History of Attempts

Imagine you have a very long piano keyboard that goes on forever.

There’s a secret piece of music hiding in it. When you press a key at the right place, the sound suddenly becomes completely quiet — that’s a zero.

Riemann’s big question is about where those quiet keys live. There’s a special “line” drawn on the keyboard, and people want to know whether lots of the quiet keys sit exactly on that line.

What Hardy’s 1914 proof says is:

“No matter how far you walk along that line, you will always be able to find more quiet keys. There isn’t just one, or ten, or a hundred — there are infinitely many.”

He does this by turning the hidden music into a kind of math recipe, and then showing that if you tried to have only a few quiet keys, the recipe would give an impossible answer. So the universe is forced to give you infinitely many of them.

Think of a graph where a curvy line goes up and down across the page. Where the curve crosses the horizontal axis (where the height is zero), we say the function has a zero.

Riemann’s zeta function is a very complicated function that lives in the complex plane (two-dimensional numbers, like coordinates \((\sigma, t)\)). There is a special vertical line called the critical line, where \(\sigma = \tfrac12\). One of the big mysteries in math is whether all the interesting zeros lie on that line (that’s the Riemann Hypothesis).

Hardy didn’t solve the whole mystery, but he did something important:

He proved that there are infinitely many zeros of the zeta function on that special line.

Roughly, he:

1. Writes down a clever integral formula for a related function \(\Xi(t)\) that has the same zeros on the critical line.
2. Tweaks this formula by turning a knob called \(\alpha\) and by taking higher and higher derivatives — this makes the integral focus more on certain parts of the graph.
3. Uses another piece of theory to show that a “messy extra term” actually shrinks to zero when the knob \(\alpha\) is turned to a particular position.
4. Now the integral must behave like a simple expression that changes sign as he changes a parameter \(p\).
5. But if \(\Xi(t)\) had only finitely many zeros, eventually it would stop crossing the axis and stay just positive or just negative — which would make the same integral keep the same sign.

Those two conclusions clash. The only way out: \(\Xi(t)\) — and hence the zeta function on the critical line — must cross the axis infinitely many times.

You can think of Riemann’s zeta function \(\zeta(s)\) as a function of a complex variable \(s = \sigma + it\). Its nontrivial zeros are the complex numbers where \(\zeta(s) = 0\) and \(0 < \sigma < 1\).

There is a special vertical line in the complex plane, called the critical line, given by \(\sigma = \tfrac12\). The Riemann Hypothesis claims that all nontrivial zeros lie on this line. Hardy’s 1914 result is a partial victory:

There are infinitely many zeros of \(\zeta(s)\) on the critical line.

Here is the idea of his proof, skipping technical details:

1. Use Riemann’s functional equation to define an entire function \[ \Xi(t) = \xi\!\left(\tfrac12 + it\right), \] whose zeros on the real axis correspond exactly to zeros \(\zeta(\tfrac12 + it) = 0\) on the critical line.
2. Starting from a Mellin-type integral involving \(\zeta(2u)\), Hardy derives a formula \[ \int_0^\infty \frac{\bigl(e^{\alpha t} + e^{-\alpha t}\bigr)\,\Xi(2t)}{\tfrac14+4t^2}\,dt \;=\; \pi\cos\Bigl(\tfrac{\alpha\pi}{4}\Bigr) - \frac{\pi}{2}e^{i\alpha\pi/4} F(q), \] where \(F(q)\) is a kind of theta series and \(q\) depends on a real parameter \(\alpha\).
3. Differentiate this identity \(2p\) times with respect to \(\alpha\). That inserts a factor \(t^{2p}\) into the integral, so for large \(p\) the integral “weights” large values of \(t\) more heavily.
4. A separate lemma (using the theory of Dirichlet series) shows that as \(\alpha \to \tfrac{\pi}{2}\), the theta-series term on the right tends to zero. In that limit the identity simplifies to \[ \int_0^\infty \frac{\bigl(e^{\alpha t} + e^{-\alpha t}\bigr)\,t^{2p}\,\Xi(2t)}{\tfrac14+4t^2}\,dt \;\longrightarrow\; \frac{(-1)^p\pi}{4^p}\cos\Bigl(\tfrac{\pi}{8}\Bigr), \] which alternates in sign as \(p\) increases.
5. Now assume, for contradiction, that \(\Xi(t)\) has only finitely many real zeros. Then beyond some point \(T\), \(\Xi(t)\) never changes sign again — say it is always positive. But in that case the integral above, for large \(p\), is a positive quantity (roughly proportional to \(T^{2p}\)), because the kernel and \(\Xi(2t)\) are both positive on a long interval.

So under this assumption, the integral should be positive for all large \(p\), but Hardy’s identity says it alternates between positive and negative values. That’s impossible, so our assumption must be wrong:

\(\Xi(t)\) — and thus \(\zeta(\tfrac12 + it)\) — has infinitely many real zeros.

