Suggested solution of HW4

Chapter 6 Q1: By the product formula of Γ, we have for all z,

∞
1 sin πz Y z
= Γ(1 − z) = eγz z (1 + )e−z/n
Γ(z) π n=1
n

where γ is the Euler’s constant.

N
Y z −z/n (z + 1)(z + 2) · ·(z + N ) −z(PN 1
eγz z(1 + )e = eγz z e n=1 n −log N ) · e−z log N
n=1
n N!

z(z + 1) · ·(z + n)
→ lim ,
n→∞ nz n!

whenever z 6= 0, −1, −2, ....

Chapter 6 Q3: Recall Walliss product formula stating that

∞  
π Y 2n 2n
= ·
2 n=1
2n − 1 2n + 1

2·2 4·4 2m · 2m
= lim · ...
m→∞ 1 · 3 3 · 5 (2m − 1)(2m + 1)
24 · 44 ...(2m)4
= lim · (2m + 1)
m→∞ 12 · 22 · 32 · ...(2m + 1)2

(m!)4 24m
= lim (2m + 1).
m→∞ [(2m + 1)!]2

Recall the Gauss formula,

ns n!
Γ(s) = lim .
n→∞ s(s + 1)...(s + n)

Direct substitution yields the following

√
n2s+1/2 (n!)2 4n+1 (n!)2 n2s n
=
s(s + 1/2)(s + 1)...(s + n)(s + 1/2 + n) (2s)(2s + 1)...(2s + 2n + 1)
 n√
(2n + 1)2s (2n + 1)! 4 2n + 1(n!)2 4n2s+1/2

= · ·
(2s)...(2s + 2n + 1) (2n + 1)! (2n + 1)2s+1/2
π 3/2−2s √ 1−2s
r
→ Γ(2s) · ·2 = π2 Γ(2s).
2

Chapter 6 Q4: For f (z) = 1/(1 − z)α = exp [−α log(1 − z)], we have

(1 − z)f 0 (z) = exp [−α log(1 − z)] = αf (z)

(1 − z)f 00 (z) = (α + 1)f 0 (z)

Inductively, we have

(1 − z)f (n) (z) = (α + n − 1)f (n−1) (z) = (α + n − 1) · (α + n − 2) · ·(α + 1)f (z).

1
 Put z = 0, we have

f (n) (0) = (α + n − 1)f (n−1) (z) = (α + n − 1) · (α + n − 2) · ·(α + 1).

Since an (α) = f (n) (0)/n! , we have

an (α) (α + n − 1) · (α + n − 2) · ·(α + 1)
α−1
= → Γ(α).
n nα−1
R∞ R∞
Chapter 6 Q10: (a) Our goal is to comparer the integral 0 e−t ts−1 dt with 0 e−it ts−1 dt. So
we consider the meromorphic function f (w) = e−w wz−1 for a fixed z ∈ C on
{z : Re(z) ≥ 0}.
Consider the contour suggested in textbook and denote it by γ = γ1 ∪ γ2 ∪ γ3 ∪ γ4
where γ1 is on the real axis, γ2 is the large circular arc, γ3 is on the imaginary
axis and γ4 is the smaller γ4 . As it is holomorphic, by cauchy theorem,
I I I I I
0= f (w) dw = + + + f (w) dw.
γ γ1 γ2 γ3 γ4

We compute them separately.

I Z R
f (w) dw = e−t tz−1 dt,
γ1 
I Z π/2
iθ
f (w) dw = e−Re (eiθ )z−1 Rz dθ.
γ2 0
I Z R Z R
f (w) dw = − e−it (it)z−1 i dt = −iz e−it tz−1 dt,
γ3  
I Z π/2
iθ
f (w) dw = − e−e (eiθ )z−1 z dθ.
γ4 0

Letting  → 0, we see that the fourth term goes to 0 if Re(z) > 0. Let R → ∞,
we see that the second term goes to 0 as illustrated below.
Z 
 π/2  Z π/2
−Reiθ iθ z−1 z
e (e ) R dθ ≤ RRe(z) · e−R cos θ dθ.
 

