Apologies for the lack of posts lately; it’s been a busy semester. This post is essentially my notes for a talk I gave in my analytic number theory class.

Our goal is to obtain bounds on the distribution of prime numbers, that is, on functions of the form {\pi(x)}. The closely related function

\displaystyle \psi(x) = \sum_{n \leq x} \Lambda(n)

turns out to be amenable to study by analytic means; here {\Lambda(n)} is the von Mangolt function,

\displaystyle \Lambda(n) = \begin{cases} \log p & \text{if } n = p^m, p \ \text{prime} \\ 0 & \text{otherwise} \end{cases}.

Bounds on {\psi(x)} will imply corresponding bounds on {\pi(x)} by fairly straightforward arguments. For instance, the prime number theorem is equivalent to {\psi(x) = x + o(x)}.

The function {\psi(x)} is naturally connected to the {\zeta}-function in view of the formula

\displaystyle - \frac{\zeta'(s)}{\zeta(s)} = \sum_{n=1}^\infty \Lambda(n) n^{-s}.

In other words, {- \frac{\zeta'}{\zeta}} is the Dirichlet series associated to the function {\Lambda}. Using the theory of Mellin inversion, we can recover partial sums {\psi(x) = \sum_{n \leq x} \Lambda(x)} by integration of {-\frac{\zeta'}{\zeta}} along a vertical line. That is, we have

\displaystyle \psi(x) = \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} -\frac{\zeta'(s)}{\zeta(s)} \frac{x^s}{s} ds ,

at least for {\sigma > 1}, in which case the integral converges. Under hypotheses on the poles of {-\frac{\zeta'}{\zeta}} (equivalently, on the zeros of {\zeta}), we can shift the contour appropriately, and estimate the integral to derive the prime number theorem.

Example 1 The fact that {\zeta} is zero-free on the line {\Re s = 1} implies the prime number theorem. The classical zero-free region for {\zeta} states that {\zeta} is zero-free for complex numbers of the form {s=\sigma + it, \sigma > 1 - \eta \frac{1 }{\log(1 + |t|)}} for a sufficiently small constant {\eta}; it implies bounds on the estimates given by the prime number theorem.

Our goal is to generalize this to the problem of primes in arithmetic progressions. Fix {q > 1} and {a \in \mathbb{Z}}, relatively prime to {q}. Instead of the function {\pi(x)}, we can consider the functions

\displaystyle \pi(x; a, q) = |\left\{p \leq x, p \equiv a \mod q\right\}| .

These count primes at most {x} in the specified arithmetic progression. We have an analog of the prime number theorem for arithmetic progressions:

Theorem 1 {\pi(x; a, q) \sim\frac{1}{\phi(q)}\frac{x}{\log x}} as {x \rightarrow \infty}. Moreover, for some {\delta > 0}

\displaystyle \pi(x; a, q) = \mathrm{li}(x) + O(x \exp( - \delta \sqrt{\log x})).

Recall that {\mathrm{li}(x)} is the logarithmic integral {\mathrm{li}(x) = \int_2^x \frac{du}{\log u}}.

The theorem implies in particular that the prime numbers equidistributed in the arithmetic progressions {\mod q} (that is, those which are relatively prime to {q}). This is thus a strengthening of Dirichlet’s theorem on primes in arithmetic progressions.

A small caveat is in order. The constant {\delta} is not independent of {a, q}, because of the possibility of “exceptional” zeros of {L}-functions on the real axis.

2. L-functions

In order to prove Theorem 1, we will need an analogous analytic theory. Given {a,q} as above, we might try studying the function

\displaystyle \psi(x; q, a) = \sum_{n \leq x, n \equiv a \mod q} \Lambda(n).

Unfortunately, the arithmetic progressions over which the sum is taken are not very “multiplicative.” A way around this is provided by the theory of {L}-functions. To a Dirichlet character {\chi} mod {q} (which, as always, is defined to be zero on integers not prime to {q}), we might study the function

\displaystyle \psi(x; \chi) = \sum_{n \leq x} \chi(n) \Lambda(n).

Then, by Fourier inversion

\displaystyle \psi(x; q, a) = \frac{1}{\phi(q)}\sum \overline{\chi(a)} \psi(x; \chi),

so that estimates for {\psi(x; \chi)} will imply estimates for {\psi(x; q, a)}.

The function {\Lambda(n) \chi(n)} has a nice Dirichlet series: namely, if one writes

\displaystyle L(s, \chi) = \sum_{n=1}^\infty \frac{\chi(n)}{n^s} = \prod_{p} (1 - \chi(p) p^{-s})^{-1}

for {\Re(s) > 1}, then

\displaystyle - \frac{L'(s, \chi)}{L(s, \chi)} = \sum_{n=1}^\infty \Lambda(n) \chi(n) n^{-s}.

