some results from analysis - ntnu...notes on the gamma function and the riemann zeta function 1....

NOTES ON THE GAMMA FUNCTION AND THE RIEMANNZETA FUNCTION

1. Some results from analysis

Lemma 1. Suppose (fn) is a sequence of functions analytic on an open subset D ofC. If (fn) converges uniformly on every compact (closed and bounded) subset of D tothe limit function f then f is analytic on D. Moreover, the sequence of derivatives(f ′n) converges uniformly on compact subsets of D to f ′.

Proof. Since fn is analytic on D we have by Cauchy’s integral formula

fn(w) =1

2πi

∫γ

fn(z)

z − wdz

where γ is any closed and positively oriented contour in D and w is any interior point.The region interior to and including γ is closed and bounded and hence compact.So (fn) converges uniformly on this region and hence we can pass to the limit underthe integral sign giving

f(w) =1

2πi

∫γ

f(z)

z − wdz.

This implies that f is analytic on the region defined by γ and hence on the whole ofD.

For the derivatives we have

f ′n(w) =1

2πi

∫γ

fn(z)

(z − w)2dz

and

f ′(w) =1

2πi

∫γ

f(z)

(z − w)2dz.

Hence

|f ′n(w)− f ′(w)| =1

2π

∣∣∣∣∫γ

fn(z)− f(z)

(z − w)2dz

∣∣∣∣≤ (length of γ)× sup

z∈γ

∣∣∣∣fn(z)− f(z)

(z − w)2

∣∣∣∣and this tends to 0 as n→∞ for any w on the interior of γ, and hence for any compactsubsets of D (just choose γ appropriately which is possible since D is open). �

1

NOTES ON THE GAMMA FUNCTION AND THE RIEMANN ZETA FUNCTION 2

Lemma 2 (Differentiating under the integral sign). Let D be an open set and let γbe a contour of finite length L(γ). Suppose ϕ : {γ}×D → C is a continuous functionand define g : D → C by

g(z) =

∫γ

ϕ(w, z)dw.

Then g is continuous. Also, if ∂ϕ∂z

exists and is continous on {γ} × D then g isanalytic with derivative

g′(z) =

∫γ

∂ϕ

∂z(w, z)dw.

Proof. Let z1, z2 ∈ D with z2 fixed. Since ϕ is continuous, given ε > 0 we can find aδ > 0 such that |z1 − z2| < δ ⇒ |ϕ(w, z1) − ϕ(w, z2)| < ε/L(γ). Hence, given ε > 0choose δ as above then by linearity of the integral and the estimation lemma

|g(z1)− g(z2)| =

∣∣∣∣∫γ

(ϕ(w, z1)− ϕ(w, z2))dw

∣∣∣∣≤ L(γ) max

w∈γ|ϕ(w, z1)− ϕ(w, z2)|

< ε.

Hence g is continuous. If ∂ϕ∂z

exists and is continuous then∣∣∣∣ϕ(w, z + h)− ϕ(w, z)

h− ∂ϕ

∂z(w, z)

∣∣∣∣→ 0

with h. Then again by linearity of the integral and the estimation lemma∣∣∣∣g(z + h)− g(z)

h−∫γ

∂ϕ

∂z(w, z)dw

∣∣∣∣=

∣∣∣∣∫γ

(ϕ(w, z + h)− ϕ(w, z)

h− ∂ϕ

∂z(w, z)

)dw

∣∣∣∣≤ L(γ) max

w∈γ

∣∣∣∣ϕ(w, z + h)− ϕ(w, z)

h− ∂ϕ

∂z(w, z)

∣∣∣∣and this tends to 0 with h. �

Corollary 3. Let D be an open set and ϕ : [a,∞] × D → C be continuous withcontinuous partial derivative ∂ϕ

∂z. If the integral

∫∞aϕ(t, z)dt converges uniformly on

compact subsets of D then it defines an analytic function there and has derivative∫∞a

∂ϕ∂z

(t, z)dt.

Proof. Let fn(z) =∫ naϕ(t, z)dt (so γ is the straight line joining a and n) . By the

above lemma each fn is analytic (with f ′n(z) =∫ na

∂ϕ∂z

(t, z)dt) and by hypothesis


fn → f =∫∞aϕ(t, z)dt uniformly on compact subsets of D. Applying Lemma 1 gives

us the result. �

2. The Gamma function

Let s = σ + it with σ, t ∈ R. We define the Γ-function for σ > 0 by

(1) Γ(s) =

∫ ∞0

e−tts−1dt.

Note that in the integral we have the two ‘bad’ points; 0 and ∞. Also, we cannotimmediately apply Corollary 3 since the integrand is not always continuous at 0. Itturns out none of this matters and we have the following.

Proposition 4. Γ(s) is analytic for σ > 0.

Proof. First, note for a > 0 the function defined by∫∞ae−tts−1dt is analytic. To see

this we only need show it is uniformly convergent (on compact blah) and then we canapply Corollary 3 since all other hypotheses are met. As expected, the exponentialdominates the tail of the integral giving∣∣∣∣∫ n

a

e−tts−1dt−∫ ∞a

e−tts−1dt

∣∣∣∣ =

∣∣∣∣∫ ∞n

e−tts−1dt

∣∣∣∣≤

∫ ∞n

e−ttσ−1dt

≤ C

∫ ∞n

e−12tdt

= 2Ce−12n

and this → 0 as n→∞ giving uniform convergence. Now for σ > 0 define,

fn(s) =

∫ ∞1n

e−tts−1dt.

By the above argument each fn is analytic. Suppose σ ≥ c > 0. For 0 < t ≤ 1 wehave e−t < 1 and tσ−1 ≤ tc−1. Hence, for n > m,∣∣∣∣∣

∫ 1m

1n

e−tts−1dt

∣∣∣∣∣ <∫ 1

m

1n

tc−1d =1

c(m−a − n−a).

