Lecture 9
RESIDUE CALCULUS, PART II
Applications:
♦ Contour integrals in the presence of branch cuts
• Combining techniques for isolated singularities
(e.g. residue theorem)
with techniques for branch points
♦ Summation of series by residue calculus
RESIDUE THEOREM
♦ Let C be closed path within and on which f is holomorphic
except for m isolated singularities. Then
I
f (z) dz = 2πi
C
m
X
Reszj f
j=1
Im z
C
z0
zm
zj
Re z
METHODS TO CALCULATE RESIDUES
♠ General method: from Laurent expansion
∞
X
1
n
cn (z − z0 ) , where cn =
f (z) =
2πi
n=−∞
I
Γ
f (ξ)
dξ
(ξ − z0 )n+1
Resz0 f = c−1
I
1
=
dξ f (ξ)
2πi Γ
♠ Method for z0 pole of order n:
1
dn−1
n
Resz0 f = lim
[(z
−
z
)
f (z)]
0
n−1
z →z0 (n − 1)! dz
For n = 1 : Resz0 f = lim [(z − z0 )f (z)]
z →z0
EXAMPLE :
I=
I
C
sin z
dz
6
z
where C is the circle of centre z = 0 and radius 1
z = 0 pole of order 5
3
5
sin z
z
1
1 1
z
1 1 1
= 6 z−
+
+ ... = 5 − 3 +
+ analytic part
6
z
z
3!
5!
z
6z
120 z
|{z}
residue
=⇒ I =
I
C
iπ
sin z
dz = 2πiResz=0 (Integrand) =
6
z
60
Note :
I
C
cos z
dz = 0
6
z
z = 0 pole of order 6 with zero residue
Integral of the square root round the unit circle
√
√ iθ/2
Take principal branch : f (z) = z = re
, 0 ≤ θ < 2π . Branch cut along R+ .
• can’t apply Cauchy theorem to |z| = 1 but can apply it to contour Γ:
Im z
I
C1
f (z) dz = 0
Γ
Γ
I
• Then write
−1
=
Γ
Z
+
C1
Z
+
L1
Z
Cε
Let ε → 0 . By Darboux inequality |
Thus
I
C1
√
z dz = −
Z
1
0
dx
√
i2π/2
xe
−
+
I
Z
Cε
1
0
Z
Cε
L2
ε
L1
1
Re z
L2
√
√ ε→0
z dz| ≤ 2πε ε −→ 0 .
√
dx x = −2
Z
0
1
dx
√
x = −4/3 .
CALCULATION OF I =
Z
∞
0
xp−1
dx 2
x +1
,
0<p<2
Principal branch : z p−1 = |z|p−1 ei(p−1)θ , 0 ≤ θ < 2π . Branch cut along R+ .
• Residue theorem applied to contour Γ ⇒
Im z
Γ
C
R
R
Cρ
ρ
⇒
I
Γ
L2
L1
Re z
πp iπ(p−1)
z p−1
dz = 2πi[Resz=+i f + Resz=−i f] = 2πi cos( )e
2
z +1
2
• Next write
I
Let R → ∞ , ρ → 0 .
⇒
I
Γ
Z
CR
+
CR
Z
+
L1
Z
+
Cρ
Z
L2
−→ 0 for R → ∞ ,
Z
Cρ
−→ 0 for ρ → 0 ⇒
Z
∞
z
1 p−1
p−1 2πi(p−1)
dz =
x
−x
e
dx 2
2
z +1
x +1
0
Z ∞
xp−1
[1 − e2πi(p−1) ] = 2ieiπ(p−1) sin(πp) I
=
dx 2
x +1
0
p−1
Γ
=
Z
Therefore
πp iπ(p−1)
I = 2πi cos( )e
/[2ieiπ(p−1) sin(πp)]
2
πp
= π cos( )/ sin(πp) = π/[2 sin(πp/2)]
2
CALCULATION OF I =
Z
Consider
ln ( z + i )
ΓR
(z
2
0
∞
ln(x2 + 1)
dx
x2 + 1
Im z
Γ
R
i
R
dz
Re z
−i
+1)
Take principal branch of log. Branch cut
I
ln 2i
iπ
ln(z + i)
ln
2
+
dz
=
2πi
Res
f
=
2πi
=
π
z=i
2+1
z
2i
2
ΓR
Z
→ 0 for R → ∞ .
