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Math 246A
Homework 4 Solutions
Due Thursday, May 3
Note the solutions are often just an outline, 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.
- Show that (^) ∫ (^) ∞
−∞
cos x x^2 + a^2
dx = π
e−a a
, for a > 0.
The poles of e
iz z^2 +a^2 are^ ±ia. Compute^
C
eiz z^2 +a^2 dz^ where^ C^ is the semicircle of radius^ R > a^ in the upper half plane. The residue at ia is e
−a 2 ia , and so^
C
eiz z^2 +a^2 dz^ =^ π^
e−a a by the residue theorem. The integral around the arc goes to 0 as R → ∞ by a simple ML-estimate (again make sure you can prove this!) and the integral along the x axis is
∫ R
−R
eix x^2 +a^2 dx. Let^ R^ → ∞^ and take real parts to get the answer.
- Show (^) ∫ (^) ∞
−∞
x sin x x^2 + a^2
= πe−a^ for all a > 0.
Compute
C
zeiz z^2 +a^2 dz^ where^ C^ is the semicircle of radius^ R > a^ in the upper half plane. The poles of zeiz z^2 +a^2 are at^ ±ai^ and are simple, the residue at^ ai^ is^
aie−a 2 ai. The integral around the outside arc is bounded by
∫ (^) π 0
|Reiθ||eiReiθ ||Ririθ^ |dθ R^2 e^2 iθ^ −a^2 ≤^
∫ (^) π 0
Re−R^ sin^ θ^ R R^2 −a^2 ≤^
2 R^2 R^2 −a^2
∫ (^) π/ 2 0 e
−R 2 θ/π (^) dθ using Jordan’s Lemma,
and this is bounded by 2 R
(^2) π 2 R(R^2 −a^2 ) (1^ −^ e
−R) → 0 as R → ∞. So letting R go to ∞ we see ∫ (^) ∞ −∞
zeiz z^2 +a^2 dz^ = 2πi^
aie−a 2 ai =^ iπe
−a (^) taking imaginary parts gives our result.
- Use contour integration to show that ∫ (^) ∞
−∞
e−^2 πixξ (1 + x^2 )^2
dx = π 2
(1 + 2π |ξ|)e−^2 π|ξ|
for all ξ real. The poles of e − 2 πizξ (1+z^2 )^2 are^ ±i^ and there is a double pole at each point. Compute^
C
e−^2 πizξ (1+z^2 )^2 dz^ where^ C^ is the semicircle of radius R in the lower half plane if ξ ≥ 0 and the semicircle of radius R in the upper half plane if ξ < 0. The residue of the integrand at i, which can be calculated by multiplying the integrand by (z − i)^2 , differentiating, and taking the limit as z → i, is 14 ie^2 πξ^ (2πξ − 1). Similarly the residue of the integrand at −i is 14 ie−^2 πξ^ (2πξ + 1). Therefore if we can prove that the integral around the appropriate circular arc of radius∫ R is zero then the residue theorem gives the result. This follows from estimating Re^2 πξR^ sin^ tdt where the integral is over [0, π] if ξ < 0 and [π, 2 π] if ξ ≥ 0. In either case a Jordan Lemma bound gives the result. A few students pointed out that the integral is even (as a function of ξ) which saves half the work.
- Show that (^) ∫ (^) ∞
−∞
dx (1 + x^2 )n+^
1 ∗ 3 ∗ 5 · · · (2n − 1) 2 ∗ 4 ∗ 6 · · · (2n)
π.
Compute
C
dz (1+z^2 )n+1^ where^ C^ is the semicircle of radius^ R >^1 in the upper half plane.^
1 (1+z^2 )n+1^ has poles of order n + 1 at ±i. The residue at z = i is limz→i (^) n^1! ( (^) dzd )n^ (z+i^1 )n+1 = (−1)
n(2n)! ∫^ (n!)^2 (2i)^2 n+1^ so C
dz (1+z^2 )n+1^ = 2πi^
(−1)n(2n)! (n!)^2 (2i)^2 n+1^ =^
(−1)n(2n)! (n!)^2 (2)^2 ni^2 n^ π^ =^
(2n)! (2∗ 4 ∗ 6 ···(2n))^2 π^ =^
1 ∗ 3 ∗ 5 ···(2n−1) 2 ∗ 4 ∗ 6 ···(2n) π.^ A simple ML-estimate completes the problem.
- Prove that (^) ∫ (^2) π
0
dθ (a + cos θ)^2
2 πa (a^2 − 1)^3 /^2
, whenever a > 1.
