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Difference between revisions of "Prime Number Theorem"

(Rewrite!)
(→‎The Bounded Integral)
 
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Now, for <math>\Re s > 1</math>,
 
Now, for <math>\Re s > 1</math>,
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
\int\limits_0^\infty [ \vartheta(e^t)e^{-t(s+1)}- e^{-s} ] dt
+
\int\limits_0^\infty [ \vartheta(e^t)e^{-t(s+1)}- e^{-st} ] dt
 
&= \int\limits_1^\infty \left[ \frac{\vartheta(x)}{x^{s+2}}
 
&= \int\limits_1^\infty \left[ \frac{\vartheta(x)}{x^{s+2}}
 
- \frac{1}{x^{s+1}} \right] dx \\
 
- \frac{1}{x^{s+1}} \right] dx \\
Line 133: Line 133:
 
<cmath> \begin{align*}
 
<cmath> \begin{align*}
 
\sum_{k=0}^\infty \vartheta(p_k)(1/p_k^{s+1} - 1/p_{k+1}^{s+1})
 
\sum_{k=0}^\infty \vartheta(p_k)(1/p_k^{s+1} - 1/p_{k+1}^{s+1})
−
&= \sum_{k=0}^\infty \sum_{i=0}^k \log p_k (1/p_k^{s+1} - 1/p_{k+1}^{s+1}) \\
+
&= \sum_{k=0}^\infty \sum_{i=0}^k \log p_i (1/p_k^{s+1} - 1/p_{k+1}^{s+1}) \\
 
&= \sum_{k=0}^{\infty} \frac{\log p_k}{p_k^{s+1}} \\
 
&= \sum_{k=0}^{\infty} \frac{\log p_k}{p_k^{s+1}} \\
 
&= \phi(s+1). \end{align*} </cmath>
 
&= \phi(s+1). \end{align*} </cmath>
−
Thus for <math>\Re s >1</math>,
+
Thus for <math>\Re s >1</math> (for the integral),
 
<cmath> \int_0^\infty [\vartheta(e^t)e^{-t} -1 ]e^{-st} dt =
 
<cmath> \int_0^\infty [\vartheta(e^t)e^{-t} -1 ]e^{-st} dt =
 
\frac{\phi(s+1)}{s+1} - \frac{1}{s} = g(s). </cmath>
 
\frac{\phi(s+1)}{s+1} - \frac{1}{s} = g(s). </cmath>
Line 156: Line 156:
 
such that there are infinitely many <math>x</math> for which
 
such that there are infinitely many <math>x</math> for which
 
<math>\vartheta(x) \ge \lambda x </math>.  Then for all such <math>x</math>,
 
<math>\vartheta(x) \ge \lambda x </math>.  Then for all such <math>x</math>,
−
<cmath>\begin{align*}
+
<cmath>
−
\int\limits_x^{\lambda{x}} \frac{\vartheta(t) -t}{t^2} dt
+
\begin{align*}
−
&\ge \int\limits_x^{\lambda{x}} \frac{\lambda x - t}{t^2}dt \\
+
\int\limits_x^{\lambda{x}} \frac{\vartheta(t) -t}{t^2} dt&\ge \int\limits_x^{\lambda{x}} \frac{\lambda x - t}{t^2}dt \\
−
&= \lambda x \left( \frac{1}{x} - \frac{1}{\lambda x} \right)
+
&= \lambda x \left( \frac{1}{x} - \frac{1}{\lambda x} \right)- \left(\log (\lambda x) - \log x\right) \\
−
- \left(\log (\lambda x) - \log x) \\
 
 
&= \lambda -1 - \log \lambda .
 
&= \lambda -1 - \log \lambda .
−
\end{align*}</cmath>
+
\end{align*}
 +
</cmath>
 
Now, <math>d(x-1 - \log x)/dx = 1 - 1/x</math>; it follows that
 
Now, <math>d(x-1 - \log x)/dx = 1 - 1/x</math>; it follows that
 
<cmath> \lambda - 1 - \log \lambda \ge 0 , </cmath>
 
<cmath> \lambda - 1 - \log \lambda \ge 0 , </cmath>
Line 171: Line 171:
 
such that there are infinitely many <math>x</math> for which
 
such that there are infinitely many <math>x</math> for which
 
<math>\vartheta(x) \le \lambda x</math>. Then for any such <math>x</math>,
 
<math>\vartheta(x) \le \lambda x</math>. Then for any such <math>x</math>,
−
<cmath>\begin{align*}
+
<cmath>
 +
\begin{align*}
 
