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Notes on analytic number theory

Kiran S. Kedlaya

Chapter 9 More on the zeroes of zeta

In this chapter, we derive some results about the location of the zeroes of the Riemann zeta function, including a bound on the number of zeroes in a box and a (small) zero-free region inside the critical strip.

Section 9.1 Order of an entire function

Definition 9.1.

For \(\alpha > 0\text{,}\) an entire function \(f: \CC \to \CC\) is said to have order \(\leq \alpha\) if for all \(\beta > \alpha\text{,}\)
\begin{equation*} f(z) = O(\exp{|z|^{\beta}}) \qquad (|z| \to \infty). \end{equation*}
We say \(f\) has order \(\alpha\) if it has order \(\leq \alpha\) but not order \(\leq \beta\) for any \(\beta \lt \alpha\text{.}\)
By the functional equation \(\xi(s) = \xi(1-s)\text{,}\) it suffices to check (9.1.1) for \(|\Real(s)|\geq 1/2\text{.}\) By (6.1.4), we have the integral representation (valid for \(\Real(s) > 0\))
\begin{equation*} \xi(s) = \frac{1}{2} + \frac{1}{2} s(s-1) \int_1^\infty (x^{s/2-1} + x^{(1-s)/2-1}) \omega(x)\,dx. \end{equation*}
It will suffice to bound the integral by \(\exp(C|s|)\) (and then throw in \(\log |s|\) to cover the factor \(s(s-1)\)). For this, simply observe that \(|x^{s/2-1} + x^{(1-s)/2-1}| \leq \exp(C|s|)\) while \(|\omega(x)|\) is bounded by the constant \(\sum_{n=1}^\infty e^{-n^2 \pi/2}\text{.}\)

Section 9.2 Product representations of entire functions

The theory of functions of finite order was introduced in order to formalize the intuition that just as the fundamental theorem of algebra allows one to do for polynomials, one should be able to factor holomorphic functions into products over their zeroes. This idea dates back to Euler, who observed that this intuition predicts the identity
\begin{equation} \sin(\pi z) = \pi z \prod_{n=1}^\infty \left(1 - \frac{z^2}{n^2} \right),\tag{9.2.1} \end{equation}
which in turn can be used to evaluate \(\zeta(s)\) for any positive even integer \(s\text{.}\)
To upgrade this intuition to a rigorous statement about the function \(f(z)\text{,}\) one must first control the number of zeroes of \(f\) in a disc. There is no harm in assuming that \(f(0) \neq 0\text{,}\) as we can arrive at this situation by dividing by a suitable power of \(z\text{.}\) We may then apply the following fact from complex analysis.
Note that multiplying two functions \(f\) corresponds to addition on both sides of the desired inequality; consequently, if we exhibit a factorization \(f = f_1 \cdots f_n\text{,}\) we may check the claim for each \(f_i\) individually.
Since the disc \(|z| \leq R\) is compact, \(f\) has only finitely many zeroes in this disc counted with multiplicity. We may thus write \(f(z) = g(z)(z-\rho_1)\cdots(z-\rho_n)\text{,}\) where \(g\) is nonzero on the disc \(|z| \leq R\) check the equality for each factor individually.
For the factor \(g(z)\text{,}\) apply the Cauchy residue formula to the contour integral \(\int \log(g(z)) \frac{dz}{z}\) around the circle \(|z| = R\text{,}\) then take real parts.
For the factor \(z-\rho_i\text{,}\) this follows from a short computation which we leave as an exercise (Exercise 9.4.8).

Remark 9.5.

The right side of (9.2.2) can also be interpreted as
\begin{equation*} \log |f(0)| + \int_0^R \#\{\rho: |\rho| \lt r\} \frac{dr}{r}. \end{equation*}

Remark 9.6.