∫Make the change of variable^ z^ =^ eiθ^ then^ dθ^ =^ dz^ iz^ and^ cosθ^ =^ z+1^2 /z. Thus our integral is equal to |z|=
4 zdz i(z^2 +2az+z+1)^2.^
4 zdz i(z^2 +2az+z+1)^2 has double poles at^ −a^ ±
a^2 − 1. Only −a +
a^2 − 1 = (^) a+√−a^12 − 1 has absolute value less than 1. The residue at this point is limz→−a+√a (^2) − 1 4 i dzd (z−(−a−z√a (^2) −1)) 2 = (^) i(a (^2) −a1) 3 / 2. Now applying the residue theorem gives the result.
- Prove that (^) ∫ (^2) π
0
dθ a + b cos θ
2 π √ a^2 − b^2
if a > |b| and a, b ∈ R.
Make the change of variable z = eiθ^ then dθ = dz iz and cosθ = z+1 2 /z. Thus our integral is equal to ∫ |z|=
2 dz i(bz^2 +2az+b).^
1 bz^2 +2az+b has simple poles at^
−a±√a^2 −b^2 b (note they’re real). Only the pole −a+ √ a^2 −b^2 b =^
−b a+√a^2 −b^2 is in the circle^ |z|^ = 1. Thus our integral is equal to 2 πi ∗ Res z= −a+
√a (^2) −b 2 b
2 i(bz^2 +2az+b) =^ √^2 π a^2 −b^2.
- Show that (^) ∫ 1
0
log(sin πx) dx = − log 2.
[Hint: Use the contour conisting of the line from i∞ to 0 followed by the line from 0 to 1 followed by the line from 1 to 1 + i∞.] Ignore the hint for the most part and follow instead the outline given in Ahlfors, Complex Analysis, p. 159. Integrate log(1 − e^2 iπz^ ) over a rectancle with vertices at 0 , 1 , 1 + iY and iY , but with quarter-circle indentations around 0 and 1. Show that the principal branch of the logarithm is well-defined in this region. Because of periodicity the integrals around the vertical sides cancel, and the integrals over the top segment and two quarter circles can be shown to go to ∫ 0. One thus obtains 1 0 log(−^2 ie
iπx (^) sin πx) dx = 0. Since log(− 2 ieiπx (^) sin πx) = log 2 + log(sin πx) + i(πx − π/2) on the principal branch, integrating one obtains the result.
- Show that if a > 0, then (^) ∫ ∞
0
log x x^2 + a^2
dx = π 2 a
log a.
[Hint: Use the indented semicircle.] Make a branch cut along the negative imaginary axis and define the appropriate branch of log. Then log z is defined on the indented semicircle C in the upper half plane. Compute
C
log z z^2 +a^2 dz^ where^ C^ is the indented semicircle; the integrand has one simple pole ia with residue log 2 ia^ ia in C; by the residue theorem ∫ C
log z z^2 +a^2 dz^ =^
π log ia a =^
π log a a +^
iπ^2 2 a. Now^
−R
log x x^2 +a^2 dx^ =^
∫ R
log x x^2 +a^2 dx^ +^
∫ R
iπ x^2 +a^2 dx, and the latter integral (by an easy application of the residue theorem and the fact that the integrand is even or using the tan−^1 ) evaluates to iπ
2 2 a as^ R^ → ∞^ and^ ^ →^0. The integrals over both arcs to go^0 as^ R^ → ∞^ and → 0 as easy ML estimates show, and the result follows.
- Show that if |a| < 1 then ∫ (^2) π
0
log | 1 − aeiθ^ |dθ = 0.
Then prove that the above result remains true if we assume only that |a| ≤ 1.
∫^ Note the statement of for^ a^ = 0^ is obvious so assume^ a^6 = 0. Make the change of variables^ z^ =^ eiθ^ to get |z|=
log | 1 −az|dz iz. Make another change of variables^ u^ = 1^ −^ az^ to get^
|u− 1 |=|a|
log |u|du i(u−1). Consider the integral
|u− 1 |=|a|
log(u)du i(u−1) where we use the branch of^ log^ with a cut on^ (−∞,^ 0]. If^ |a|^ <^1 then B(1, |a| + ) is contained in the domain of this branch of the log. (^) ilog((u−u1)) has its only pole inside the circle at u = 1 so the integral is equal to 2 πi Resu=1 (^) ilog((u−u1)) = 2π log(1) = 0 and taking real parts we’re done.
If |a| = 1 then we need to consider the integral
C
log(u)du i(u−1) where^ C^ is the circle^ |u^ −^1 |^ = 1^ with a radius semicircle around the origin. By the same argument
C
log(u)du i(u−1) = 0^ More precisely define^ C^ by the curves u = 1 + eiθ^ θ = − cos−^1 (^2 / 2 − 1)... cos−^1 (^2 / 2 − 1) and u = eiθ