\int\limits_{\lambda x}^x \frac{\vartheta{x} - t}{t^2}dt
 
\int\limits_{\lambda x}^x \frac{\vartheta{x} - t}{t^2}dt
 
\le \int\limits_{\lambda x}^x \frac{\lambda x -t}{t^2}dt
 
\le \int\limits_{\lambda x}^x \frac{\lambda x -t}{t^2}dt
Line 177: Line 178:
 
- ( \log x - \log(\lambda x) ) \\
 
- ( \log x - \log(\lambda x) ) \\
 
&= 1 - \lambda + \log \lambda \le 0.
 
&= 1 - \lambda + \log \lambda \le 0.
−
\end{align*} </cmath>
+
\end{align*}
 +
</cmath>
 
Again, by theorem 1, this quantity must equal zero in absolute
 
Again, by theorem 1, this quantity must equal zero in absolute
 
value; it follows that <math>\lambda = 1</math>.
 
value; it follows that <math>\lambda = 1</math>.
Line 198: Line 200:
 
\sum_{x^{1-\epsilon} \le p \le x} \log p
 
\sum_{x^{1-\epsilon} \le p \le x} \log p
 
&\ge \sum_{x^{1-\epsilon}\le p \le x} (1-\epsilon) \log x \\
 
&\ge \sum_{x^{1-\epsilon}\le p \le x} (1-\epsilon) \log x \\
−
&\ge (1-\epsilon) \log x ( \pi(x) - x^{1-\epsilon} ) , </cmath>
+
&\ge (1-\epsilon) \log x ( \pi(x) - x^{1-\epsilon} ) ,
 +
\end{align*}
 +
</cmath>
 
so
 
so
 
<cmath> \pi(x) \log x \le \frac{\vartheta(x)}{1-\epsilon} + x^{1-\epsilon}\log x
 
<cmath> \pi(x) \log x \le \frac{\vartheta(x)}{1-\epsilon} + x^{1-\epsilon}\log x
 
= x \left( \frac{\vartheta(x)}{(1-\epsilon)x} +
 
= x \left( \frac{\vartheta(x)}{(1-\epsilon)x} +
−
\frac{\log x}{\epsilon} \right) . </cmath>
+
\frac{\log x}{x^\epsilon} \right) . </cmath>
 
Again, since <math>\vartheta(x) \sim x</math>, it follows that for
 
Again, since <math>\vartheta(x) \sim x</math>, it follows that for
 
any <math>\epsilon > 0</math>,
 
any <math>\epsilon > 0</math>,

Latest revision as of 18:52, 21 October 2023

The Prime Number Theorem (PNT) is one of the most celebrated results in analytic number theory. Indeed, it is possibly the most famous major result in all of number theory, with the exception of Fermat's Last Theorem. (Fortunately, the proof is easier, though still non-trivial!) It gives an asymptotic formula for the distribution of the prime numbers; specifically, it states that the functions $\pi(x)$ and $x/\log x$ are asymptotically equivalent, where $\pi(x)$ is the number of primes less than or equal to $x$. In other words, it states that \[\lim_{x\to \infty} \frac{\pi(x) \log x}{x} = 1 .\]

History

First Conjectures

Gauss conjectured the theorem as early as 1793, in terms of the logarithmic integral, which is asymptotically equivalent to $x / \log x$. Legendre conjectured in 1798 that for some constants $A$ and $B$, \[\pi(x) \sim \frac{x}{A \log x - B} .\] In 1808 he refined his conjecture to \[\pi(x) = \frac{x}{\log x - A(x)} ,\] with $A(x)$ tending to some constant number around 1.08366. (In fact, $A(x)$ does not seem to tend to this value, but its actual asymptotic behavior is apparently unknown.)

Early Results

In 1850, Chebyshev proved that for sufficiently large $x$, there existed reals $A, B$ such that \[A < \frac{\pi(x) \log x}{x} < B ,\] and he was able to give \[A = \frac{\log 2}{2} + \frac{\log 3}{3} + \frac{\log 5}{5} - \frac{\log 30}{30} \approx 0.921292 ,\] and $B = \frac{6A}{5} \approx 1.10555$.

