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Solution of Homework Assigned on February 5, 2009 due February 17, 2009
Problem 1 (from Stein & Shakarchi, pp.30-31, #25).
(a) Evaluate the integral (^) ∫
γ
z n dz
for all integers n (positive, negative, or zero). Here γ is any circle centered at the origin with the positive (counterclockwise) orientation.
(b) Same question as before, but with γ any circle not containing the origin.
Hint: Use the parametrization z = a + re iθ for 0 ≤ θ ≤ 2 π for the circle of center a ∈ C and radius r > 0. Find a complex-valued function F (θ) of the real variable θ for 0 ≤ θ ≤ 2 π such that
d
dθ
F (θ) =
a + re iθ )n d
dθ
a + re iθ
for 0 ≤ θ ≤ 2 π. Consider F (2π) − F (0). Distinguish between the case where |a| < r and the case where |a| > r.
(c) Show that if |a| < r < |b|, then ∫
γ
(z − a)(z − b)
dz =
2 πi
a − b
where γ denotes the circle centered at the origin, of radius r, with the positive orientation.
Hint: Use the decomposition of 1 (z−a)(z−b) into partial fractions^
A z−a +^
B z−b (where A and B are complex numbers) and use Part (b).
Solution of Problem 1. (a) Use the parametrization
z = e iθ for0 ≤ θ ≤ 2 π
for γ to compute the integral
∫
γ
z n dz =
∫ (^2) π
θ=
e inθ d
e iθ
∫ (^2) π
θ=
ie i(n+1)θ dθ
∫ (^2) π
θ=
sin ((n + 1)θ) dθ + i
∫ (^2) π
θ=
cos ((n + 1)θ) dθ
which yields 2πi for n = −1 and 0 for n 6 = −1.
(b) We use the parametrization z = a + re iθ for 0 ≤ θ ≤ 2 π for the circle γ.
Then (^) ∫
γ
zndz =
∫ (^2) π
θ=
a + reiθ
)n ireiθdθ.
For n 6 = −1 we can use the explicit primitive
F (θ) =
n + 1
a + reiθ
)n+
for
a + reiθ
)n ireiθ^ as a function of θ. Since F (2π) = F (0), it follows that
the integral
∫
γ
z n dz =
∫ (^2) π
θ=
a + re iθ )n ire iθ dθ = F (2π) − F (0) = 0
for n 6 = −1. For n = −1, we can use the primitive
F (θ) = log
a + re iθ)^ def = log
∣a + reiθ
∣ (^) + i tan−^1 Im
a + re iθ
Re (a + reiθ)
where the numerical value of
tan − 1 Im^
a + re iθ
Re (a + reiθ)
is chosen as a continuous function of θ for 0 ≤ θ ≤ 2 π. The integral
∫
γ
dz
z
= F (2π) − F (0)
is equal to 2πi for |a| < r and to 0 if |a| > r. Instead of using this primitive
for n = −1 we can also compute by brute force the integral
∫
γ
dz
z
∫ (^2) π
θ=
ireiθdθ
a + reiθ
as follows by changing the interval of the parameter to −π ≤ θ ≤ π and
using the rational parametrization of the unit circle
cos θ =
1 − t^2
1 + t^2
and the partial fraction decomposition
(z − a)(z − b)
b − a
z − b
z − a
we get
∫
γ
dz
(z − a)(z − b)
b − a
γ
dz
z − b
γ
dz
z − a
2 πi
a − b
Problem 2 (from Stein & Shakarchi, p.64, #1). Prove that
∫ (^) ∞
x=
sin
x 2
dx =
x=
cos
x 2
dx =
2 π
4
These are the Fresnel integrals. Here,
0 is interpreted as limR→∞
∫ R
Hint: Integrate the holomorphic function f (z) = e −z^2 over the path which is
the boundary of
{ z = reiθ
∣ 0 < r < R,^0 < θ <
π
4
and use
−∞ e
−x^2 dx =
π which can be derived by squaring and using polar
coordinates in R^2.
Solution of Problem 2. Denote by CR the arc
{ z = Reiθ
∣ 0 ≤^ θ^ ≤^
π
4
in the counter-clockwise sense. Let f (z) = e−z
2
. We are going to check that
lim R→∞
CR
f (z)dz = 0.