 0  0

Let t = cos θ, we have

π/2 1
e−Rt
Z Z
e−R cos θ dθ = √ dt → 0.
0 0 1 − t2
R1 R 1/2
The convergence can be seen by splitting it into two parts, 1/2
, 0
, or simply
using Lebesgue’s monotone convergence theorem. Thus, if Re(z) ∈ (0, 1), we
have Z 
 π/2 
−Reiθ iθ z−1 z
e (e ) R dθ → 0
 

 0 

as R → ∞. So we yield the following result.

Z ∞ Z ∞
e−it tz−1 dt = i−z e−t tz−1 dt = e−iπz/2 Γ(z).
0 0

Part (a) follows from taking real part and imaginary part of this expression.

2
 (b) For t ∈ (0, 1), we have sin t ≤ t. So, whenever Re(z) ∈ (−1, 1),
Z 1  
 sin t 
Z 1
 dt ≤ 1
 dt < ∞.
0
 t1−z 
0 t−Re(z)

On the other hand,

Z ∞ Z ∞
sin t
dt = − cos 1 + (z − 1) cos (t)tz−2 dt.
1 t1−z 1

Thus, M (sin)(z) define a analytic function on −1 < Re(z) < 1 . By analytic

continuation, the second equality holds on the large strip. Put z = 0 and z =
−1/2 to obtain the conclusion.

Chapter 6,Q14: (a) First part follows from the Fundamental theorem of calculus. And the second
part follows from integrating log x.
(b) By (a), and monotonicity of Γ(x), for large n, we have
Z n+1
log n + log Γ(n) = log Γ(n + 1) ≥ log Γ(t) dt = n log n − n + c ≥ log Γ(n).
n

So, we have the following inequality,

n log n − n + c ≥ log Γ(n) ≥ (n − 1) log n − n + c

1 c log Γ(n) n−1 1 c

1− + ≥ ≥ − + .
log n n log n n log n n log n n log n

Take n → ∞ to obtain the asymptotic behaviour. Indeed, one can observe

log Γ(n) ∼ n log n + O(n) from the first inequality.

chapter 6, Q15:
Z ∞ Z ∞
Γ(s) = e−t ts−1 dt = ns e−nt ts−1 dt;
0 0

That is
Z ∞
n−s Γ(s) = e−nt ts−1 dt.
0

Summing over all n yield

∞ Z
X ∞
ζ(s)Γ(s) = e−nt ts−1 dt
n=1 0

Z ∞ ∞
X
= ts−1 e−nt dt
0 n=1
∞
ts−1
Z
= dt.
0 et − 1

Chapter 6,Q16: You may follow the step suggested in the tutorial class on 3/11 by proving the Rie-
mann’s Functional Equation. Or use the hints suggested in the textbook as stated in

3
 the following.

By Q15, we can write

∞ 1 ∞
ts−1 ts−1 ts−1
Z Z Z
Γ(s)ζ(s) = dt = dt + dt
0 et − 1 0 et − 1 1 et − 1
As the function in the second integral is of exponential decay, it defines a entire
function. Now we take a look on the first integral. By expanding the power series of
z/(ez − 1) at z = 0, we have
∞
1 X Bm m−1
= z
ez − 1 m=0 m!

where
 Bm 1/m
 
1
lim sup 
  = . (1)
m→∞ m!  2π

by Cauchy Hadamard theorem and the fact that radius of convergence of z/(ez − 1)
is 2π. Using this, we are able to represent the first integral in the following form.
1 ∞
ts−1
Z X Bm
t
dt = .
0 e −1 m=0
m!(s + m − 1)

By the control of (1), the series converges uniformly on any compact set away from
{1, 0, −1, ...}. It gives a meromorphic function on C with simple pole at {1, 0, −1, ...}.
But since 1/Γ(s) is a entire funtion with simple zero at {0, −1, −2, ...}. So
Z 1 s−1
1 t
dt
Γ(s) 0 et − 1
define a meromorphic function with a simple pole at s = 1.

Chapter 7,Q1: By summation by part formula, we can reformulate the Dirichlet series in the following
way,
N N −1  
X an AN X 1 1
= s − An − s .
n=1
ns N n=1
(n + 1)s n

Noted that in general, the mean value theorem does not hold for holomorphic function.
So we argue in the following way.

For a fixed s ∈ C, define a holomorphic function f (z) = z −s on the right half plane.
Consider the straight line γ from n + 1 to n.
 