• By Jordan lemma
• By residue theorem
SR
Z
Z
ln(x + i)
dx =
2
x +1
R
Z
ln(−x + i)
dx+
2
x +1
R
Z
R
ln(x + i)
ln(x2 + 1) + iπ
dx =
dx
•
2
2
x +1
x +1
0
0
0
−R
Z ∞
ln(x2 + 1)
• Equating real parts for R → ∞ =⇒
= π ln 2
dx
2
x +1
0
R
.
SUMMATION OF SERIES BY RESIDUE CALCULUS
∞
X
n=1
g(n) ,
∞
X
(−1)n g(n) with g given function
n=1
In certain cases the sum of these series can be calculated
by exploiting the structure of poles and residues
of complex-valued functions.
P∞
n=1
g(n)
♠ cot(πz) has poles of order 1 at z = n, n ∈ Z, with residue 1
⇒ Consider ϕ(z) = π cot(πz)g(z). If g has no poles at n,
Resz=n ϕ = lim π(z − n) cot(πz)g(z) = g(n)
z →n
Im z
ΓN
(z) dz
ΓN
cot(z) bounded on Γ
N
−N−1/2
N+1/2
Re z
On vertical sides : | cot πz| = | cot π(N + 1/2 + iy)| = | tan iπy| = | tanh πy| ≤ 1
e−πy + eπy
eiπz + e−iπz
| ≤ −πy
≤ coth π(N +1/2) ≤ coth 3π/2
On horizontal sides : | cot πz| = | iπz
e
− e−iπz
|e
− eπy |
H
♦ If g vanishes sufficiently fast at N → ∞ for ΓN → 0, then
P
n g(n) = - sum of residues of ϕ at the poles of g
Example :
∞
X
1
2
n
n=1
• g(z) = 1/z 2 . This has a pole at z = 0.
X
π cot(πz)
1
= −Resz=0 ϕ , where ϕ(z) =
.
• Then
2
2
n
z
n=±1,±2,...
π 1 − (πz)2 /2! + (πz)4 /4! + . . .
π cot(πz)
= 2
• Laurent expansion :
2
z
z πz − (πz)3 /3! + (πz)5 /5! + . . .
1 π2 1
= 3 −
+ ...
z |{z}
3 z
residue
2
∞
X
X
π
π2
1
1
1
1
1
−
=
=
= − Resz=0 ϕ = −
=⇒
2
2
n
2 n=±1,±2,... n
2
2
3
6
n=1
Example :
∞
X
n=1
1
n2 + 3
ΓN
π cot π z
( z) =
z 2 +3
Im z
−N−1/2
N+1/2
Re z
(z) dz
ΓN
Resz=n ϕ = 1/(n2 + 3) , n = 0, ±1, ±2, . . .
√
√
√
√
√
Resz=i 3 ϕ = π cot(iπ 3)/(2i 3) = −π coth(π 3)/(2 3)
√
√
√
√
√
Resz=−i 3 ϕ = π cot(−iπ 3)/(−2i 3) = −π coth(π 3)/(2 3)
Then
I
I
ΓN
ΓN
"
N
X
√
1
π
1
ϕ(z) dz = 2πi − √ coth(π 3) + 2
+
2
n +3 3
3
n=1
→ 0 for N → ∞ because |ϕ(z)| → 0 like |z|−2 .
Thus
N
X
n=1
√
1
π
1
= √ coth(π 3) − .
2
n +3
6
2 3
#
♠ Alternating-sign series
P∞
n
(−1)
g(n) can be summed using
n=1
π
g(z)
ϕ(z) =
sin(πz)
1
g(z) = (−1)n g(n)
⇒ Resz=n ϕ = lim π(z − n)
z →n
sin(πz)
♦
PUnder nthe same hypotheses as in the previous discussion
n (−1) g(n) = - sum of residues of ϕ at the poles of g
♠ Series of the form
P∞
n
(−1)
g(2n + 1) can be summed using
n=1
π
ϕ(z) =
g(z)
cos(πz)
• ϕ has poles at z = n + 1/2, n ∈ Z
Example :
∞
X
(−1)n
n=1
π
,
• With ϕ(z) = 2
z sin(πz)
n2
(−1)n
= Resz=0 ϕ .
we have
2
n
n=±1,±2,...
X
π
1
π
= 2
• Laurent expansion : 2
z sin(πz)
z πz − (πz)3 /3! + (πz)5 /5! + . . .
1
π2 1
= 3+
+ ...
z
6 z
|{z}
residue
=⇒
∞
X
(−1)n
n=1
n2
1 X (−1)n
1
1 π2
π2
=
= − Resz=0 ϕ = −
=−
2
2 n=±1,±2,... n
2
2 6
12