By Mellin inversion again, we can expect analytic properties of the logarithmic derivative {-\frac{L'(s, \chi)}{L(s, \chi)}} to yield information about the partial sums {\psi(x; \chi)}.

3. Bounds on the L-function

The {L}-function {L(s, \chi)}, initially defined for {\Re(s)>1}, has an meromorphic continuation to complex plane. If {\chi } is not the trivial character {\chi_0}, then {L(s, \chi)} is entire; if {\chi} is the trivial character, then {L(s, \chi)} has a simple pole at {s = 1}. We will need this in the following, as the strategy behind the proof of the prime number theorem is to shift the contour of integration in Mellin inversion as far inside the strip {0 < \Re s < 1} as possible.

There are a number of basic estimates for {L(s, \chi)} analogous to those for {\zeta}. The notation {s = \sigma + it}, {\tau = |t| + 4}, will be in use below.

Proposition 2 Let {\chi } be nontrivial. In any region {\delta \leq \sigma \leq 2} ({\delta > 0}), we have

\displaystyle L(s, \chi) \ll (1 + ( q \tau)^{1 - \sigma}) \min( (\sigma - 1)^{-1}, \log q \tau).

The bounds are uniformly in {\chi}, though later bonds probably will not be. Proof: We can use partial summation (or integration). Let {S(t, \chi) = \sum_{n \leq t} \chi(n)}. Then clearly, for any {X},

\displaystyle L(s, \chi) = \sum_{1}^X \frac{\chi(n)}{n^s} + \int_{X^+}^\infty t^{-s} d S(t, \chi) = \sum_{1}^X \frac{\chi(n)}{n^s} + \int_{X^+}^\infty t^{-s} d S(t, \chi) - X^{-s}S(X, \chi) + s \int_{X^+}^\infty t^{-s-1} S(t, \chi) dt.

This follows by integration by parts. The function {S(\cdot, \chi)} is uniformly bounded by {q}. Take {X = q \tau}; a similar method as in the bound for {\zeta} can now be applied. \Box

I’ve omitted some details; they are explained (for the ordinary zeta function) in Montgomery and Vaughn’s book, or in Ingham’s older book.

This bound implies, in particular, that {L(s, \chi)} grows sublinearly in the imaginary part in the strip {\delta \leq \sigma \leq 2}.

Using this, one can prove a similar factorization as for {\xi}. Let {\kappa = \kappa(\chi)} be {0} if {\chi} is even, and {1} if {\chi} is odd. Define the function

\displaystyle \xi(s, \chi) = L(s, \chi) \Gamma((s + \kappa)/2) \left( \frac{q}{\pi}\right)^{(s + \kappa)/2} .

Then this function can be shown to be entire. One has the functional equation

\displaystyle \xi(s, \chi) = \epsilon(\chi) \xi(1 - s, \overline{\chi}),

where {\epsilon(\chi)} is a complex number of modulus one.

Using Stirling’s formula and these estimates, one finds that

\displaystyle \xi(s, \chi) \ll \exp( |s| \log q|s| )

when {|s| \geq 2, \Re(s) \geq \frac{1}{2}}. By the functional equation, this is true in general. Hadamard’s factorization theorem allows us to express

\displaystyle \xi(s, \chi) = e^{as + b} \prod_{\rho} \left( 1 - \frac{s}{\rho}\right)e^{s /\rho}.

In particular, we get a partial fraction decomposition for {\frac{\xi'(s, \chi)}{\xi(s, \chi)}}. As in the case for {\zeta}, one can obtain bounds for the number of zeros of {L(s, \chi)} in a rectangle (using Jensen’s inequality).

4. The approximate partial fraction decomposition

We will next need an “approximate” partial fraction decomposition for the logarithmic derivative {L(s, \chi)} in terms of its roots in the critical strip (always denoted {\rho}). {Henceforth}, we will always be working in the case where {|\Im(s)| \geq 4}. The analysis when one works with all {s} is more complicated and is explained in the textbook.

As always, we use the notation

\displaystyle s = \sigma + it.

The decomposition is given by:

Proposition 3 For {5/6 \leq \sigma \leq 2} (and {|t| \geq 4}), we have

\displaystyle \frac{L'(s, \chi)}{L(s, \chi)} = \sum_{\rho}\frac{1}{s - \rho} + O( \log q \tau). \ \ \ \ \ (1)

The sum is taken over roots {\rho = \beta + i \gamma} satisfying {|\rho - (3/2 + i \gamma)| \leq 5/6}.

In order to prove this, we will need a general fact from complex analysis on the “approximate” expansion of logarithmic derivatives in partial fractions. Given a holomorphic function {f} on a disk {D_R(z_0)} of radius {R}, with zeros {w_1, \dots, w_N} in the disk (counted with multiplicities), one might try to approximate {\frac{f'}{f}(z) = \sum \frac{1}{z-w_i}}.