Given ε > 0 we can choose 0 < δ < 1 such that 1c(m−a − n−a) < ε whenever

|m−1−n−1| < δ. Hence the fn satisfy the Cauchy condition for uniform convergencein compact subsets of the halfplane σ > 0. Applying Lemma 1 we see that thegamma function is analytic for σ > 0. �


We can show the Γ function is an extension of factorial function to complex argu-ments, via the following functional equation

Proposition 5. For σ > 0 we have

(2) Γ(s+ 1) = sΓ(s).

Proof. Integration by parts gives∫ ∞0

e−ttsdt = −e−tts∣∣∣∞0

+ s

∫ ∞0

e−tts−1dt = sΓ(s)

�

By direct computation we see Γ(1) = 1 and hence by induction Γ(n+1) = n! for allpositive integers n. The functional equation also gives us the analytic continuationof Γ.

Theorem 1. The Γ function can be extended over the whole complex plane to a mero-morphic function with simple poles at the negative integers and zero. The residuesof these poles are given by

(3) Ress=−n(Γ(s)

)=

(−1)n

n!

Proof. By (2) we have

(4) Γ(s) =Γ(s+n)

s(s+1)(s+2) · · · (s+n−1)

for any positive integer n. Now Γ(s+n) is analytic for σ > −n so the function onthe right is meromorphic for σ > −n with simple poles at 0,−1,−2, . . . ,−(n − 1).Since n is arbitrary we’re done. By construction this extension of Γ satisfies (2). Tocalculate the residues we rewrite (4) as

Γ(s) =Γ(s+n+ 1)

s(s+1)(s+2) · · · (s+n)

and proceed directly viz:

Res(Γ; −n) = lims→−n

(s+ n)Γ(s+n+ 1)

s(s+1)(s+2) · · · (s+n)

= lims→−n

Γ(s+n+ 1)

s(s+1)(s+2) · · · (s+n− 1)

=(−1)n

n!

where we have used Γ(1) = 1 in the numerator. �


From now on when we refer to the Γ-function we mean the meromorphic contin-uation. Heuristically we can think of this as the limit in n of the right hand side of(4), and this is in fact not too far from the truth.

2.1. Product Representations of Γ(s). Since ex = limn→∞(1 + x

n

)nit is not

unreasonable to expect

(5) Γ(s) = limn→∞

∫ n

0

(1− t

n

)nts−1dt.

We first prove this and then use it to give an alternative representation of Γ(s), whichcan be thought of as the limit in n of (4).

Lemma 6. Formula (5) holds for σ > 0.

Proof. Denote

fn(s) =

∫ n

0

(1− t

n

)nts−1dt.

Then

Γ(s)− fn(s) =

∫ n

0

(e−t −

(1− t

n

)n)ts−1dt+

∫ ∞n

e−tts−1dt.

The second integral is just the tail of the Γ-function which → 0 as n → ∞. Wewant to show the first integral also → 0. This seems feasible since the first factor ofthe integrand gets arbitrarily small for increasing n. Now, for 0 ≤ y ≤ 1 we have1 + y ≤ ey ≤ (1− y)−1. For n large set y = t/n then(

1− t

n

)n≤ e−t ≤

(1 +

t

n

)−n.

Hence

0 ≤ e−t −(

1− t

n

)n= e−t

(1− et

(1− t

n

)n)≤ e−t

(1−

(1 +

t

n

)n(1− t

n

)n)= e−t

(1−

(1− t2

n2

)n).

Now, if 0 ≤ a ≤ 1 then (1− a)n ≥ 1− na when na < 1. Letting a = t2/n2 then forlarge n

1−(

1− t2

n2

)n≤ t2

n.


Therefore

0 ≤ e−t −(

1− t

n

)n≤ n−1t2e−t.

This gives

|Γ(s)− fn(s)| ≤ 1

n

∫ n

0

e−ttσ+1dt <1

nΓ(σ + 2)→ 0

as n→∞ since Γ(σ + 2) is finite. �

We deduce the alternative representation of Γ(s) as follows. Substituting u = t/nwe have∫ n

0

(1− t

n

)nts−1dt = ns

∫ 1

0

(1− u)nus−1du

= ns(

1

sus(1− u)n

∣∣∣10

+n

s

∫ 1

0

(1− u)n−1usdu

)= ns

(n

s

∫ 1

0

(1− u)n−1usdu

)= · · · · · ·

= nsn(n− 1) . . . 1

s(s+ 1) . . . ((s− 1) + n)

∫ 1

0

u(s−1)+ndu

=n!

s(s+ 1) · · · (s+ n)ns.

Taking the limit as n→∞ gives

Proposition 7. For s 6= 0,−1, . . .

(6) Γ(s) = limn→∞

n!

s(s+ 1) · · · (s+ n)ns.

This converges for all other s so gives us another meromorphic continuation of Γ.

This formula is quite useful and has a few consequences. The first of which is

Corollary 8 (Weierstrass Product). For s 6= 0,−1, . . . we have

(7) Γ(s) =e−γs

s

∞∏k=1

(1 +

s

k

)−1ez/k

where γ is the Euler-Mascheroni constant.


Proof. For s 6= 0,−1, . . . we have

Γ(s) = limn→∞

n!

s(s+ 1) · · · (s+ n)ns

= limn→∞

1

s(1 + s)(1 + s/2) · · · (1 + s/n)es logn

= limn→∞

es(logn−1−12−···− 1

n)

s

es(1+12+···+ 1

n)

(1 + s)(1 + s/2) · · · (1 + s/n)

=e−γs

slimn→∞

n∏k=1

(1 +

s

k

)−1ez/k.

�

This formula clearly demonstrates the poles as well as giving us the fact that Γ(s)has no zeros. Also, taking the logarithm of the product, differentiating and thenevaluating at s = 1 gives Γ′(1) = −γ.

2.2. The Reflection and Duplication Formulae. We can use (6) to prove thefamous reflection and duplication formulae. For the first of these we need a lemma.

