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Math 246A
Homework 2 Solutions
Due Thursday, April 19
Note the solutions are often just an outline, but I have tried to address all of the key points and you should
be able to fill in the details. If you have questions about them feel free to ask me; also if there are any
errors in my solutions please let me know. Of course there are usually many other correct ways to solve
each problem as well.
- Prove that ∫ ∞
0
sin(x
2 ) dx =
∞
0
cos(x
2 ) dx =
2 π
These are the Fresnel integrals. Here
∞ 0
is interpreted as limR→∞
R 0
[Hint: Integrate the function e
−z 2 over the path that goes from 0 to R along the real axis followed by
R to Re
i π (^4) along a circular arc and then returns to 0 via a straight line. Recall that ∫ (^) ∞
−∞
e −x 2 dx =
π.]
Parameterize the three paths as α(t) = t for t ∈ [0, R], β(t) = Re
it for t ∈ [0, π/4], and γ
− (t) = te
iπ/ 4
for t ∈ [0, R]. Since e −z
2 is entire, the integral around the closed path is 0 , and we have
R
0
e
−it 2
dt =
R
0
e
−t 2 dt +
π/ 4
0
e
−R 2 e i 2 t iRe
it dt.
As R → ∞ the second integral on the right goes to 0 as will be shown below and the first integral on the
right goes to
π/ 2 by the hint (since e −t 2 is even). Now use e −it 2 = cos t 2 − i sin t 2 and equate real and
imaginary parts.
To show the integral around the circular arc goes to zero as R → ∞, note that ∣ ∣ ∣
π/ 4 0 e
−R 2 e i 2 t iRe
it dt
∣ ≤ R
π/ 4 0 e
−R 2 cos 2t dt and now use a variant of Jordan’s Lemma (see the solution
to exercise 2 below): For 0 ≤ t ≤ π/ 4 , we have cos 2t ≥ 1 − 4 t/π and so e
−R 2 cos 2t ≤ e
R 2 (4t/π) e
−R 2 .
Since
π/ 4 0
e
R^2 (4t/π)
π 4 R 2 (e
R^2 − 1) we have R
π/ 4 0
e
−R^2 cos 2t dt ≤
π 4 R
(1 − e
−R^2 ) which goes to 0 as
R → ∞. Some people chose to integrate over e iz 2 rather than e −z 2 which also works and allows one to
use the bound on sin 2t instead; the calculation is almost identical.
- Show that
∞ 0
sin x x
dx = π 2
[Hint: The integral equals
1 2 i
∞ −∞
e ix − 1 x
dx. Use the indented semicircle.]
Ignore the hint and instead follow the method used for Example 2 in Stein and Shakarchi. Integrate e
iz /z
over the indented semicircle of radius R whose intendation has radius . As → 0 the integrals along the
x axis add up to
−∞
e ix
x dx. As R → ∞ the integral around the outer circle goes to 0 , as we will prove
momentarily. Since e iz
z
1 z
- E(z) where E(z) is bounded as z → 0 , we have the integral around the
small circle equal to −
π 0 (i + ie
it E(e
it )) dt → −iπ as → 0. The result follows from Cauchy’s theorem
by taking imaginary parts and noting that
sin x x is even.
The proof that integral around the outer circle γR goes to 0 as R → ∞ requires a more careful estimate,
which is referred to in Gamelin’s text as Jordan’s Lemma. Since sin t is concave down on 0 ≤ t ≤ π/ 2 , we
have sin t ≥ 2 t/π on this interval, and so
∫ (^) π
0 e
−R sin t dt = 2
∫ (^) π/ 2
0 e
−R sin t dt ≤ 2
∫ (^) π/ 2
0 e
− 2 Rt/π dt < π/R.
From this it follows
γR
e iz
z dz
∣ < π/R and thus goes to 0 as R → ∞.
- 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.
Let γ be the path from the origin to R along the positive real axis, then the circular arc from R to Re
iω ,
then the line segment returning to the origin. The integral along the real axis is
∫ R
0
e
−Ax dx which goes to
1 /A as R → ∞. The integral along the circular arc is bounded by
∫ (^) ω
0
e
−AR cos t R dt which goes to 0 as
a/A ≤ cos t ≤ 1 in the range of integration. The final integral is
−
0
e
−At(cos ω+i sin ω) e
iω dt = −e
iω
0
e
−ax cos bx dx − i
0
e
−ax sin bx dx
where sin ω = b/A (you
get the same answer if sin ω = −b/A). Using Cauchy’s theorem and equating real and imaginary parts
results in
0
e
−ax cos bx dx =
a a^2 +b^2 and
0
e
−ax sin bx dx =
b a^2 +b^2
- Prove that for all ξ ∈ C we have e
−πξ 2 =
−∞
e
−πx 2 e
2 πixξ dx.