The following criterion gives a criterion to the extent to which one can do this:

Proposition 4 Let {f} be a holomorphic function in a neighborhood of the disk {D(0, R)}. Suppose {|f(z)| \leq M}. Then for any {r < R} and {z \in D_r(0)}, we have

\displaystyle \frac{f'(z)}{f(z)} = \sum \frac{1}{z - \omega_i} + O( \log \frac{M}{|f(0)|}),

where the {\omega_i} are the zeros in {D_R(0)}.

The proof this result requires a little work; it is explained in Montgomery and Vaughn’s book, and if I get time I will try to blog about it separately.

Applying it, and using the fact that {L(s, \chi)} grows at most linearly in the strip {\left\{0 \leq \Re s \leq 1\right\}} (for {|\Im(s)| \geq 4} in the principal case), we can get the result.

As a result, we can bound below the real part of the logarithmic derivative.

Corollary 5 For {\sigma > 1}, we have

\displaystyle \Re \left( \frac{L'(s, \chi)}{L(s, \chi)}\right) \geq -C \log q \tau,

for a constant {C > 0}.

Proof: In fact, the terms occurring in the sum all have positive real part when {\sigma > 1}. \Box

5. The trigonometric inequality

Suppose that {L(s, \chi)} had a zero at {\rho = \beta + i \gamma } where {\beta} was very close to {1}, say {|\beta - 1| < \frac{ \epsilon}{1+\log \gamma}} for {\epsilon \ll 1}. Then (1) shows that at a value {s = 1 + i\gamma + \eta}, we have that

\displaystyle \Re \left( \frac{L'(1 + i\gamma + \eta, \chi)}{L(1 + i\gamma + \eta, \chi)} \right) \ \ \ \ \ (2)

is very large for {\eta } close to zero. This is not too surprising: if there were a zero at {1 + i\gamma} itself, then (2) would actually tend to {\infty} as {\epsilon' \rightarrow 0}. The partial fraction expansion will allow us to say something a little weaker when there are roots close to {\sigma = 1}.

Nonetheless, there will be a problem with the largeness of (2). This is shown by the following:

Proposition 6 Let {\sigma > 1}. Then for any {t},

\displaystyle \Re \left( 3\frac{L'(\sigma, \chi_0)}{L(\sigma, \chi_0)} + 4 \frac{L'(\sigma + it, \chi)}{L(\sigma + it, \chi)} + \frac{L'(\sigma + 2it, \chi^2)}{L(\sigma + 2it, \chi^2)} \right) \leq 0. \ \ \ \ \ (3)

Proof: This is established using a trigonometric argument. Namely,

Adding gives

This is {\leq 0} because, for any complex number {z} of modulus {1}, {3 + 4z + z^2} has nonnegative real part. This is precisely the trigonometric inequality {3 + 4 \cos \theta + \cos (2 \theta) \geq 0}. \Box

In view of this, the earlier inequalities, and the fact that {\frac{L'(\cdot, \chi_0)}{L(\cdot, \chi_0)}} has a simple pole with residue {1} at {1}, we can write

 In other words, we can get a bound on the real part

\displaystyle \Re \left( \frac{L'(\sigma + it, \chi)}{L(\sigma + it, \chi)} \right) \leq \frac{3}{4(\sigma - 1)} + C \log q\tau. \ \ \ \ \ (4)

This valid for {\sigma \in (1, 2]} and {|t| \geq 4}.

6. The zero-free region

We can now see directly that {L(s, \chi)} is zero-free on {\Re(s) = 1}. In fact, this follows from (3). Suppose there were a zero at {1 + it}, we could take {\sigma } close to {1}, and observe that the term {3 \frac{L'(\sigma, \chi_0)}{L(\sigma, \chi_0)}} is asymptotically {-\frac{1}{\sigma -1}} as {\sigma \rightarrow 1}. The term {4 \frac{L'(\sigma + it, \chi)}{L(\sigma + it, \chi)}} is asymptotically {\frac{4c}{\sigma - 1}} if {c} is the order of the zero at {1 + it} (so {c \geq 1} clearly). The last term is bounded below as {\sigma \rightarrow 1}, since {L} has at worst a zero at {1 + 2it} (not a pole). In total, we get

\displaystyle \Re \left( 3\frac{L'(\sigma, \chi_0)}{L(\sigma, \chi_0)} + 4 \frac{L'(\sigma + it, \chi)}{L(\sigma + it, \chi)} + \frac{L'(\sigma + 2it, \chi^2)}{L(\sigma + 2it, \chi^2)} \right) \geq -3\frac{1}{\sigma - 1} + \frac{4c}{\sigma - 1} + O(1)

as {\sigma \rightarrow 1}. This clearly contradicts (3).