Lemma 9. We have the following expansions

(8) π cot(πs) =1

s+ 2s

∞∑n=1

1

s2 − n2

and

(9)sin(πs)

πs=∞∏n=1

(1− s2

n2

).

Proof. Let

F (s) =1

s+ 2s

∞∑n=1

1

s2 − n2.

with s ∈ C\Z. If s is near an integer we expect to see some fairly large terms inthe series but these will die out as n increases. This is enough to guarantee absoluteconvergence: for n > 1√

2|s| we have |s2 − n2| ≥ n2 − |s|2 > n2/2 and hence∑

n> 1√2|s|

1

|s2 − n2|<

∑n> 1√

2|s|

1

n2

which converges. Adding in the finite number of other terms gives that the seriesin (8) converges absolutely. Note this also implies the series converges uniformly


on compact subsets and hence defines an analytic function on C\Z by Lemma 1.Splitting the summand into partial fractions we see F(s) is periodic in σ with period1. The pole at 0 is simple and has residue 1. By periodicity every poles is simplewith residue 1. Therefore, the function defined by f(s) = π cot(πs)− F (s) is entireand periodic in σ with period 1. We show f(s) is bounded then apply Liouville’stheorem. By periodicity it suffices to show f is bounded when 0 ≤ σ < 1 and sincef is entire we need to show it’s bounded as t = =(s)→ ±∞. Now,

π cot(πs) = πieπis + e−πis

eπis − e−πis= πi+

2πi

e2πis − 1.

Since |e2πis| = e−2πt we have limt→±∞ π cot(πs) = ∓πi. For F (s) note that inthe region 0 ≤ σ < 1 we have |t| ≤ |s| < |t| + 1. We also have |s2 − n2| =|σ2− t2− n2 + 2iσt| ≥ |σ2− t2− n2| = |σ2− (t2 + n2)| ≥ |t2 + n2| − σ2 > t2 + n2− 1.Hence

|F (s)| ≤ 1

|t|+ 2(|t|+ 1)

∞∑n=1

1

n2 + t2 − 1

≤ 1

|t|+ 2(|t|+ 1)

∫ ∞0

dx

x2 + t2 − 1

=1

|t|+ 2(|t|+ 1)

tan−1(x/√t2 − 1

)√t2 − 1

∣∣∣∣∣∞

0

=1

|t|+ π

|t|+ 1√t2 − 1

.

So F (s) is bounded. Therefore f(s) is bounded and hence constant by Liouville. Ats = 1/2 we have π cot(π/2) = 0 and

F (1/2) = 2−∞∑n=1

(1

n− 1/2− 1

n+ 1/2

)= 0

hence f ≡ 0.To see (9) consider g(s) = sin(πs)/(πs

∏n≥1(1−s2/n2)). The product is absolutely

convergent so g exists for s ∈ C\Z. g(s) tends to 1 as s tends to 0 and g has period1 implying g(s) tends to 1 as s tends to any integer. The logarithmic derivative isgiven by

g′(s)

g(s)= π cot(πs)−

(1

s+ 2s

∞∑n=1

1

s2 − n2

)= 0

hence g is constant and since g(0) = 1 we have g(s) = 1 for all s. �


Proposition 10 (Reflection Formula).

(10) Γ(s)Γ(1− s) =π

sin πs.

Proof. By (6) and (9) we have

Γ(s)Γ(−s) = limn→∞

n!

s(s+ 1) · · · (s+ n)ns

n!

−s(−s+ 1) · · · (−s+ n)n−s

= limn→∞

1

−z2∏n

k=1

(1 + s

n

) (1− s

n

)= − π

s sinπs.

So by (2)

Γ(s)Γ(1− s) = Γ(s)(−s)Γ(−s) =π

sin πs.

�

Setting s = 1/2 in the reflection formula gives Γ(1/2) =√π.

Proposition 11 (Duplication Formula). We have the following formula

(11) Γ(s)Γ

(s+

1

2

)= 21−2sπ1/2Γ(2s)

Proof. The trick here is to use a convenient expression for Γ(2s). By (6) we have

Γ(s)Γ(s+ 1/2)

Γ(2s)= lim

n→∞

{n!ns

s(s+ 1) · · · (s+ n)

n!ns+1/2

(s+ 12)(s+ 3

2) · · · (s+ 1

2+ n)

×2s(2s+ 1) · · · (2s+ 2n)

(2n)!(2n)2s

}

=1

22slimn→∞

{(n!)2n1/222n+1

(2n)!(z + n+ 12)

}

=1

22slimn→∞

{(n!)222n+1

(2n)!n1/2(1 + z/n+ 12n

)

}

=1

22slimn→∞

{(n!)222n+1

(2n)!n1/2

}=

1

22sC

Setting s = 1/2 gives C = 2Γ(1/2) = 2√π and we’re done. �


We finish this section on the Γ function with a formula that is very useful forestimating Γ(s).

2.3. Stirling’s Formula. The following theorem characterises the Γ function uniquelyand will prove useful.

Theorem 2 (Uniqueness Theorem). Let F be analytic in the right half-plane A ={s ∈ C : σ > 0}. Suppose F (s + 1) = sF (s) and that F is bounded in the strip1 ≤ σ < 2. Then F (s) = aΓ in A with a = F (1).

Proof. Consider f(s) = F (s) − aΓ(s). The equation f(s + 1) = sf(s) holds in Aand so we can extend f meromorphically to the whole plane as we did for Γ. Ifany poles occur these must be at the negative integers. Since f(1) = 0 we havelims→0sf(s) = 0, hence f doesn’t have a pole, or anything worse, at 0 and we canthus continue f analytically to 0. This gives the analytic continuation of f to thenegative integers via f(s+ 1) = sf(s).

