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(Princeton lectures in analysis 2) Elias M Stein, Rami Shakarchi - Complex analysis Vol 2 -Princeton University Press (2003)

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c.
(b) Show as a consequence that log Γ(n) ∼ n log n as n→∞. In fact, prove
that log Γ(n) ∼ n log n+ O(n) as n→∞. [Hint: Use the fact that Γ(x) is
monotonically increasing for all large x.]
15. Prove that for Re(s) > 1,
ζ(s) =
1
Γ(s)
∫ ∞
0
xs−1
ex − 1 dx.
[Hint: Write 1/(ex − 1) =∑∞n=1 e−nx.]
16. Use the previous exercise to give another proof that ζ(s) is continuable in the
complex plane with only singularity a simple pole at s = 1.
[Hint: Write
ζ(s) =
1
Γ(s)
∫ 1
0
xs−1
ex − 1 dx +
1
Γ(s)
∫ ∞
1
xs−1
ex − 1 dx.
4. Problems 179
The second integral defines an entire function, while∫ 1
0
xs−1
ex − 1 dx =
∞∑
m=0
Bm
m!(s + m− 1) ,
where Bm denotes the m
th Bernoulli number defined by
x
ex − 1 =
∞∑
m=0
Bm
m!
xm.
Then B0 = 1, and since z/(e
z − 1) is holomorphic for |z| < 2π, we must have
lim supm→∞ |Bm/m!|1/m = 1/2π.]
17. Let f be an indefinitely differentiable function on R that has compact support,
or more generally, let f belong to the Schwartz space.4 Consider
I(s) =
1
Γ(s)
∫ ∞
0
f(x)x−1+s dx.
(a) Observe that I(s) is holomorphic for Re(s) > 0. Prove that I has an analytic
continuation as an entire function in the complex plane.
(b) Prove that I(0) = 0, and more generally
I(−n) = (−1)nf (n+1)(0) for all n ≥ 0.
[Hint: To prove the analytic continuation, as well as the formulas in the second
part, integrate by parts to show that I(s) = (−1)
k
Γ(s+k)
∫∞
0
f (k)(x)xs+k−1 dx.]
4 Problems
1. This problem provides further estimates for ζ and ζ′ near Re(s) = 1.
(a) Use Proposition 2.5 and its corollary to prove
ζ(s) =
∑
1≤n<N
n−s − N
s−1
s− 1 +
∑
n≥N
δn(s)
for every integer N ≥ 2, whenever Re(s) > 0.
(b) Show that |ζ(1 + it)| = O(log |t|), as |t| → ∞ by using the previous result
with N =greatest integer in |t|.
4The Schwartz space on R is denoted by S and consists of all indefinitely differentiable
functions f , so that f and all its derivatives decay faster than any polynomials. In other
words, supx∈R |x|m|f(�)(x)| < ∞ for all integers m, � ≥ 0. This space appeared in the
study of the Fourier transform in Book I.
180 Chapter 6. THE GAMMA AND ZETA FUNCTIONS
(c) The second conclusion of Proposition 2.7 can be similarly refined.
(d) Show that if t 
= 0 and t is fixed, then the partial sums of the series∑∞
n=1 1/n
1+it are bounded, but the series does not converge.
2.∗ Prove that for Re(s) > 0
ζ(s) =
s
s− 1 − s
∫ ∞
1
{x}
xs+1
dx
where {x} is the fractional part of x.
3.∗ If Q(x) = {x} − 1/2, then we can write the expression in the previous problem
as
ζ(s) =
s
s− 1 −
1
2
− s
∫ ∞
1
Q(x)
xs+1
dx.
Let us construct Qk(x) recursively so that∫ 1
0
Qk(x) dx = 0,
dQk+1
dx
= Qk(x), Q0(x) = Q(x) and Qk(x + 1) = Qk(x).
Then we can write
ζ(s) =
s
s− 1 −
1
2
− s
∫ ∞
1
(
dk
dxk
Qk(x)
)
x−s−1 dx ,
and a k-fold integration by parts gives the analytic continuation for ζ(s) when
Re(s) > −k.
4.∗ The functions Qk in the previous problem are related to the Bernoulli polyno-
mials Bk(x) by the formula
Qk(x) =
Bk+1(x)
(k + 1)!
for 0 ≤ x ≤ 1.