Alternatively, we could use (4) (which we’ve established only for {|t| \geq 4}, though); if we had a zero at {1 + it}, then at {\sigma + it}, we have {\frac{L'(\sigma + it)}{L(\sigma + it)} \sim \frac{c}{\sigma - 1}} for {c} the order of the zero. This contradicts (4).

A key component in this argument was that {4 > 3}; this will be used below in the refinement.

Theorem 7 Fix {\chi}. Then there exists a {c} such that {L(s, \chi)} has no roots in {\sigma > 1 - \frac{c}{\log q \tau}}.

The constants here will depend on the choice of {\chi}, though not if we restrict to points with {|t| \geq 4}.

Proof: In fact, we will use a similar argument, except we will use (1) to bound below {\Re\left( \frac{L'(s, \chi)}{L(s, \chi)}\right)}. Fix {\epsilon > 0}. Suppose that there exists a root {\rho = \beta + i \gamma}. We assume {|\gamma| \geq 4}.

Let us recall (1):

\displaystyle \frac{L'(s, \chi)}{L(s, \chi)} = \sum_{\rho'} \frac{1}{s - \rho'} + O(\log q\tau)

where the sum is over all {\rho'} sufficiently close to {s}. If {\sigma = \Re(s) > 1}, then we observe that all the terms in the sum have positive real part. It follows that

\displaystyle \Re \left( \frac{L'(s, \chi)}{L(s, \chi)} \right) \geq \Re\frac{1}{s - \rho} - C \log (q \gamma),

for some constant {C }, if {s} is sufficiently close to {\rho} (so that {\rho} occurs in the sum). In other words, we have just taken one of the terms in the sum.

If we take {s = 1 + \eta + i \gamma} where {\eta} is sufficiently small, we get

\displaystyle \Re \left( \frac{L'(1 + \eta + i\gamma, \chi)}{L(1 + \eta + i \gamma, \chi)} \right) \geq \frac{1}{1 + \eta - \beta} - C \log (q \gamma) .

Let {\beta = 1-r} for {r> 0}; then we have

\displaystyle \Re \left( \frac{L'(1 + \eta + i\gamma, \chi)}{L(1 + \eta + i \gamma, \chi)} \right) \geq \frac{1}{ \eta + r} - C \log (q \gamma) .

We also know that, by (4),

\displaystyle \Re \left( \frac{L'(1 + \eta + i\gamma, \chi)}{L(1 + \eta + i \gamma, \chi)} \right) \leq \frac{3}{4 \eta} + C \log \gamma .

This means that the lower bound is at most the upper bound:

\displaystyle \frac{1}{ \eta + r} - C \log ( \gamma) \leq \frac{3}{4\eta} + C \log \gamma,

for any {\eta \in (0, 1]}. Take a large constant {D}, and set {\eta = Dr}. We get

\displaystyle \frac{1}{(D+1) r} \leq \frac{3}{4D r} + C \log \gamma,

or

\displaystyle r^{-1} \left( \frac{1}{D+1} - \frac{3}{4D}) \right) \leq C \log \gamma,

which clearly means that {r \geq \frac{c}{\log \gamma}} for some constant {c}.

In general, we have

\displaystyle \Re \left( \frac{L'(1 + \eta, \chi_0)}{L(1 + \eta , \chi_0)} \right) \geq -\frac{1}{\eta} - C

and

\displaystyle \Re \left( \frac{L'(1 + \eta + 2 i \gamma, \chi)}{L(1 + \eta + 2 i \gamma , \chi)} \right) \geq - C \log \gamma.

Combining these, we will get a contradiction if {\epsilon } is sufficiently small. In fact, (3) gives here

\displaystyle - 3\left( \frac{1}{\eta} - C\right) + 4 \left( \frac{1}{\eta + \epsilon/ \log \gamma} - C\log \gamma \right) - C \log \gamma \leq 0.

This inequality has to hold for all {\eta \leq 2}. If {\epsilon} is sufficiently small, this will be impossible. Pick a large constant {D}, and let {\eta = D ( \epsilon/\log \gamma)}. Then we have

\displaystyle -3 \left( \frac{\log \gamma}{D \epsilon} - C \right) + 4 \left( \frac{\log \gamma}{(D+1) \epsilon} - C\log \gamma \right) - C \log \gamma \leq 0.

If {D} is chosen such that {\frac{4}{D+1} > \frac{3}{D}}, then it is easy to see that this is impossible for {\epsilon} sufficiently small (independently of {\gamma}). \Box

7. Remarks

Something stronger is proved in the textbook “Multiplicative Number Theory” by Montgomery and Vaughn; there is an absolute constant such that most of the conclusion about the zero-free region for {L(s, \chi)} holds simultaneously for every {\chi}. The caveat is that there is allowed to be one real zero in the region when {\chi} is quadratic. (No example is known, though.) This requires more careful estimates.

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