 1 1 

 (n + 1)s − = |f (n + 1) − f (n)|
ns 
Z n+1 
f 0 (t) dt
 
= 
n
Z n+1
≤ |f 0 (t)| dt ≤ sup{|f 0 (t)| : t ∈ [n, n + 1]}.
n

4
 Within [n, n + 1], for Re(s) > 0,

|s| |s| |s|

|f 0 (x)| = ≤ Re(s)+1 ≤ Re(s)+1
|xs+1 | x n

Combine this to the series, we obtain

N −1   N −1
X  1 1  X M |s|
|An | 
 − s ≤ .
n=1
(n + 1)s n n=1
nRe(s)+1

∞
X an
The series is absolutely convergent. So converge. And the convergence is
n=1
ns
uniform on any compact set due to the above estimate.
P P
Chapter 7,Q2: (a) We claim that for two absolutely convergent series an , bn ,

∞
X ∞
X ∞ X
X
ak · bm = ak bm .
k=1 m=1 n=1 mk=n

Let  > 0, there exists N such that,

∞
X ∞
X
|ak | + |bm | < .
k=N m=N

On the other hand,

 
N2 X N2 N2
N N
X X X  X X X X


 ak · bm − ak bm  ≤
 |ak bm | + |ak bm | .
k=1 m=1 n=1 mk=n  n=N +1 mk=n;m>N n=N +1 mk=n;k>N

Due to the symmetry, it suffices to estimate one of terms in the right hand side.
2 2
N
X X N
X N
X
|ak bm | ≤ |ak | · |bm |
n=N +1 mk=n;k>N k=N +1 m=1

∞ ∞ ∞
!
X X X
≤ |ak | · |bm | <  |bm | .
k=N m=1 m=1

∞ ∞
X 1 X 1 X 1
(b) Since (ζ(s))2 = s
· s
= .
k m=1 m (mk)s
k=1 k,m

For each n ∈ N, the number of divisors is equal to the number of ways of repre-
senting n by mk. Thus,
∞
X 1 X d(n)
= .
(mk)s n=1
ns
k,m

Similarly,
X ma
ζ(s)ζ(s − a) =
(mk)s
k,m

5
 For a given n ∈ N, we sum over all ma where m is a divisor of n. Thus

∞
X σa (n)
ζ(s)ζ(s − a) = .
n=1
ns

Chapter 7-Q3: (a) By Euler’s identity,

∞
1 Y 1 X an
= (1 − s ) = .
ζ(s) p
p n=1
ns

If n = 1, a1 = 1 clearly. If n = p1 p2 · ·pk , where pj are all distinct prime. Then

an = (−1)k . Result follows from unique factorization of integer.
1
(b) Since 1 = ζ(s) · ζ(s) ,

∞ ∞ ∞
X 1 X µ(k) X an
1= · =
m=1
ms ks n=1
ns
k=1

P P
where an = mk=n µ(k) = k|n µ(k). By comparing coefficient, we show the
claim.

Chapter 7-Q5: (a) This is a special case of Q1 in Ch7.

(b)
∞ ∞
X 1 1 X 1
ζ(s) − ζ̃(s) = s
= s s
= 2−s ζ(s).
n=1
(2n) 2 n=1
n

Rearrange it to obtain the desired equation.

(c) For s ∈ (0, 1), since
1 1
− >0
(2k − 1)s (2k)s

for all k ∈ N, we have ζ̃(s) > 0. By analytic continuation, equation in part (b)
holds on the whole complex plane. So, ζ(s) has no zero on (0, 1).
According to the functional equation,

Γ((1 − s)/2)
ζ(s) = π s−1/2 ζ(1 − s) .
Γ(s/2)

At s = 0, the simple pole at s = 0 for ζ(1 − s) is cancelled out with the simple
zero of Γ(s/2). Thus, ζ has no zero at s = 0.