Now, |Γ(s)| ≤ Γ(σ) and this is bounded for 1 ≤ σ < 2. Since F is bounded hereby hypothesis, f is also. Now consider the region with 0 ≤ σ ≤ 1. If t = =(s) ≤ 1then f is bounded since it’s analytic here. If t > 1 then f is bounded here sincef(s) = f(s+ 1)/s and f is bounded for 1 ≤ σ < 2. Since f(s) and f(1− s) assumethe same values for 0 ≤ σ ≤ 1 we have that g(s) = f(s)f(1 − s) is bounded andanalytic. By Liouville g(s) ≡ g(1) = 0 and hence f ≡ 0. �

Our goal is to use the uniqueness theorem to prove

Γ(s) =√

2πss−12 e−seµ(s)

for s ∈ C− = C\R≤0 where

µ(s) := −∫ ∞0

P1(x)

s+ xdx

and P1(x) = x − bxc − 12. We need to show µ is analytic and that it posseses an

appropriate functional equation (so that the above representation of Γ satisfies thefunctional equation (2)). For this we need a lemma.

Lemma 12 (‘Twisted’ 4 -inequality). For s = reiθ and x ≥ 0 we have

(12) |s+ x| ≥ (|s|+ x) cos(θ/2).

This gives

(13) |s+ x| ≥ (|s|+ x) sin(δ/2)

when |θ| ≤ π − δ, 0 < δ ≤ π (since cos(θ/2) ≥ sin(δ/2)).


Proof. Using cos θ = 1− 2 sin2(θ/2) and (r + x)2 ≥ 4rx we have

|s+ x|2 = r2 + 2rx cos θ + x2 = (r + x)2 − 4rx sin2(θ/2) ≥ (r + x)2 cos2(θ/2)

�

Proposition 13. µ(s) is analytic in C−.

Proof. Define Q(x) = 12(x − bxc − (x − bxc)2). Then Q(t) is an antiderivative of

−P1(x) (so continuous) and 0 ≤ Q(x) ≤ 1/8. We have

(14) −∫ β

α

P1(x)

s+ xdx =

Q(x)

s+ x

∣∣∣∣βα

+

∫ β

α

Q(x)

(s+ x)2dx

for 0 < α < β < ∞. Now let 0 < δ ≤ π and ε > 0. Then for x ≥ 0, s = reiθ withr > ε and |θ| ≤ π − δ we have (by the above lemma)∣∣∣∣ Q(x)

(s+ x)2

∣∣∣∣ ≤ |Q(t)|sin2(δ/2)(ε+ x)2

≤ 1

8 sin2(δ/2)(ε+ x)2.

Hence the integral ∫ ∞0

Q(x)

(s+ x)2dx

is uniformly convergent in compact subsets of C− and therefore defines an analyticfunction by Corollary 3. But by (14) we have

(15) µ(s) =

∫ ∞0

Q(x)

(s+ x)2dx.

�

Note by (15) and (12), (13) we have

|µ(s)| ≤ 1

8 cos2(θ/2)|s|(16)

|µ(s)| ≤ 1

8 sin2(δ/2)|s|(17)

for s = reiθ, |θ| ≤ π − δ, 0 < δ ≤ π.

Proposition 14. For s ∈ C− we have

(18) µ(s)− µ(s+ 1) =

(s+

1

2

)log

(s+ 1

s

)− 1.


Proof. Since P1(x+ 1) = P1(x) we have

µ(s+ 1) = −∫ ∞0

P1(x)

(s+ 1) + xdx = −

∫ ∞0

P1(x+ 1)

s+ x+ 1dx

= −∫ ∞1

P1(x)

s+ xdx = µ(s)−

(−∫ 1

0

x− bxc − 1/2

s+ xdx

)= µ(s) +

∫ 1

0

x− 1/2

s+ xdx

= µ(s) +

∫ 1

0

(1− s+ 1/2

s+ x

)dx.

�

Theorem 3 (Complex Stirling’s Formula). For s ∈ C− we have

(19) Γ(s) =√

2πss−12 e−seµ(s).

Proof. By Proposition 13 the function

F (s) = ss−12 e−seµ(s)

is analytic in C− (we define ss−12 = e(s−

12) log s where log is the principal branch of

the logarithm). By Proposition 14 we have

F (s+ 1) = (s+ 1)s+1/2e−s−1eµ(s)−(s+ 12) log( s+1

s )+1 = ss+1/2e−seµ(s) = sF (s).

Also F (s) is bounded in the region A = {s ∈ C : σ > 0}: Clearly, eµ(s) is bounded by

(16). Writing s = σ + it = |s|eiθ ∈ C we have |ss− 12 e−s| = |s|σ−1/2e−θte−σ. Then for

s ∈ A with |t| ≥ 2 we have σ − 1/2 ≤ 2, |s| ≤ 2t and −tθ ≤ −π|t|/2. Hence, in this

region we have |ss− 12 e−s| ≤ 4t2e−π|t|/2e−1 → 0 as |t| → ∞. Hence F (s) is bounded in

A. By the Uniqueness Theorem we must have

Γ(s) = ass−12 e−seµ(s)

for some a. Substituting this into the duplication formula (11) gives

a2ss−1/2e−seµ(s)(s+ 1/2)se−s−1/2eµ(s+1/2) = a21−2s√π(2s)2s−1/2e−2seµ(2s)

= a√

2πs2s−1/2e−2seµ(2s).

Hencea(1 + 1/2s)seµ(s)+µ(s+1/2) =

√2πeeµ(2s).

Now let s be real and approach infinity. By (16) the exponentials tend to 1 and sincelimx→∞(1 + 1/2x)x =

√e we have a =

√2π. �

This result immediately gives the original form of stirling’s formula


Corollary 15. As x→∞ in R we have Γ(x) ∼√

2πxx−1/2e−x.

Another consequence is the following.

Corollary 16 (Rapid decay in vertical strips). Let σ ∈ R be fixed. Then as |t| → ∞we have

(20) |Γ(σ + it)| ∼√

2π|t|σ−1/2e−π|t|/2.

