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Math 259: Introduction to Analytic Number Theory The contour integral formula for ψ(x)
We now have several examples of Dirichlet series, that is, series of the form^1
F (s) =
∑^ ∞
n=
ann−s^ (1)
from which we want to extract information about the growth of
n<x an^ as x→∞. The key to this is a contour integral. We regard F (s) as a function of a complex variable s = σ + it. For real y > 0 we have seen already that |y−s| = y−σ^. Thus if the sum (1) converges absolutely^2 for some real σ 0 , then it converges uniformly and absolutely to an analytic function on the half-plane Re(s) ≥ σ 0 ; and if the sum converges absolutely for all real s > σ 0 , then it converges absolutely to an analytic function on the half-plane Re(s) > σ 0. Now for y > 0 and c > 0 we have
2 πi
∫ (^) c+i∞
c−i∞
ys^
ds s
1 , if y > 1; 1 2 ,^ if^ y^ = 1; 0 , if y < 1 ,
in the following sense: the contour of integration is the vertical line Re(s) = c, and since the integral is then not absolutely convergent it is regarded as a principal value: (^) ∫ c+i∞
c−i∞
f (s) ds := lim T →∞
∫ (^) c+iT
c−iT
f (s) ds.
Thus interpreted, (2) is an easy exercise in contour integration for y 6 = 1, and an elementary manipulation of log s for y = 1. So we expect that if (1) converges absolutely in Re(s) > σ 0 then
∑
n<x
an =
2 πi
∫ (^) c+i∞
c−i∞
xsF (s)
ds s
for any c > σ 0 , using the principal value of the integral and adding ax/2 to the sum if x happens to be an integer. But getting from (1) and (2) to (3) involves interchanging an infinite sum with a conditionally convergent integral, which is not in general legitimate. Thus we replace
∫ (^) c+i∞ c−i∞ by^
∫ (^) c+iT c−iT , which legitimizes the manipulation but introduces an error term into (2). We estimate this error term as follows:
Lemma. For y, c, T > 0 we have
1 2 πi
∫ (^) c+iT
c−iT
ys^
ds s
1 + O(yc^ min(1, (^) T | log^1 y| )), if y ≥ 1; O(yc^ min(1, (^) T | log^1 y| )), if y ≤ 1 ,
(^1) As noted by Serre, everything works just as well with “Dirichlet series” ∑∞ k=0 ak n−k s, where nk are positive reals such that nk →∞ as k→∞. In that more general setting we would seek to estimate ∑ 2 nk^ <x^ ak^ as^ x→∞. We shall see later that the same results hold if absolute convergence is replaced by con- ditional convergence throughout. For example, for every nonprincipal character χ the series for L(s, χ) converges uniformly in the half-plane Re(s) > σ 0 for each positive σ 0.
the implied O-constant being effective and uniform in y, c, T.
(In fact the error’s magnitude is less than both yc^ and yc/πT | log y|. Of course if y equals 1 then the error term is regarded as O(1) and is valid for both approximations 0, 1 to the integral.)
Proof : Complete the contour of integration to a rectangle extending to real part −M if y ≥ 1 or +M if y ≤ 1. The resulting contour integral is 1 or 0 respectively by the residue theorem. We may let M →∞ and bound the horizontal integrals by (πT )−^1
0 y
c±r (^) dr; this gives the estimate yc/πT | log y|. Using a circular
arc centered at the origin instead of a rectangle yields the same residue with a remainder of absolute value < yc.
This Lemma will let us approximate
n<x an^ by (2πi)
− 1 ∫^ c+iT c−iT x
sF (s) ds/s.
We shall eventually choose some T and exploit the analytic continuation of F to shift the contour of integration past the region of absolute convergence to obtain nontrivial estimates.
The next question is, which F should we choose? Consider for instance ζ(s). We have in effect seen already that if we take F (s) = log ζ(s) then the sum of the resulting an over n < x closely approximates π(x). Unfortunately, while ζ(s) continues meromorphically to σ ≤ 1, its logarithm does not: it has essential logarithmic singularities at the pole s = 1, and at zeros of ζ(s) to be described later. So we use the logarithmic derivative of ζ(s) instead, which at each pole or zero of ζ has a simple pole with a known residue and thus a predictable effect on our contour integral.
What are the coefficients an for this logarithmic derivative? It is convenient to use not ζ′/ζ but −ζ′/ζ, which has positive coefficients. Using the Euler product we find
ζ′(s) ζ(s)
p
d ds
log(1 − p−s) =
p
log p
p−s 1 − p−s^
p
log p
∑^ ∞
k=
p−ks.
That is,
−
ζ′ ζ
(s) =
∑^ ∞
n=
Λ(n)n−s.
So the coefficient of n−s^ is none other than the von Mangoldt function which arose in the factorization of x!. Hence our contour integral
1 2 πi
∫ (^) c+iT
c−iT
ζ′ ζ
(s) xs^ ds s
(c > 1)
approximates ψ(x). The error can be estimated by our Lemma (4): since |Λ(n)| ≤ log n, the error is of order at most
∑^ ∞
n=
(x/n)c^ log n · min(1,
T | log(x/n)|
which is O(T −^1 xc^ log^2 x) provided 1 < T < x. (See the Exercises below.)
- Use (4) to show that nevertheless ψ(x) is given by the principal value integral
ψ(x) = lim T →∞
2 πi
∫ (^) c+iT
c−iT
ζ′ ζ
(s) xs^
ds s
for all x, c > 1.
- Show that
n=1 μ(n)n
−s (^) = 1/ζ(s), with μ being the M¨obius function defined
in the previous set of exercises. Deduce an integral formula for
n<x μ(n) analogous to (6), and an approximate integral formula analogous to (5) but with error only O(T −^1 x log x) instead of O(T −^1 x log^2 x).
References
[Newman 1980] Newman, D.J.: Simple Analytic Proof of the Prime Number Theorem, Amer. Math. Monthly 87 (1980), 693–696.
[Zagier 1997] Zagier, D.: Newman’s Short Proof of the Prime Number Theorem, Amer. Math. Monthly 104 (1994), 705–708.