Chapter 7-Q6: If a = 1,
Z c+iN  
1 c + iN
ds = log → log(−1) = iπ.
c−iN s c − iN

If a ∈ [0, 1), Since as /s has no pole on the right half plane of the vertical strip,

as
I
1
ds = 0,
2πi γ s

6
 where γ is the curve composed of the vertical strip from c − iN to c + iN and a semi-
circle γ 0 from c + iN to c − iN in clockwise direction.
Thus,
c+iN
as as
Z Z
1 1
lim ds = lim ds.
N →∞ 2πi c−iN s N →∞ 2πi −γ 0 s

But
 Z π/2 c+N cos t
as 
Z
 |a |
 ds ≤ it |
· N dt

−γ 0 s −π/2 |c + N e
Z π/2 c+N cos t
|a |
≤ · N dt
−π/2 N −c
Z π/2
c
≤ 4a aN cos t dt,
0

for all large N . Via change of coordinate cos t = x, we get

Z 1 Z 1
Nx 1 aN x
a dx = √ dx → 0 as N → ∞.
0 sin t 0 1 − x2
R1 R 1/2
You can show this by splitting the integral into 1/2
and 0
. And followed by direct
integration. Or use dominated convergence theorem. Combine everything to conclude
the case of a ∈ [0, 1).

If a > 1, consider the contour γ composed of the vertical strip from c − iN to c + iN ,

the horizontal strip from c + iN to y-axis, the semi-circle from iN to −iN oriented
in the anticlockwise direction, and then the horizontal strip from −iN to the vertical
strip c + it, t ∈ R. Denote the contour γ by γ1 + γ2 + γ3 + γ4 respectively.

For γ3 , we have
π/2 π/2
as aN cos t
Z Z Z
| ds| ≤ N dt = aN cos t dt.
γ3 s 3π/2 |N eit | 3π/2

Argue similar to above, we have when N → ∞

Z π/2
aN cos t dt → 0.
3π/2

Similarly, the integral over γ2 , γ4 also tends to 0, as N → ∞. By residue theorem, the

integral over γ is 1 which yield our result.

Chapter 7-Q8: (a)

F (−s) = ξ(1/2 − s) = ξ(1 − 1/2 + s) = ξ(1/2 + s) = F (s).

The second equality follows from functional equation of ξ. By expanding its

P∞ n
power series, F (s) = n=0 an s . We conclude that an = 0 if n is odd. So
P∞
G = n=0 a2n sn .

7
(b) The only pole of ζ is at s = 1, so it remains to show the order of growth of
(s − 1)ζ(s). Make use of the following functional equation, for Re(s) < 0,

Γ((1 − s)/2)
ζ(s) = π s−1/2 ζ(1 − s) .
Γ(s/2)
1
Noted that , π s is of order 1.
Γ(s)
Z ∞ Z ∞
−t (−s−1)/2
|Γ((1 − s)/2)| ≤ e |t | dt ≤ e−t t(−Re(s)−1)/2 dt
0 0

≤ Γ((1 − Re(s))/2)

≤ (−Re(s))!

By Stirling formula, it is of order 1. So it suffices to consider Re(s) ≥ 0. But it

has already been done in Proposition 2.7 of ch6.

It remains to show the lower bound. Substitute s = −2n − 1, we have

Γ(n + 1)
|ζ(−2n − 1)| = π −2n−3/2 ζ(2 + 2n)
Γ(−n − 1/2)
∞
n! X 1
= π −2n−3/2 .
Γ(−n − 1/2) k 2+2n
k=1

On the other hand,

√
π
|Γ(−n − 1/2)| =
(n + 1/2)(n + 1/2 − 1) · · · (3/2)(1/2)
√
2n π
=
(2n + 1)(2n − 1) · · · (3)(1)
√
22n n! π
= ·
(2n)!
Also, by comparing to the integral , we have
∞
X 1 1
≈ .
k 2+2n 1 + 2n
k=1

Combine all this, we get

(2n)! 1
|ζ(−2n − 1)| ≈ π
(2π)2n 1 + 2n
Taking log yield
log |ζ(−2n − 1)| ≈ log π + log[(2n)!] − 2n log (2π) − log (1 + 2n)

Since log(2n)! ≈ 2n log(2n) + O(n), we conclude that ζ(−2n − 1) is of order 1.

(c) By Q14 in ch5, G has infinitely many zeros. Since

ξ(s + 1/2) F (s)

ζ(s + 1/2) = π 1/4+s/2 = π 1/4+s/2 ,
Γ(s/2 + 1/4) Γ(s/2 + 1/4)

ζ(s) has infinitely many zeros.