First we observe that
sin ϕ ≥
ϕ
for 0 ≤ ϕ ≤ π 3 , because if we set^ g(ϕ) = sin^ ϕ^ −^
1 2 ϕ, then from^ g(0) = 0 and
g ′ (ϕ) = cos ϕ −
≥ cos
π
3
≥ 0 for 0 ≤ ϕ ≤
π
3
it follows by the mean-value theorem that g(ϕ) ≥ 0 for 0 ≤ ϕ ≤ π
- We now estimate
∣ ∣ ∣ ∣
CR
f (z)dz
∫ π 4
θ=
e −R^2 cos(2θ) Rdθ
∫ π 4
θ=
e −R^2 cos( (^2) (π 4 −θ)) Rdθ =
∫ π 4
θ=
e −R^2 sin(2θ) Rdθ
∫ π 4
θ= π 6
e −R^2 sin(2θ) Rdθ +
∫ π 6
θ=
e −R^2 sin(2θ) Rdθ
∫ π 4
θ= π 6
e − 12 R^2 Rdθ +
∫ π 6
θ=
e −R^2 θ Rdθ
∫ π 4
θ= π 6
e − 12 R^2 Rdθ +
[
R^2
e −R^2 θ R
]π 6
θ=
which approaches 0 as R → ∞.
Next we observe that ∫ (^) ∞
x=−∞
e −x^2 dx =
π,
because
(∫ (^) ∞
x=−∞
e −x^2 dx
x=−∞
e −x^2 dx
y=−∞
e −y^2 dy
(x,y)∈R^2
e −x^2 −y^2 dxdy =
∫ (^2) π
θ=
r=
re −r^2 drdθ
= 2π
r=
re −r^2 dr = 2π
[
−e−r
2
]∞
r=
= π.
By applying Cauchy’s theorem to the integration of f (z) = e −z^2 over the
path which is the boundary of
{ z = re iθ
∣ 0 < r < R,^0 < θ <
π
4
and letting R → ∞, we conclude that
∫ (^) ∞
t=
e −((1+i)t)^2 (1 + i) dt =
π
2
because
eiz^ − 1
z
∑^ ∞
n=
in+
(n + 1)!
z n
is holomorphic at z = 0. By integrating by parts with the integration of the
factor eiz^ , we get
∫
CR
e iz
z
dz =
e iz
iz
]z=R
z=−R
CR
e iz
iz^2
dz
which approaches 0 as R → ∞. By using the parametrization of CR by
z = Reiθ^ for 0 ≤ θ ≤ π, we get
∫
CR
z
dz = πi.
Hence (^) ∫ ∞
x=−∞
e ix − 1
x
dx = πi
and by taking the imaginary parts of both sides, we get
∫ (^) ∞
x=
sin x
x
π
2
Problem 4 (from Stein & Shakarchi, p.64, #3). Evaluate the integrals
∫ (^) ∞
0
e −ax cos bx dx and
0
e −ax sin bx dx, a > 0
by integrating e −Az , A =
a^2 + b^2 , over an appropriate sector with angle ω,
with cos ω = a A.
Solution of Problem 4. By replacing b by −b if necessary, we can assume
without loss of generality that b > 0, because the two integrals can readily be
computed in the case of b = 0. We can write a + ib = A eiω^ with 0 < ω < π 2 ,
where A =
a^2 + b^2 , so that cos ω = (^) Aa and sin ω = (^) Ab.
The integral of the entire function f (z) = e −Az over the boundary of the
sector (^) {
z = re iθ
∣ 0 < r < R, 0 < θ < ω
is zero. Let CR be the arc
{ z = Re iθ
∣ 0 ≤^ θ^ ≤^ ω
in the counter-clockwise sense. Then
∫
CR
e −Az dz =
−e −Az
A
]z=Reiω
z=R
A
e −AR − e −AR(cos ω+i sin ω)
A
e −AR − e −R(a+ib)
→ 0 as R → ∞.
Hence
∫ (^) ∞
t=
e−At(cos^ ω+i^ sin^ ω)^ (cos ω + i sin ω) dt =
t=
e−Atdt =
A
Using cos ω = a A and sin ω = b A , we get
∫ (^) ∞
t=
e −at (cos (bt) − i sin (bt))
a
A
b
A
dt =
A
Equating the real and imaginary parts of the above equation, we get
a
t=
e −at cos (bt) dt + b
t=
e −at sin (bt) dt = 1,
b
t=
e −at cos (bt) dt − a
t=
e −at sin (bt) dt = 0.
Multiplying the first equation by a and adding the result to b times the
second equation, we get
∫ (^) ∞
t=
e −at cos (bt) dt =
a
a^2 + b^2
and (^) ∫ (^) ∞
t=
e −at sin (bt) dt =
b
a^2 + b^2
