Solutions to HW 5
Karol Koziol
February 21, 2011
p. 327
15. (c) To evaluate the integral, we use the method of Problem 14, which tells us that if f (z) is a rational
function that is summable (meaning, for example, Rthat f (z) = P (z)/Q(z) with deg(Q) ≥ 2 + deg(P )),
P
then we may evaluate N
k=−N f (k) by evaluating ΓN πf (z) cot(πz)dz . In the integral, we have ΓN as
the square contour that surrounds the integer points −N, −N +1, . . . , N −1, N . Note that the function
π cot(πz) has simple poles at each integer, with residue 1. Also, note that
R π cot(πz) is uniformly
bounded by a constant (independent of N ) on ΓN , and therefore the integral ΓN πf (z) cot(πz)dz goes
to 0 as N → ∞.
Now, let f (z) = 1/z 2 . The f evidently meets the requirements of problem 14, and therefore by the
residue theorem
Z
ΓN
π cot(πz)
dz
z2
=
Res
X
2πi
poles zj ∈ΓN

=
2πi 
N
X
k=−N,k6=0
π cot(πz)
; zj
z2
1
+ Res
k2

π cot(πz)
;0 .
z2
Since the left side goes to 0 as N goes to innity, we are left with
∞
X
k=−∞,k6=0
1
= −Res
k2
π cot(πz)
;0 ,
z2
so it suces to compute this residue. Note that f (z) has a triple pole at z = 0. We compute the rst
several terms of the Laurent expansion of π cot(πz)
at z = 0 to determine the residue. First, note that
z2
we may obtain the expansion for 1/ sin(z) by setting 1/ sin(z) = a−1 z −1 + a0 + a1 z + . . ., and solving
1
the relation (1/ sin(z)) sin(z) = 1 recursively. We get
1
1 5
= z − z3 +
z − ...
6
120
1 1
7 3
1
=
+ z+
z + ...
sin(z)
z
6
360
1
1
π
7π 3 3
=
+ z+
z + ...
sin(πz)
πz
6
360
π2 2 π4 4
z +
z + ...
cos(πz) = 1 −
2
24
1 π2
π cot(πz) =
−
z + ...
z
3
1
π2
π cot(πz)
=
−
+ ...
2
3
z
z
3z
P∞
2
2
= π3 , and k=1 k12 = π6 .
sin(z)
Therefore,
P∞
1
k=−∞,k6=0 k2
16. The Bernoulli numbers Bn are dened by the series Taylor series
∞
X Bk
z
=
zk .
ez − 1
k!
k=0
We compute the rst several terms of this expansion. Note that by l'Hôpital's rule (denoted by ∗), we
have
z
∗
lim z
=
e −1
d
z
lim
=
z→0 dz
ez − 1
z→0
1
ez − 1 − zez
z→0 (ez − 1)2
−z
∗
= lim
z→0 2(ez − 1)
1
= − .
2
z
P
+1)
B` `
Now, consider the function g(z) = ezz−1 + z2 = z(e
`6=1 `! z . We have g(−z) = g(z), so we
2(ez −1) =
have
that Bn = 0 for every odd n greater than 1. Therefore, we may rewrite the series for g(z)
Pshown
B2k 2k
as ∞
k=0 (2k)! z .
We now need the expansion of cot in terms of the exponential. We have
cot(z)
lim
eiz + e−iz
eiz − e−iz
e2iz + 1
i 2iz
,
e −1
=
i
=
2
and therefore
2πiz(e2πiz + 1)
2(e2πiz − 1)
= g(2πiz)
∞
X
B2k
=
i2k
(2πz)2k .
(2k)!
πz cot(πz)
=
k=0
We now may compute the sum. We proceed exactly as in problem 15(c), and we obtain
∞
X
1
`=−∞,`6=0
`2n
= −Res
π cot(πz)
;
0
.
z 2n
The coecient of z −1 in π cot(πz)
is exactly the coecient of z 2n in πz cot(πz), which by our previous
z 2n
2n
computation is (−1)n (2π)
(2n)! B2n . Dividing by 2 gives
∞
X
1
22n−1
= (−1)n−1 π 2n
B2n .
2n
`
(2n)!
`=1
1
17. (a) Let a be a real number, and set f (z) = (z+a)
2 . Assume a is not an integer (the integer case
π cot(πz)
is similar to problem 15(c)). Then the function (z+a)2 has simple poles at all integers, with residue
1/(k + a)2 , and a double pole at z = −a. Let ΓN be the contour from problem 15(c), so that
Z
ΓN
π cot(πz)
dz
(z + a)2
=
2πi
Res
X
poles zj ∈ΓN
=
2πi
N
X
k=−N
π cot(πz)
; zj
(z + a)2
1
+ Res
(k + a)2
!
π cot(πz)
; −a
.
(z + a)2
We also have
Res
π cot(πz)
; −a
(z + a)2
d
(π cot(πz))
dz
−π 2
= lim
z→−a sin(πz)2
=
lim
z→−a
= −π 2 csc(πa)2 .
Therefore,
P∞
1
k=−∞ (k+a)2
= π 2 csc(πa)2 .
π cot(πz)
1
(b) Let a be any (non-zero) real number, and let f (z) = z2 +a
has
2 . Then the function
z 2 +a2
simple poles at all integers, with residue 1/(k2 + a2 ), and two simple poles at z = ±ai. We have, as
3
before
Z
N
X
1
π cot(πz)
dz = 2πi
+ Res
2
2
2
k + a2
ΓN z + a
k=−N
π cot(πz)
π cot(πz)
Res
;
ai
=
|z=ai
2
2
z +a
2z
π cot(πai)
=
2ai
π coth(πa)
= −
2a
π coth(πa)
π cot(πz)
; −ai
= −
.
Res
z 2 + a2
2a
Putting everything together gives
P∞
1
k=−∞ k2 +a2
=
4
!
π cot(πz)
π cot(πz)
; ai + Res
; −ai
z 2 + a2
z 2 + a2
π coth(πa)
.
a