Proof. Assume t ≥ 0. By (19) and (16) we have

log |Γ(σ + it)| = <(log(Γ(σ + it)))

= <((σ + it− 1/2) log(σ + it)− σ − it) +1

2log 2π +O(1/t)

= <((σ + it− 1/2)(log(σ2 + t2)/2 + i arg(σ + it)))

−σ +1

2log 2π +O(1/t)

= (σ − 1/2)(

log(|t|√

1 + σ2/t2))− t arg(σ + it)

−σ +1

2log 2π +O(1/t)

= (σ − 1/2)(log |t|+ o(1))− t(π/2− tan−1(σ/t))

−σ +1

2log 2π + o(1)

= (σ − 1/2) log |t| − πt/2 +1

2log 2π + o(1).

Doing similar stuff for t ≤ 0 gives the result. �

3. The Riemann zeta function

3.1. Analytic continuation and functional equation. Let s = σ+ it with σ, t ∈R and let

(21) ζ(s) =∞∑n=1

1

ns=∏p

(1− 1

ps

)−1.

The series is absolutely and locally uniformly convergent for σ > 1 and hence ζ(s)is analytic in this region. We plan to extend the domain on which ζ can be defined(hopefully this will allow us to exploit its connection with the primes). More precisely,we wish to find a function which agrees with the zeta function on the domain σ > 1,


but that also makes sense outside of this region. This process is known as analyticcontinuation.

As an example of analytic continuation, consider the two functions

f(z) =∞∑n=0

zn,

and

g(z) =1

1− z.

Now, f is analytic in the domain D = {z ∈ C : |z| < 1} and g is analyic in the largerdomain z ∈ C\{1}. However, the restriction of g to D, g|D, is equal to f and so g isan analytic continuation of f . Also, suppose we have another analytic continuationof f i.e. an analytic function h : C\{1} → C for which h|D = f . Then on D wemust have g(z) − h(z) ≡ 0. Hence g(z) = h(z) for all z ∈ C\{1} since holomorphicfunctions are essentially determined by their local behaviour. Generalising theseideas we see that an analytic continuation of a given function is unique, and so wemay talk of THE analytic continuation.

Theorem 4 (Analytic continuation and functional equation). The Riemann zetafunction can be continued to a function analytic on all of C with the exception of asimple pole at s = 1 with residue 1. The continuation satisfies the functional equation

(22) π−s2 Γ

(s

2

)ζ(s) = π−

1−s2 Γ

(1− s

2

)ζ(1− s).

Remark. Here we have denoted the analytic continuation of the zeta function alsoby ζ(s), really, this is the zeta function proper. When referring to the zeta functionfrom now on we will mean its analytic continuation.

A key tool in our proof of Theorem 4 is the Poisson summation formula. In order toprove this, amongst other things, we need a suitable condition under which integralsand sums can be interchanged. This is given by a special case of Fubini’s Theorem.

Proposition 17. Suppose fn ∈ L1(R) for all n ∈ Z and that fn(t) ∈ `1 for all t ∈ R.If either ∫

R

∑n∈Z

|fn(t)|dt <∞ or∑n∈Z

∫R|fn(t)|dt <∞

then ∫R

∑n∈Z

|fn(t)|dt =∑n∈Z

∫R|fn(t)|dt.


Recall that the Schwartz space consists of smooth functions (infinitely differen-tiable) of rapid decay (derivatives of all order decay faster than any negative powerof x: for all k ≥ 0 and m ≥ 1, xmf (k)(x)→ 0 as |x| → ∞ ).

Theorem 5 (Poisson summation). Suppose f is in Schwartz space. Define thefourier transform of f by

f̃(t) =

∫Rf(x)e−2πitxdx.

Then ∑n∈Z

f(n) =∑n∈Z

f̃(n).

Proof. We first periodise f . Let

S(t) =∑n∈Z

f(n+ t).

Since f is in Schwarz space, S(t) is absolutely convergent. Also notice S(t) is periodicof period ≤ 1. Hence we can represent S as a fourier series

S(t) =∑k∈Z

cke2πikt.

The fourier coefficients ck are given by

ck =

∫ 1

0

S(x)e−2πikxdx

=

∫ 1

0

(∑n∈Z

f(n+ x)

)e−2πikxdx

=∑n∈Z

∫ 1

0

f(n+ x)e−2πikxdx

=∑n∈Z

∫ n+1

n

f(t)e−2πik(t−n)dt

=

∫ ∞−∞

f(t)e−2πiktdt

= f̃(k).

Note the interchange of integration and summation is valid since S(x) is absolutelyconvergent. Hence ∑

n∈Z

f(n+ t) =∑n∈Z

f̃(n)e2πint

and setting t = 0 gives the result. �


Proof of Theorem 4. Our starting point is an identity involving the gamma function.Recall that

Γ(s) =

∫ ∞0

e−tts−1dt.

Letting s 7→ s/2 and substituting t = πn2x gives∫ ∞0

xs2−1e−πn

2xdx = n−sπ−s/2Γ(s/2).

Hence for σ > 1,

π−s/2ζ(s)Γ(s/2) =∞∑n=1

∫ ∞0

xs2−1e−πn

2xdx =

∫ ∞0

xs2−1

∞∑n=1

e−πn2xdx

if the interchange of summation and integration can be justified. This is clear sincefor σ > 1

∞∑n=1

∫ ∞0

|xs2−1e−πn

2x|dx = π−σ/2ζ(σ)Γ(σ/2) <∞.

We define

ψ(x) =∞∑n=1

e−πn2x

so that

π−s/2ζ(s)Γ(s/2) =

∫ ∞0

xs2−1ψ(x)dx

=

∫ 1

0

xs2−1ψ(x)dx+

∫ ∞1

xs2−1ψ(x)dx.