In Theorem 9.4, if \(\log |f(z)| \lt r(|z|)\) for some function \(r\text{,}\) then the left side of (9.2.2) is bounded by \(2r(R)\text{,}\) whereas the right side is at least
\begin{equation*} \log |f(0)| + \log(2) \#\{\rho: |\rho| \leq R/2\}. \end{equation*}
Consequently, if \(r(R) = O(R^\alpha)\text{,}\) then the number of roots of \(f\) in the disc \(|\rho| \leq R\) is also \(O(R^\alpha)\text{.}\)
By a similar consideration, the bound \(\log |\xi(s)| = O(|s| \log |s|)\) from Lemma 9.3 implies that the number of zeroes of \(\zeta\) with \(|\Imag(s)| \leq T\) is \(O(T \log T)\text{,}\) thus proving Theorem 8.6.
Define
\begin{equation} h(z) := \prod_{n=1}^\infty (1 - z/\rho_n) e^{z/\rho_n};\tag{9.2.3} \end{equation}
the product converges absolutely uniformly on any closed disc because the multiplicand is
\begin{equation*} 1 + \frac{1}{2} \left(\frac{z}{\rho_n} \right)^2 + O\left(\left(\frac{z}{\rho_n} \right)^3\right) \end{equation*}
and the fact that the number of roots of norm \(\leq R\) is \(O(R^{1+\epsilon})\) (by Remark 9.6) implies that \(\sum 1/|\rho_n|^2\) converges (by partial summation).
We now argue (following [3], section 11) that \(f/h\) is also of order \(\leq 1\text{.}\) It will suffice to establish an upper bound on \(|1/h(z)|\) of the form \(\exp(R^{1+\epsilon})\) when \(|z| = R\) for some infinite divergent sequence of values of \(R\text{,}\) as then this will imply a corresponding bound for the entire function \(f/h\) and we can use the maximum modulus principle to fill in other values of \(R\text{.}\) (Note that we cannot take this last step directly for \(1/h\) because it is not entire.) Put \(r_n := |\rho_n|\text{.}\) Since \(\sum 1/r_n^2\) converges, the total lengths of the intervals \([r_n - r_n^{-2}, r_n + r_n^{-2}]\) is finite, so we can choose an infinite divergent sequence of values of \(R\) for each of which we have
\begin{equation*} |R-r_n| > r_n^{-2} \qquad (n=1,2,\dots). \end{equation*}
To prove the claimed bound for a particular such \(R\text{,}\) we factor \(h = h_1 h_2 h_3\) by partitioning the product (9.2.3) according to which range \(r_n\) lies in:
\begin{equation*} h_1\colon [0, R/2), \qquad h_1\colon [R/2, 2R], \qquad h_3: (2R, \infty). \end{equation*}
It will now suffice to prove that
\begin{equation} |z| = R \Longrightarrow |h_i(z)| > \exp(-R^{1+\epsilon}) \qquad (i=1,2,3);\tag{9.2.4} \end{equation}
we leave this step as an exercise (Exercise 9.4.10).
At this point, we know that \(f/h\) is entire of order \(\leq 1\) with no zeroes at all. Consequently, \(g(z) = \log (f(z)/h(z))\) is entire and satisfies \(|g(z)| = O(|z|^{1+\epsilon})\text{.}\) This in turn means that
\begin{equation*} g_2(z) = \frac{g(z) - g(0) - g'(0)z}{z^2} \end{equation*}
is entire and bounded, hence constant by Liouville's theorem. This yields the desired result.
By combining Lemma 9.3 with Theorem 9.7, we obtain a product representation for \(\xi\text{.}\) Taking a logarithmic derivative yields the following (where \(\rho\) runs over zeroes of \(\xi\)):
\begin{equation} \frac{\xi'(s)}{\xi(s)} = B + \sum_\rho \left( \frac{1}{s - \rho} + \frac{1}{\rho} \right).\tag{9.2.5} \end{equation}

Section 9.3 A zero-free region for \(\zeta\)