In 1859, Riemann established the relation between the distribution of the zeros of the Riemann zeta function and the distribution of the prime numbers; in this same paper, he posed the Riemann Hypothesis, namely that the zeta function's nontrivial zeros all lie on the line $\Re z = 1/2$. To this day, it remains unsolved.

In 1892, Sylvester was able to improve Chebyshev's bounds with $A = .956$, $B=1.045$. However, his methods did not seem likely to yield better bounds.

Proof and Refinement

Finally, in 1896, Jacques Hadamard and Charles-Jean de la ValÎée Poussin independently proved that the zeta function has no zeros on the line $\Re s = 1$, and from this deduced the prime number theorem. Their proofs were somewhat long; Hadamard's paper was some 20 pages long. De la Vallée Poussin's proof that $\zeta(1+ri)$ has no zeros was about 25 pages long; Hadamard's proof was essentially the modern version, though de la Vallée Poussin and Mertens later simplified it. The proof that this statement implied the prime number theorem remained long for some time.

In 1948, Alte Selberg and Paul Erdős simultaneously found "elementary" proofs of the prime number theorem. Unfortunately, these proofs are still much longer than the shortest proofs of today that use complex analysis.

Finally, in 1980, D.J. Newman found a theorem with a short proof that provided a much simpler link between the zeta function and the prime number theorem. This is essentially the proof given here.

Outline

The major results are the fact that the Riemann zeta function has no zeros on the line $\Re s = 1$, and the Tauberian theorem due to Newman. The rest of the theorem's proof is comparatively straightforward, though still non-trivial. We do not prove those results in this article, but instead refer to their proofs here and Newman's Tauberian Theorem.

Proof

We use the Riemann zeta function, which is defined as \[\zeta(s) = \sum_{n\ge 1} \frac{1}{n^s} = \prod_{p \text{ prime}} ( 1 - p^{-s} )^{-1} .\] This function has an analytic continuation to the entire complex plane except $s= 1$, where it has a simple pole of residue 1.

We define \[\phi(s) = \sum_{p} \frac{\log p}{p^s} .\] As discussed here, the function $\phi(s)$ extends to the set $\Re s > 1/2$ by the relation \[\phi(s) = - \frac{\zeta'(s)}{\zeta(s)} - \sum_p \frac{\log p}{ p^s (p^s- 1)} .\]

Thus we may define the function \[g(z) = \frac{\phi(z+1)}{z+1} - \frac{1}{z} .\] Since $\zeta(s)$ has no zeros on the line $\Re s =1$, the function $g(z)$ is holomorphic on the set $\Re z \ge 0$.

The Bounded Integral

Theorem 1. The integral \[\int\limits_1^{\infty} \frac{\vartheta(x) - x}{x^2} dx\] converges to $g(0)$.

Proof. We rely on a tauberian theorem due to Newman.

Let $x = e^t$. We note that \[\int\limits_1^{e^T} \frac{\vartheta(x) -x}{x^2} dx = \int\limits_0^T [\vartheta(e^t)e^{-t} - 1 ]dt.\]

Now, for $\Re s > 1$, \begin{align*} \int\limits_0^\infty [ \vartheta(e^t)e^{-t(s+1)}- e^{-st} ] dt &= \int\limits_1^\infty \left[ \frac{\vartheta(x)}{x^{s+2}} - \frac{1}{x^{s+1}} \right] dx \\ &= \sum_{k=0}^{\infty} \int\limits_{p_k}^{p_{k+1}} \frac{\vartheta(x)} {x^{s+2}} - \frac{1}{s} \\ &= \sum_{k=0}^{\infty} \vartheta(p_k) \int\limits_{p_k}^{p_{k+1}} \frac{dx}{x^{s+2}} - \frac{1}{s}\\ &= \frac{1}{s+1}\sum_{k=0}^\infty \vartheta(p_k) (1/p_k^{s+1} - 1/p_{k+1}^{s+1}) - \frac{1}{s} . \end{align*} Now, by the Abel Summation Technique, we have \begin{align*} \sum_{k=0}^\infty \vartheta(p_k)(1/p_k^{s+1} - 1/p_{k+1}^{s+1}) &= \sum_{k=0}^\infty \sum_{i=0}^k \log p_i (1/p_k^{s+1} - 1/p_{k+1}^{s+1}) \\ &= \sum_{k=0}^{\infty} \frac{\log p_k}{p_k^{s+1}} \\ &= \phi(s+1). \end{align*} Thus for $\Re s >1$ (for the integral), \[\int_0^\infty [\vartheta(e^t)e^{-t} -1 ]e^{-st} dt = \frac{\phi(s+1)}{s+1} - \frac{1}{s} = g(s).\] Now, by a theorem due to Chebyshev, the function $\vartheta(x)/x - 1$ is bounded above (by 1). The function $f(t) = \vartheta(e^t)e^{-t} -1$ thus satisfies the conditions of Newman's Tauberian Theorem, and the result follows. $\blacksquare$