Since ψ(x) = O(e−πx) we see by the usual argument that the second integral isabsolutely convergent for all values of s, and uniformly convergent on compact subsetsof C. Since the integrand is differentiable the integral defines an analytic functionby Corollary 3. We seek similar behaviour in the first integral so that we can givemeaning to ζ(s) for σ < 1. To this end we introduce

θ(x) =∑n∈Z

e−πn2x = 2ψ(x) + 1


Writing f(u) = e−πu2x we see

f̃(n) =

∫ ∞−∞

e−πu2xe−2πinudu

=

∫ ∞−∞

e−πx(u+in/x)2

e−πn2/xdu

= e−πn2/x

∫ ∞−∞

e−πxu2

du

=1√xe−πn

2/x.

Since f is a Schwartz function we can apply Poisson summation to give

θ(x) =1√xθ

(1

x

)and hence

2ψ(x) + 1 =1√x

(2ψ(1/x) + 1).

Therefore for σ > 1 we obtain

π−s/2ζ(s)Γ(s/2) =

∫ 1

0

xs2−1(

1√xψ(1/x) +

1

2√x− 1

2

)dx+

∫ ∞1

xs2−1ψ(x)dx

=1

s− 1− 1

s+

∫ 1

0

xs2− 3

2ψ(1/x)dx+

∫ ∞1

xs2−1ψ(x)dx

=1

s(s− 1)+

∫ ∞1

y1−s2−1ψ(y)dy +

∫ ∞1

xs2−1ψ(x)dx

=1

s(s− 1)+

∫ ∞1

(xs2−1 + x

1−s2−1)ψ(x)dx.

(23)

Once again we see the integral represents an entire function and so the right handside is analytic on C\{0, 1}. Since the factor πs/2Γ(s/2) never vanishes we may usethe above expression to define ζ(s) for C\{0, 1}. Since Γ(s/2)−1 has a simple zeroat s = 0 we conclude the resulting expression for ζ(s) is analytic at s = 0. Alsonote that the expression on the right is invariant under s 7→ 1 − s and hence thefunctional equation follows. Finally, to find the residue at s = 1 we use (23) tocalculate lims→1(s− 1)ζ(s) and note Γ(1/2) =

√π. �

The nature of the functional equation

π−s2 Γ

(s

2

)ζ(s) = π−

1−s2 Γ

(1− s

2

)ζ(1− s)


allows us to infer the behaviour of ζ(s) for σ < 0 from its behaviour when σ > 1. Inparticular, the left hand side is non-zero and holomorphic for σ > 1. Since Γ(s) hassimple poles at the negative integers, ζ(s) must have simple zeros at the negativeeven integers, and these are the only zeros for σ < 0. We refer to these as the ‘trivial’zeros of ζ(s). If ζ(s) were to have any other zeros then these must lie in the region0 ≤ σ ≤ 1, known as the critical strip.

3.2. The Hadamard Product. Similarly to the product expression for sin:

sinπz

πz=∞∏n=1

(1− z2

n2

),

we wish to express the zeta function as a product over its zeros. Thus, we must firstinvestigate infinite products in general.

3.2.1. Infinite products.

Definition 1. Let (zn) be a sequence of complex numbers and let

P (N) =N∏n=1

zn.

Then if the limit limN→∞ P (N) exists it is called the infinite product of (zn) and itis denoted

∏∞j=1 zn.

If no term in the sequence (zn) is zero then P (N) is non-zero for all N . IflimP (N) 6= 0 then

lim zN = limP (N)

P (N − 1)= 1.

Therefore, a necessary condition for the convergence of a non-zero infinite product isthat the nth term goes to 1. Consequently, we may assume throughout the followingthat <(zn) > 0 for all n. This allows us to take logarithms of the zn (using theprincipal branch) and hence convert products into sums.

Proposition 18. Suppose <(zn) > 0 for all n. Then the infinite product∏∞

n=1 znconverges to a non-zero number if and only if

∑∞n=1 log zn converges.

Proof. If∑N

n=1 log zn converges to some limit, z say, then by continuity

∞∏n=1

zn = limN→∞

exp

( N∑n=1

log zn

)= exp

(limN→∞

N∑n=1

log zn

)= ez 6= 0.


Suppose∏∞

n=1 zn = z = reiθ with −π < θ ≤ π and r > 0. Let

P (N) =N∏n=1

zn, S(N) =N∑n=1

log zn

and let LogP (N) = log |P (N)| + iθN with −π + θ < θN ≤ π + θ. Since P (N) =exp(S(N)) we must have S(N) = LogP (N) + 2πikN for some integer kN . Now,since P (N)→ z, we have zN → 1 and hence S(N)− S(N − 1) = log zN → 0. Also,LogP (N)−LogP (N − 1)→ 0 which implies θN − θN−1 → 0 and so kN − kN−1 → 0.Therefore, since kN is an integer there exists some fixed integer k for which kn = kfor all n after some point n0. Consequently, S(N)→ log r + iθ + 2πik. �

Definition 2. If <(zn) > 0 for all n we say the product∏∞

n=1 zn converges absolutelyiff the sum

∑∞n=1 log zn converges absolutely. We may always refer to finite products

as absolutely convergent (no restriction on <(zn) needed).

The following inequality will prove useful. By the power series expansion for thelogarithm we have for |z| < 1∣∣∣∣1− log(1 + z)

z

∣∣∣∣ =|z/2− z2/3 + · · · |

≤1

2(|z|+ |z|2 + · · · )

=1

2

|z|1− |z|

.

If |z| < 1/2 then this last quantity is < 1/2. Hence, for |z| < 1/2 we have

(24) 12|z| ≤ | log(1 + z)| ≤ 3

2|z|.

This immediately gives the following.

Proposition 19. Suppose <zn > −1. Then the series∑∞

n=1 log(1 + zn) convergesabsolutely if and only if

∑∞n=1 zn converges absolutely.

Examples.(i) The Euler product ∏

p

(1− 1

ps

)−1converges absolutely for σ > 1: This follows from

∑p |p−s| =

∑p p−σ<∑

n n−σ <∞.


