln p = x - \sum_{\rho} \frac{x^\rho}{\rho}-\ln(2 \pi)- \frac{1}{2}\ln(1-x^{-2})$$

suppose there is a finite number of non-trivial zeros, then $\frac{d}{dx}\psi(x)$ could not be the distribution : $$\sum_{p \in \mathcal{P}} \sum_{k=1}^\infty \ \delta(x-p^k) \ \ln p $$

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