Write ξ = re it
. Then
∞ −∞
e −πx
2 e 2 πixξ dx =
∞ −∞
e −πx
2 e 2 πixr cos t e − 2 πxr sin t dx = ∫ ∞ −∞
e −π(x 2 +2xr sin t) e 2 πixr cos t dx = e π(r sin t)
2 ∫^ ∞
−∞
e −π(x+r sin t) 2 e 2 πixr cos t dx. Change variables letting
y = x + r sin t. This results in
e π(r 2 sin 2 t− 2 ir sin t cos t))
∞ −∞
e −πy 2 e 2 πiyr cos t dy = e π(r 2 sin 2 t− 2 ir sin t cos t)) e −πr 2 cos 2 t = e −πξ 2 where we
have used Example 1 (the real variable version of this problem) of the text and lectures.
Another idea which works is basically do the same proof as in the real case. Assume ξ is not real (Really
only ξ = 0 is exceptional) and integrate e
−πz 2 around the parallelogram with vertices
R, R + iξ, −R + iξ, −R. Note to make sure things make sense when ξ is purely imaginary.
- Suppose f is continuously complex differentiable on Ω, and T ⊂ Ω is a triangle whose interior is also
contained in Ω. Apply Green’s threorem to show that
T
f(z) dz = 0. This provides a proof of
Goursat’s theorem under the additional assumption that f ′ is continuous.
[Hint: Green’s theorem says that if (F, G) is a continuously differentiable vector field, then
T
F dx + G dy =
Interior of T
∂G
∂x
∂F
∂y
dx dy.
For appropriate F and G, one can then use the Cauchy-Riemann equations.]
Let f(z) = f(x, y) = u(x, y) + iv(x, y). Then ∫
T f(z) dz =
T (u(x, y) + iv(x, y))(dx + idy) = F dx + G dy where F = u + iv and G = −v + iu. If f
′
is continuous then the partial derivatives of u and v exist and are continuous, and so (F, G) is
continuously differentiable. Now Gx = −vx + iux = uy + ivy = Fy by the Cauchy-Riemann equations,
and so by Green’s theorem
T f(z) dz = 0.
- Let Ω be an open subset of C and let T ⊂ Ω be a triangle whose interior is also contained in Ω.
Suppose that f is a function holomorphic in Ω except possibly at a point w inside T. Prove that if f
is bounded near w, then
T f(z) dz = 0.
For > 0 sufficiently small, let γ(t) be the boundary of D (w), and let f(z) ≤ M for all z ∈
D(w).
Connect the vertices T to the closest points of γ, and note that the three indented triangles created are
toy countours with f holomorphic on open sets containing them, and so by Cauchy’s theorem it follows ∫
T f(z) dz =
γ f(z) dz. Furthermore
γ f(z) dz
∣ ≤ M 2 π → 0 as → 0 , and so
T f(z) dz = 0.
- Suppose f : D → C is holomorphic. Show that the diameter d = supz,w∈D |f(z) − f(w)| of the image
of f satisfies 2 |f
′ (0)| ≤ d. Moreover, it can be shown that equality holds precisely when f is linear,
f(z) = a 0 + a 1 z.
[Hint: 2f
′ (0) =
1 2 πi
|ζ|=r
f (ζ)−f (−ζ) ζ^2
dζ whenever 0 < r < 1.]
For 0 < r < 1 we have f ′ (0) = 1 2 πi
|ζ|=r
f (ζ) ζ^2
dζ and reparameterizing using −ζ gives
f
′ (0) =
1 2 πi
|ζ|=r
−f (−ζ) ζ^2 dζ; adding the two gives 2 f
′ (0) =
1 2 πi
|ζ|=r
f (ζ)−f (−ζ) ζ^2 dζ as in the hint. Now
use the ML estimate to see that 2 |f
′ (0)| ≤
1 r |f(ζ) − f(−ζ)| ≤
d r
. Taking the limit as r → 1 gives the
result. It is easy to see that when f is linear there is equality; I didn’t try to prove the converse.
R
2 − |z|
2
- i 2 R(−Re (z) sin γ + Im (z) cos γ)
R
2 − 2 RRe (ze −iγ ) + |z| 2
Taking the real part gives the result.
Problem 1a. Here is an example of an analytic function on the unit disc that cannot be extended
analytically past the unit circle. The following definition is needed. Let f be a function defined in the unit
disc D, with boundary circle C. A point w on C is said to be regular for f if there is an open neighborhood
U of w and an analytic function g on U , so that f = g on D ∩ U. A function f defined on D cannot be
continued analytically past the unit circle if no point of C is regular for f.
Let f =
n= z 2 n for |z| < 1. Notice that the radius of convergence of the above series is 1. Show that f
cannot be continued analytically past the unit disc. [Hint: Suppose θ = 2πp/ 2 k , where p and k are positive
integers. Let z = re iθ ; then
∣f(reiθ^ )
∣ (^) → ∞ as r → 1.]
Following the hint, we have f(re
iθ ) =
∑k− 1
n= r
2 n e
iθ 2 n
n=k r
2 n
. It follows that
f(re
iθ )
→ ∞ as r → 1. It
follows that e
iθ cannot be regular since f(e
iθ ) cannot have a finite value and still be the limit of points f(re
iθ ).
Now for any x ∈ [0, 1] and any > 0 , there exists some p, k ∈ Z
such that
x − p/ 2
k
< (let 2
−k < and
divide [0, 1] into the 2
k parts of equal length to see this). It follows no point on the unit circle can be regular.