(ii) The product∏∞

n=1

(1− z2

n2

)converges absolutely for all z ∈ C: Fix z ∈

C and let |z| = R. Then the product∏2R

n=1

(1− z2

n2

)is finite and hence

absolutely convergent. In the remaining product∏∞

n=2R+1

(1− z2

n2

)we have

|z2/n2| < 1/2 and hence <(1 − z2/n2) > 0. The result now follows from∑n>2R z

2/n2 � R2.

(iii) The product∏∞

n=1

(1− z

n

)does not converge for any non-zero z ∈ C: This

follows from the divergence of the harmonic series.

(iv) The product∏∞

n=1

(1− z

n

)ez/n converges absolutely for all z ∈ C: Fix z and

choose n0 such that |z/n| < 1/2 and <((1 − z/n)ez/n) > 0 for all n ≥ n0.Then

∞∑n=n0

log((1− z/n)ez/n) =∞∑

n=n0

log(1− z/n) + z/n

=−∞∑

n=n0

z2

2n2+

z3

3n3+ · · ·

�|z|2∞∑

n=n0

1

n2

(1 +|z|n

+|z|2

n2+ · · ·

)=|z|2

∞∑n=n0

1

n2

1

1− |z|/n< 2ζ(2)|z|2.

Similarly, for any sequence (zn) which satisfies∑∞

n=1 |zn|−2 <∞ the product∏∞n=1(1− z/zn)ez/zn is absolutely convergent for all z ∈ C. This is important

because it allows us to create a function whose only zeros are at the pointszn. The main goal of this section is to show that this function is also analytic.

Lemma 20. Let K be a compact (closed and bounded) subset of C and let f, fn :K → C be a set of functions such that fn(z) → f(z) uniformly for z ∈ K. If thereis a constant A such that <f(z) ≤ A for all z ∈ K then exp(fn(z)) → exp(f(z))uniformly for z ∈ K.

Proof. Given ε > 0 choose δ > 0 such that |ez − 1| < εe−A for |z| < δ. Also, chooseN such that |fn(z)− f(z)| < δ for all z ∈ K whenever n > N . With these choices

εe−A > | exp(fn(z)− f(z))− 1| = | exp(fn(z))exp(f(z))

− 1|.


Therefore,

| exp(fn(z))− exp(f(z))| < εe−A| exp(f(z))| < ε.

�

Proposition 21. Let (zn) be a sequence of non-zero complex numbers and suppose∑∞n=1 |zn|−2 converges. Then the product

(25) P (z) =∞∏n=1

(1− z

zn

)ez/zn

is absolutely and uniformly convergent on compact subsets of C and is thereforeanalytic (by Lemma 1).

Proof. We have already shown the absolute convergence. It remains to show that

Sm(z) =m∑n=1

log((1− z/zn)ez/zn)

converges uniformly to

S(z) =∞∑n=1

log((1− z/zn)ez/zn)

on compact subsets K and that also in such regions there exists an A such that<S(z) ≤ A. So fix a compact K and let maxz∈K |z| = R. Since we’ve already shownabsolute convergence then it must follow that <S(z) is bounded for z ∈ K. Also,following a similar analysis to before

|S(z)− Sm(z)| =∣∣∣ ∞∑n=m+1

log((1− z/zn)ez/zn)∣∣∣� R2

∞∑n=m+1

1

|zn|2→ 0

as m→∞ and clearly this estimate holds uniformly for z ∈ K. �

So given a sequence (zn) satisfying∑|zn|−2 < ∞ we can construct an analytic

function whose only zeros are the points zn. Conversely, given an analytic func-tion f(z) with zeros (zn), we would like to know what conditions on f guarantee∑|zn|−2 <∞. Assuming that f satisfies these conditions we could then write

f(z) = F (z)P (z)

where P (z) is given by (25) and where F (z) is manifestly some analytic functionwith no zeros. It turns out that the behaviour of a functions zeros is closely relatedto its order of growth.


3.2.2. Entire functions of order 1.

Definition 3. A function is called entire if it is analytic on all of C. An entirefunction f(z) is said to be of finite order if there exists an α such that

f(z) = O(e|z|α

), |z| → ∞The order of f is defined as the infimum of all such α.

The behaviour of an analytic functions zeroes and its order of growth are relatedthrough the following result of Jensen, which we quote without proof.

Theorem 6 (Jensen’s formula). Suppose f is analytic on the disc |z| ≤ R andsuppose z1, . . . , zn are its zeros on the interior |z| < R. Also, suppose f does nothave any zeros on the boundary |z| = R. Then

(26)1

2π

∫ 2π

0

log |f(Reiθ)|dθ − log |f(0)| = logRn

|z1| · · · |zn|Corollary 22. Let f be an entire function of order α and let n(R) denote the numberof zeros of f in |z| < R. Then

(27) n(R)� Rβ

for any β > α.

Proof. Let z1, . . . , zn denote the number of zeros of f in |z| < r and let |zi| = ri.Then the right hand side of (26) can be written as∫ R

0

n(r)

rdr

since this is equal to

log r2/r1 + 2 log r3/r2 + · · ·+ n logR/rn = log(Rn/r1 · · · rn).

Suppose β > α. Thenlog |f(Reiθ| � Rβ

and hence ∫ R

0

n(r)

rdr � Rβ − log |f(0)| � Rβ.

On the other hand ∫ 2R

R

n(r)

rdr ≥ n(R)

∫ 2R

R

r−1dr = n(R) log 2.

�

Corollary 23. Suppose f is an entire function of order 1 with non-zero zeros (zn).Then the series

∑∞n=1 |zn|−1−ε converges for any ε > 0.


Proof. By partial summation we have∞∑n=1

|zn|−1−ε = (1 + ε)

∫ ∞0

r−2−εn(r)dr �∫ ∞0

r−1−ε/2dr <∞.

�

A consequence of this is that for an entire function f of order 1 with zeros (zn),the product

(28) P (z) =∞∏n=1

(1− z

zn

)ez/zn

is an entire function. Writing

f(z) = F (z)P (z)

we see that F (z) is an entire function without zeros.