We next use the product representation (9.2.5) for \(\xi\) to obtain a zero-free region for \(\zeta\) which falls far short of what the Riemann hypothesis predicts, but is enough to obtain some control on the error term in the prime number theorem. The idea is to squeeze a bit of extra information out of Exercise 2.6.3, which we used for nonvanishing on the line \(\Real(s) = 1\text{.}\)
In this proof, we adopt the convention that each appearance of the symbol \(*\) represents a different positive constant.
For the benefit of the reader that may not have worked out Exercise 2.6.3, we spell out the argument: by writing
\begin{equation*} \Real(\log(\zeta(s)) = \sum_p \sum_{n=1}^\infty \frac{1}{n} \cos(\Imag(s) \log p^n) p^{-n \Real(s)} \end{equation*}
and observing that
\begin{equation*} 3 + 4 \cos \theta + \cos 2 \theta \geq 0 \qquad (\theta \in \RR), \end{equation*}
we deduce that
\begin{equation*} 3 \Real (\log \zeta(\sigma)) + 4 \Real (\log \zeta(\sigma + it)) + \Real(\log \zeta(\sigma + 2it)) \geq 0 \qquad (\sigma > 1, t \in \RR). \end{equation*}
By contrast, if \(\zeta(1 + it)\) were zero for some \(t\text{,}\) then the sum would tend to \(-\infty\) as \(\sigma \to 1^+\) (because \(\zeta(\sigma + 2it)\) does not diverge and \(4 > 3\)).
We can apply the same argument with \(\log \zeta\) replaced by its negative derivative
\begin{equation*} - \Real \zeta'(s)/\zeta(s) = \sum_{n=1}^\infty \Lambda(n) n^{-\Real(s)} \cos(\Imag(s) \log n) \end{equation*}
to obtain an analogous inequality
\begin{equation} -3 \Real \frac{\zeta'(\sigma)}{\zeta(\sigma)} -4 \Real \frac{\zeta'(\sigma+it)}{\zeta(\sigma+it)} - \Real \frac{\zeta'(\sigma+2it)}{\zeta(\sigma+2it)} \geq 0 \qquad (\sigma > 1, t \in \RR).\tag{9.3.1} \end{equation}
Suppose now that \(t\) is the imaginary part of a zero \(\rho\) of \(\zeta\text{.}\) To use (9.3.1) to get information about \(t\text{,}\) we bound each term on the left-hand side for \(\sigma\) slightly bigger than 1. For starters, to control the terms of (9.3.1) involving \(\sigma\text{,}\) note that since \(\zeta\) has a simple pole at \(s=1\text{,}\)
\begin{equation} -\frac{\zeta'(\sigma)}{\zeta(\sigma)} \lt \frac{1}{\sigma - 1} + *.\tag{9.3.2} \end{equation}
To control the terms of (9.3.1) involving \(s = \sigma+2it\text{,}\) we adjust (9.2.5) to get rid of the Gamma factors:
\begin{equation} -\frac{\zeta'(s)}{\zeta(s)} = \frac{1}{s-1} - B - \frac{1}{2} \log \pi + \frac{1}{2} \frac{\Gamma'((s+1)/2)}{\Gamma((s+1)/2)} - \sum_\rho \left(\frac{1}{s - \rho} + \frac{1}{\rho} \right).\tag{9.3.3} \end{equation}
For \(1 \leq \Real(s) \leq 2\) and \(|\Imag(s)| \geq 1\text{,}\) everything on the right-hand side of (9.3.3) other than the sum over \(\rho\) is dominated by \(* \log |\Imag(s)|\text{.}\) Hence taking real parts, we obtain
\begin{equation} - \Real \frac{\zeta'(s)}{\zeta(s)} \lt * \log |\Imag(s)| - \sum_\rho \Real \left( \frac{1}{s - \rho} + \frac{1}{\rho} \right).\tag{9.3.4} \end{equation}
Since \(\Real(\rho) > 0\) and \(\Real(s-\rho) > 0\text{,}\) we also have \(\Real(1/\rho) > 0\) and \(\Real(1/(s-\rho)) > 0\text{,}\) so the sum over \(\rho\) is positive. Hence
\begin{equation} - \Real \frac{\zeta'(s)}{\zeta(s)} \lt * \log |\Imag(s)|.\tag{9.3.5} \end{equation}
To control the terms of (9.3.1) involving \(s = \sigma+it\text{,}\) we bound \(-\Real \frac{\zeta'(s)}{\zeta(s)}\) by keeping only the summand of (9.3.4) corresponding to \(\rho\) (which is valid by the previous argument). Writing \(\rho = \beta + it\text{,}\) we have
\begin{equation} - \Real \frac{\zeta'(s)}{\zeta(s)} \lt * \log |t| - \frac{1}{\sigma - \beta}.\tag{9.3.6} \end{equation}
Combining (9.3.1) with (9.3.2), (9.3.5) (taking \(s = \sigma+2it\)), and (9.3.6) (taking \(s = \sigma+it\)), we obtain
\begin{equation*} \frac{4}{\sigma - \beta} \lt \frac{3}{\sigma - 1} + * \log |t|. \end{equation*}
For \(\sigma = 1 + */(\log |t|)\text{,}\) this yields
\begin{equation*} \beta \lt 1 - \frac{*}{\log |t|} \end{equation*}
as desired.