Also, if k is a positive integer, then
2ζ(2k) = (−1)k+1 (2π)
2k
(2k)!
B2k,
where Bk = Bk(0) are the Bernoulli numbers. For the definition of Bk(x) and Bk
see Chapter 3 in Book I.
7 The Zeta Function and Prime
Number Theorem
Bernhard Riemann, whose extraordinary intuitive pow-
ers we have already mentioned, has especially reno-
vated our knowledge of the distribution of prime num-
bers, also one of the most mysterious questions in
mathematics. He has taught us to deduce results in
that line from considerations borrowed from the in-
tegral calculus: more precisely, from the study of a
certain quantity, a function of a variable s which may
assume not only real, but also imaginary values. He
proved some important properties of that function,
but enunciated two or three as important ones with-
out giving the proof. At the death of Riemann, a note
was found among his papers, saying “These properties
of ζ(s) (the function in question) are deduced from an
expression of it which, however, I did not succeed in
simplifying enough to publish it.”
We still have not the slightest idea of what the
expression could be. As to the properties he simply
enunciated, some thirty years elapsed before I was able
to prove all of them but one. The question concern-
ing that last one remains unsolved as yet, though, by
an immense labor pursued throughout this last half
century, some highly interesting discoveries in that di-
rection have been achieved. It seems more and more
probable, but still not at all certain, that the “Rie-
mann hypothesis” is true.
J. Hadamard, 1945
Euler found, through his product formula for the zeta function, a
deep connection between analytical methods and arithmetic properties
of numbers, in particular primes. An easy consequence of Euler’s for-
mula is that the sum of the reciprocals of all primes,
∑
p 1/p, diverges,
a result that quantifies the fact that there are infinitely many prime
numbers. The natural problem then becomes that of understanding
how these primes are distributed. With this in mind, we consider the
182 Chapter 7. THE ZETA FUNCTION AND PRIME NUMBER THEOREM
following function:
π(x) = number of primes less than or equal to x.
The erratic growth of the function π(x) gives little hope of finding a
simple formula for it. Instead, one is led to study the asymptotic behavior
of π(x) as x becomes large. About 60 years after Euler’s discovery,
Legendre and Gauss observed after numerous calculations that it was
likely that
(1) π(x) ∼ x
log x
as x→∞.
(The asymptotic relation f(x) ∼ g(x) as x→∞ means that
f(x)/g(x)→ 1 as x→∞.) Another 60 years later, shortly before Rie-
mann’s work, Tchebychev proved by elementary methods (and in partic-
ular, without the zeta function) the weaker result that
(2) π(x) ≈ x
log x
as x→∞.
Here, by definition, the symbol ≈ means that there are positive constants
A < B such that
A
x
log x
≤ π(x) ≤ B x
log x
for all sufficiently large x.
In 1896, about 40 years after Tchebychev’s result, Hadamard and de
la Valle´e Poussin gave a proof of the validity of the relation (1). Their
result is known as the prime number theorem. The original proofs of
this theorem, as well as the one we give below, use complex analysis.
We should remark that since then other proofs have been found, some
depending on complex analysis, and others more elementary in nature.
At the heart of the proof of the prime number theorem that we give
below lies the fact that ζ(s) does not vanish on the line Re(s) = 1. In
fact, it can be shown that these two propositions are equivalent.
1 Zeros of the zeta function
We have seen in Theorem 1.10, Chapter 8 in Book I, Euler’s identity,
which states that for Re(s) > 1 the zeta function can be expressed as an
infinite product
ζ(s) =
∏
p
1
1− p−s .
1. Zeros of the zeta function 183
For the sake of completeness we provide a proof of the above identity.
The key observation is that 1/(1− p−s) can be written as a convergent
(geometric) power series
1 +
1
ps
+
1
p2s
+ · · ·+ 1
pMs
+ · · · ,
and taking formally the product of these series over all primes p, yields
the desired result. A precise argument goes as follows.
Suppose M and N are positive integers with M > N . Observe now
that, by the fundamental theorem of arithmetic,1 any positive integer
n ≤ N can be written uniquely as a product of primes, and that each
prime that occurs in the product must be less than or equal to N and
repeated less than M times. Therefore
N∑
n=1
1
ns
≤
∏
p≤N
(
1 +
1
ps
+
1
p2s
+ · · ·+ 1
pMs
)
≤
∏
p≤N
(
1
1− p−s
)
≤
∏
p
(