Proposition 24. Let f be an entire function of order 1 with zeros (zn) and letf(z) = F (z)P (z) where P (z) is given by (28). Then F (z) is a non-zero entirefunction of order at most 1.

Proof. We need to show that F has order at most 1. We do this by lower boundingP (z) on a sequence of circles |z| = R. These must be kept away from the valuesrn = |zn|, since otherwise we may obtain trivial lower bounds e.g. from an occurrenceof 1 − |z|/|zn| = 0. However, since

∑∞n=1 r

−2n converges, these points cannot occur

too densely. Indeed, the combined length of the intervals (rn− r−2n , rn + r−2n ) is finiteand hence we may always choose an arbitrarily large R such that

(29) |R− rn| > r−2n

for all n.Write P (z) = P1(z)P2(z)P3(z) where the subproducts are extended over the fol-

lowing sets of n:

P1 : rn < R/2

P2 : R/2 ≤ rn ≤ 2R

P3 : rn > 2R.

(30)

Throughout the following, ε’s may vary from line to line. For P1, on |z| = R we have

|(1− z/zn)ez/zn| ≥ (|z/zn| − 1)e−|z|/|zn| > e−R/rn

and since ∑r≤R/2

r−1n =∑r≤R/2

rεnr−1−εn ≤ (R/2)ε

∞∑n=1

r−1−εn � Rε


it follows thatP1(z)� exp(−R1+ε).

For P2 we have

|(1− z/zn)ez/zn| ≥ |z/zn − 1|e−|z|/|zn| ≥ e−2|z − zn|/2R� R−3

by (29). By (27), the number of factors in P2 is � R1+ε and hence

|P2(z)| � (R−3)R1+ε � exp(−R1+2ε).

Finally, In P3 we have |z/zn| < 1/2 and hence

|(1− z/zn)ez/zn| = |1− z2

2z2n+O((z/zn)3)| > e−c(R/rn)

2

for some constant c > 0. Also,∑n>2R

r−2n =∑n>2R

r−1+εr−1−ε < (2R)−1+ε∞∑n=1

r−1−εn � R−1+ε.

Therefore,P3(z)� exp(−cR1+ε)� exp(−R1+2ε).

Consequently, for |z| = R we have

P (z)� exp(−R1+ε)

and henceF (z)� exp(R1+ε)

since f is of order 1. �

Proposition 25. Let F (z) be an entire function of order α with no zeros. ThenF (z) = eG(z) for some polynomial G. The order of F is the degree of G and hence isan integer.

Proof. Since F is entire and non-zero the function G(z) := logF (z) is itself entireand can be defined to be single valued. It satisfies

<(G(z)) = log |F (z)| � Rα

for |z| = R. Writing

G(z) =∞∑n=0

(an + ibn)zn

then

<(G(z)) =∞∑n=0

anRn cosnθ −

∞∑n=1

bnRn sinnθ

for z = Reiθ. Multiplying the above by 1 ± cosmθ and integrating over θ ∈ [0, 2π]gives π(2a0 ± amr

m) on the right side and O(Rα) on the left side. Consequently,


am = O(Rα−m) and hence it follows, by taking R arbitrarily large, that am = 0 ifm > α. A similar argument gives the same for bm and hence G(z) is a polynomialof degree at most α. Since F is of order α the degree is in fact equal to α, whichconsequently is an integer. �

Combining the above two Propositions with our previous results gives the follow-ing.

Theorem 7. Let f(z) be an entire function of order 1 with zeros (zn). Then

(31) f(z) = ea+bz∞∏n=1

(1− z

zn

)ez/zn

for some constants a, b. Moreover, the series∑∞

n=1 |zn|−1−ε converges for all ε > 0.

Note that if∑∞

n=1 |zn|−1 converges then f(z) ≤ ec|z| for some constant c. Thisfollows on applying the inequality

|(1− z)ez| = |1− z2

2− z3

3− z4

8− · · · | ≤ e2|z|

to the product representation (31). Consequently, if the inequality f(z) ≤ ec|z| doesnot hold, then the series

∑∞n=1 |zn|−1 diverges and so f must have infinitely many

zeros.

3.3. The Hadamard product for the Riemann zeta function. Let

(32) ξ(s) = 12s(s− 1)π−s/2Γ(s/2)ζ(s).

Then this is an entire function whose only zeros are the non-trivial zeros of the zetafunction. We will show that ξ(s) is of order 1. Since ξ(s) = ξ(1− s) we only need tolook at the case σ ≥ 1/2. Clearly,

12s(s− 1)π−s/2 � exp(c|s|)

and by Stirling’s formula we have

Γ(s/2)� exp(c|s| log |s|).On applying partial summation to the series representation of ζ(s) when σ > 1 wefind

ζ(s) =s

s− 1− s

∫ ∞1

(x− bxc)x−s−1dx.

Note that the integral converges for σ > 0 and hence this representation acts as ananalytic continuation of ζ(s) to σ > 0. In particular, the integral is bounded forσ ≥ 1/2 and therefore

ζ(s)� |s|


for |s| large. Putting these together gives

(33) ξ(s)� exp(c|s| log |s|)as |s| → ∞. Note that since log Γ(s) ∼ s log s and ζ(s) → 1 the above inequalityfor ξ(s) is essentially best possible. Therefore ξ(s) = O(e|s|

α) for all α > 1 and the

infimum of all such α is 1 i.e. ξ(s) is of order 1. In particular, ξ(s) is not � ec|z| forany c. On applying the results of the previous section we have the following.

Theorem 8. The zeta function has an infinity of non-trivial zeros, ρ1, ρ2, . . ., suchthat the series

∞∑n=1

|ρn|−1−ε

converges for any ε > 0 and∞∑n=1

|ρn|−1

diverges. It satisfies the product representation

(34) ζ(s) =eA+Bs

(s− 1)Γ( s2

+ 1)

∏ρ

(1− s

ρ

)es/ρ

for some constants A,B.

some results from analysis - ntnu...notes on the gamma function and the riemann zeta function 1....

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