Exercises 9.4 Exercises

1.

Derive the estimate
\begin{equation*} |\Gamma(s/2)| \lt \exp(C_2|s| \log |s|) \qquad (\Real(s) \geq 1/2) \end{equation*}
by first proving a suitably strong version of Stirling's formula, e.g.,
\begin{equation*} \log \Gamma(s) = \left( s - \frac{1}{2} \right) \log s - s + \frac{1}{2} \log 2\pi + O(|s|^{-1}) \qquad (|s| \to \infty, \Real(s) \geq 1/2). \end{equation*}

2.

Prove that \(1/\Gamma\) is entire of order \(\leq 1\text{.}\) Then prove that
\begin{equation*} \frac{1}{s \Gamma(s)} = e^{\gamma s} \prod_{n=1}^\infty (1+s/n) e^{-s/n} \qquad (s \neq 0, -1, -2, \dots), \end{equation*}
where \(\gamma\) is Euler's constant, by applying Hadamard's theorem.

3.

Prove that \(\Gamma\) satisfies the additional functional equation
\begin{equation} \Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin (\pi z)} \qquad (z \notin \ZZ).\tag{9.4.1} \end{equation}
Hint.
Show that the ratio of the two sides is entire of order \(\leq 1\) with no zeroes, and therefore constant. Alternatively, compare factorizations of both sides; for \(\sin (\pi z)\text{,}\) see (9.2.1).

4. (Legendre duplication formula).

Prove the identity
\begin{equation*} \Gamma(z) \Gamma(z+\frac{1}{2}) = \frac{\sqrt{\pi}}{2^{2z-1}} \Gamma(2z). \end{equation*}
(This is a special case of the Gauss multiplication formula.)
Hint.
Apply Exercise 9.4.2. Alternatively, the original proof uses the beta integral representation of \(\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\text{.}\)

5.

Prove that
\begin{equation*} \frac{\Gamma'(s)}{\Gamma(s)} = \log(s) + O(|s|^{-1}) \qquad (|s| \to \infty, \Real(s) \geq 1/2). \end{equation*}
Hint.
Differentiate the expression obtained in Exercise 9.4.2.

6.

Prove that a function of order \(\leq \alpha\) need not satisfy \(|f(z)| = O(\exp(|z|^\alpha))\text{.}\)
Hint.
Look at \(\zeta\) on the positive real axis.

7.

Adapt the proof of Lemma 9.3 to show that for any Dirichlet character \(\chi\text{,}\) the function \(L(s,\chi)\) is of order \(\leq 1\text{.}\)

8.

Complete the proof of Theorem 9.4 by computing both sides of the claimed inequality in the case \(f(z) = z-\rho\) for some \(\rho \in \CC\) with \(0 \lt |\rho| \lt R\text{.}\)
Hint.
If you prefer not to compute the left side directly, note that it is the real part of a holomorphic function away from \(\rho>\text{,}\) so we can change contours to a circle centered at \(\rho\text{.}\)

9.

Prove that for \(T>0\text{,}\)
\begin{equation*} \sum_{\rho} \frac{1}{1 + (T - \Imag(\rho))^2} = O(\log T), \end{equation*}
where \(\rho\) runs over nontrivial zeroes of \(\zeta\text{.}\)

10.

Prove (9.2.4).
Hint.
A crude bound will suffice for \(h_1\) and \(h_3\text{.}\) For \(h_2\text{,}\) use Remark 9.6 to control the number of factors in the product.

11.

Find the constants \(A\) and \(B\) in the product representation for \(\xi\) given by Hadamard's theorem. Then deduce as a corollary that \(\frac{\zeta'(0)}{\zeta(0)} = \log 2\pi\text{.}\)

12. (Hadamard).

By adapting the proof of Theorem 9.7, prove the following. Let \(m\) be a positive integer and define
\begin{equation*} Q_m(z) = z + \frac{z^2}{2} + \cdots + \frac{z^m}{m}. \end{equation*}
Let \(f(z)\) be an entire function of order \(\leq m\) with zeroes \(\rho_1,\rho_2,\dots\text{.}\) Then
\begin{equation*} f(z) = e^{P(z)} \prod_{n=1}^\infty (1-z/\rho_n) e^{Q_m(z/\rho_n)} \end{equation*}
for some polynomial \(P(z)\) of degree \(\leq m\text{.}\)
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