End of Proof

The rest of the theorem is more simple.

Theorem 2. The functions $\vartheta(x)$ and $x$ are asymptotically equivalent.

Proof. Suppose that $\lambda \ge 1$ is a number such that there are infinitely many $x$ for which $\vartheta(x) \ge \lambda x$. Then for all such $x$, \begin{align*} \int\limits_x^{\lambda{x}} \frac{\vartheta(t) -t}{t^2} dt&\ge \int\limits_x^{\lambda{x}} \frac{\lambda x - t}{t^2}dt \\ &= \lambda x \left( \frac{1}{x} - \frac{1}{\lambda x} \right)- \left(\log (\lambda x) - \log x\right) \\ &= \lambda -1 - \log \lambda . \end{align*} Now, $d(x-1 - \log x)/dx = 1 - 1/x$; it follows that \[\lambda - 1 - \log \lambda \ge 0 ,\] with equality exactly when $\lambda =1$. But by theorem 1, this quantity must equal 0 in absolute value, so $\lambda = 1$.

Analogously, suppose that $\lambda \le 1$ is a number such that there are infinitely many $x$ for which $\vartheta(x) \le \lambda x$. Then for any such $x$, \begin{align*} \int\limits_{\lambda x}^x \frac{\vartheta{x} - t}{t^2}dt \le \int\limits_{\lambda x}^x \frac{\lambda x -t}{t^2}dt &= \lambda x \left( \frac{1}{x} - \frac{1}{\lambda x} \right) - ( \log x - \log(\lambda x) ) \\ &= 1 - \lambda + \log \lambda \le 0. \end{align*} Again, by theorem 1, this quantity must equal zero in absolute value; it follows that $\lambda = 1$.

If follows that $\limsup \vartheta(x)/x = \liminf \vartheta(x)/x =1$. $\blacksquare$

Theorem 3 (Prime Number Theorem). The functions $\pi(x)\log x$ and $x$ are asymptotically equivalent.

Proof. We note that \[\vartheta(x) = \sum_{p \le x} \log p \le \sum_{p\le x} \log x = \pi(x) \log x .\] Since $\vartheta(x) \sim x$, it follows that \[\liminf \frac{\pi(x) \log x}{x} \ge 1.\]

On the other hand, for any $\epsilon > 0$, \begin{align*} \vartheta(x) = \sum_{p\le x} \log p \ge \sum_{x^{1-\epsilon} \le p \le x} \log p &\ge \sum_{x^{1-\epsilon}\le p \le x} (1-\epsilon) \log x \\ &\ge (1-\epsilon) \log x ( \pi(x) - x^{1-\epsilon} ) , \end{align*} so \[\pi(x) \log x \le \frac{\vartheta(x)}{1-\epsilon} + x^{1-\epsilon}\log x = x \left( \frac{\vartheta(x)}{(1-\epsilon)x} + \frac{\log x}{x^\epsilon} \right) .\] Again, since $\vartheta(x) \sim x$, it follows that for any $\epsilon > 0$, \[\limsup \frac{\pi(x) \log x}{x} \le \limsup \left(\frac{1}{1-\epsilon} + \frac{\log x}{x^\epsilon} \right) = \frac{1}{1 -\epsilon} .\] Thus \[\limsup \frac{\pi(x)\log x}{x} \le 1 .\] Therefore \[\lim_{x\to \infty} \frac{\pi(x)\log x}{x} = 1. \qquad \blacksquare\]

Bibliography

See also

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