→ Read the full technical proof

Hardy’s 1914 result shows that the zeta function \[ \zeta\!\left(\tfrac12 + it\right) \] has infinitely many zeros on the critical line. To understand the proof at an undergraduate level, it helps to see the overall shape of the argument rather than every technical detail.

1. Move to the completed function \(\Xi(t)\)

The functional equation lets us package \(\zeta(s)\) into an entire function \[ \xi(s) = \tfrac12 s(s-1)\,\pi^{-s/2}\Gamma\!\left(\tfrac{s}{2}\right)\zeta(s), \] and on the critical line we set \[ \Xi(t) = \xi\!\left(\tfrac12 + it\right). \] Then \(\Xi(t)\) is real for real \(t\), and \(\Xi(t) = 0 \iff \zeta(\tfrac12 + it) = 0\). Thus it is enough to show \(\Xi(t)\) has infinitely many real zeros.

2. A key integral identity

Hardy begins with a Mellin transform identity (a standard tool in complex analysis) and rewrites it using the functional equation. After a change of variables, he arrives at a parameter-dependent equality of the form

\[ \int_0^\infty \frac{\bigl(e^{\alpha t}+e^{-\alpha t}\bigr)\,\Xi(2t)} {\tfrac14 + 4t^2}\,dt = \pi \cos\!\left(\tfrac{\alpha\pi}{4}\right) -\frac{\pi}{2}e^{i\alpha\pi/4}F(q), \tag{1} \]

where

• \(\alpha\) is a real parameter you can “turn”,
• \(F(q)\) is a theta-type series,
• \(q = e^{-\pi e^{4\alpha}}\).

Think of this as a weighted average of \(\Xi(2t)\) expressed in a clean, structured way.

3. Higher moments: differentiating in \(\alpha\)

When we differentiate (1) with respect to \(\alpha\), the exponential factors introduce powers of \(t\). After \(2p\) derivatives we get

\[ \int_0^\infty \frac{\bigl(e^{\alpha t}+e^{-\alpha t}\bigr)\,t^{2p}\,\Xi(2t)} {\tfrac14+4t^2}\,dt = \frac{(-1)^p\pi}{4^p} \cos\!\left(\tfrac{\alpha\pi}{4}\right) -\cdots, \tag{2} \]

which is now a moment of \(\Xi\): for larger \(p\), the integrand places more weight on larger values of \(t\), but the kernel is still a smooth, positive function for real \(\alpha\).

4. A limit where the extra term disappears

A central technical step (proved using tools from Dirichlet-series theory) is:

• as \(\alpha \to \tfrac{\pi}{2}\),
• the theta series term \(F(q)\) becomes negligible for each fixed \(p\).

In this limit, equation (2) simplifies to

\[ \int_0^\infty \frac{\bigl(e^{\alpha t}+e^{-\alpha t}\bigr)\,t^{2p}\,\Xi(2t)} {\tfrac14+4t^2}\,dt \xrightarrow[\alpha\to\pi/2]{} \frac{(-1)^p\pi}{4^p} \cos\!\left(\tfrac{\pi}{8}\right). \tag{3} \]

The crucial point: the right-hand side alternates in sign as \(p\) increases.

5. The contradiction if \(\Xi\) had only finitely many zeros

Assume for contradiction that \(\Xi(t)\) eventually keeps one sign—say \(\Xi(t) > 0\) for all sufficiently large \(t\).

Then for large \(p\), because the kernel is positive and heavily weighted toward large \(t\), the left side of (3) would also be strictly positive.

But the right side of (3) flips sign with \(p\). Thus the identity cannot hold for all large \(p\). The assumption must be false.

Therefore \(\Xi(t)\) changes sign infinitely often, which means \(\zeta(\tfrac12+it)\) has infinitely many zeros on the critical line.

→ Read the full technical exposition with all identities

Hardy’s 1914 result is best viewed as a sign-change argument applied to the completed zeta function \[ \Xi(t) = \xi\!\left(\tfrac12 + it\right), \] whose real zeros correspond exactly to zeros of \(\zeta(s)\) on the critical line.

The proof uses:

1. A Mellin transform identity expressing a theta series in terms of \(\zeta(2u)\).
2. The functional equation to rewrite this in terms of \(\Xi(2t)\).
3. A tunable parameter \(\alpha\) and repeated differentiation, which produces the weighted moments \[ \int_0^\infty K_{p,\alpha}(t)\,\Xi(2t)\,dt \] with \(K_{p,\alpha}(t) \sim t^{2p}\) for large \(p\).
4. Bohr–Riesz summability (via the Dirichlet series \((1-2^{1-s})\zeta(2s)\)) to show that the theta-series contribution vanishes when \(\alpha \to \pi/2\).
5. A resulting asymptotic moment formula whose right side alternates in sign with \(p\).
6. A contradiction: if \(\Xi(t)\) eventually kept one sign, the same moments would be eventually positive.

This forces \(\Xi(t)\) to change sign infinitely often, giving infinitely many zeros of \(\zeta(\tfrac12+it)\) on the critical line.

For a full, structured exposition — including all the identities and the limiting argument — see the detailed proof: → Hardy (1914) full technical proof

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