Math 311 - Spring 2014
Solutions to Assignment # 12
Completion Date: Thursday June 12, 2014
Question 1. [p 267, #2]
Use residues to evaluate the improper integral
Z
∞
0
dx
.
(x2 + 1)2
Ans: π/4.
Solution: Let f (z) =
1
, and for R > 1, consider the integral of f over the contour CR shown
(1 + z 2 )2
below.
y
CR
i
0
−R
We have
that is,
Now,
Z
CR
dz
+
(1 + z 2 )2
Z
CR
Z
0
−R
dx
+
(1 + x2 )2
dz
+2
(1 + z 2 )2
Z
R
0
Z
R
0
R
x
dx
= 2πi Res
z=i
(1 + x2 )2
dx
= 2πi Res
z=i
(1 + x2 )2
1
(1 + z 2 )2
1
(1 + z 2 )2
,
.
1
Φ(z)
=
(1 + z 2 )2
(z − i)2
1
1
is analytic at z = i and Φ(i) = − 6= 0, so that f has a pole of order m = 2 at z = i,
where Φ(z) =
2
(z + i)
4
with
2
i
−2
=−
Res(f (z)) = Φ0 (i) =
=− .
z=i
(z + i)3 z=i
(2i)3
4
f (z) =
Therefore,
Z
CR
dz
+2
(1 + z 2 )2
Z
R
0
dx
i
π
=
2πi
−
=
(1 + x2 )2
4
2
However, on CR , we have z = Reiθ , and
|1 + z 2 |2 ≥ |z|2 − 1
if R > 1, so that
as R → ∞.
Z
CR
2
≥ (R2 − 1)2
2πR
dz
→0
≤
(1 + z 2 )2
(R2 − 1)2
(∗)
Letting R → ∞ in (∗), we have
∞
Z
2
0
that is,
Z
∞
0
π
dx
= ,
(1 + x2 )2
2
π
dx
= .
(1 + x2 )2
4
Question 2. [p 267, #5]
Use residues to evaluate the improper integral
Z ∞
0
(x2
x2 dx
.
+ 9)(x2 + 4)2
Ans: π/200.
Solution: Let f (z) =
(z 2
z2
, and for R > 3, consider the integral of f over the contour CR
+ 9)(z 2 + 4)2
shown below.
y
CR
3i
2i
0
−R
R
x
We have
Z
CR
z2
dz + 2
2
(z + 9)(z 2 + 4)2
Z
R
(x2
0
h
i
x2
dx = 2πi Res (f (z)) + Res (f (z)) .
2
2
z=2i
z=3i
+ 9)(x + 4)
Now, f (z) has a simple pole at z = 3i with
Res (f (z)) =
z=3i
(3i)2
3
=−
,
(3i + 3i)((3i)2 + 4)2
50i
and f (z) has a pole of order 2 at z = 2i with
Res (f (z)) =
z=2i
d
z2
dz (z 2 + 9)(z + 2i)
=
z=2i
13
.
200i
Therefore,
Z
CR
z2
dz + 2
(z 2 + 9)(z 2 + 4)2
Z
R
0
2π
x2
12
13
=
dx = 2πi −
+
,
2
2
2
(x + 9)(x + 4)
200i 200i
200
that is,
Z
CR
z2
dz + 2
(z 2 + 9)(z 2 + 4)2
On CR , we have
f (z) ≤
so that
as R → ∞.
Z
CR
(z 2
Z
0
R
x2
2π
.
dx =
2
2
2
(x + 9)(x + 4)
200
R2
,
(R2 − 9)(R2 − 4)2
z2
πR3
dz ≤
→0
2
2
2
+ 9)(z + 4)
(R − 9)(R2 − 4)2
(∗∗)
Letting R → ∞ in (∗∗), we get
2
Z
∞
0
that is,
Z
x2
2π
dx =
,
(x2 + 9)(x2 + 4)2
200
∞
0
(x2
π
x2
dx =
.
+ 9)(x2 + 4)2
200
Question 3. [p 267, #8]
Use residues and the contour shown in Fig. 95, where R > 1, to establish the integration formula
Z ∞
dx
2π
= √ .
3
x +1
3 3
0
y
R exp ( i 2 π / 3)
CR
R x
0
Solution: Let f (z) =
above.
1
, and consider the integral of f around the contour CR with R > 1, as shown
z3 + 1
Since f is analytic inside and on this contour except for a simple pole at z = eiπ/3 , then
Z R
Z 2π/3
Z 0 2πi/3
Z
dx
iReiθ dθ
e
dr
1
dz
,
=
+
+
= 2πi Res
3
3
3 2πi + 1
R3 e3iθ + 1
z3 + 1
z=eπi/3
0 x +1
0
R r e
CR z + 1
that is,
and therefore
Z
R
0
Z
2π/3
0
2π/3
R
iReiθ dθ
−
R3 e3iθ + 1
Z
iReiθ dθ
2πi/3
+
1
−
e
R3 e3iθ + 1
Z
dx
+
3
x +1
Z
0
0
R
0
e2πi/3 dr
= 2πi Res
r3 + 1
z=eπi/3
dx
= 2πi Res
x3 + 1
z=eπi/3
1
3
z +1
1
z3 + 1
,
.
Now, we have
Res
z=eπi/3
so that
Z
2π/3
0
1
z3 + 1
=
1
3z 2
iReiθ dθ
2πi/3
+
1
−
e
R3 e3iθ + 1
=
z=eπi/3
Z
R
0
e−2πi/3
,
3
e−2πi/3
dx
= 2πi ·
.
+1
3
x3
On the circular arc z = Reiθ , 0 ≤ θ ≤ 2π/3, and we have
Z 2πi/3
2πR
R
iReiθ dθ
≤
· 3
→0
3 e3iθ + 1
R
3
R
−1
0
as R → ∞, so that from (∗ ∗ ∗) we get
Z
1 − e2πi/3
that is,
Z
∞
0
∞
0
dx
e−2πi/3
=
2πi
·
,
x3 + 1
3
√
dx
2πi · e−2πi/3
2πi(−1 − 3 i)
2π
=
√
=
= √ .
x3 + 1
3 1 − e2πi/3
3(3 − 3 i)
3 3
(∗ ∗ ∗)
Question 4. [p 276, #5]
Use residues to evaluate the improper integral
Z ∞
x sin ax
dx
4
−∞ x + 4
Ans:
(a > 0).
π −a
e sin a.
2
zeiaz
Solution: Let f (z) = 4
, and consider the integral of f around the contour shown below, where
z +4
√
R > 2.
y
CR
0
−R
Now, f is analytic inside and on the contour except at
√
and
z1 = 2eiπ/4 = 1 + i
R
z2 =
x
√ 3iπ/4
2e
= −1 + i
and f has simple poles at these points with
Res (f (z)) =
z1 eiaz1
eia(1+i)
1
=
= e−a · eia ,
3
4z1
4(1 + i)2
8i
Res (f (z)) =
eia(−1+i)
1
z2 eiaz2
=
= − e−a · e−ia .
3
4z2
4(−1 + i)2
8i
z=z1
and
z=z2
Therefore,
Z
R
−R
xeiax
dx +
x4 + 4
Now, on CR we have
eiaRe
Z
CR
iθ
−a
zeiaz
πi
e
ia
−ia
= e−a sin a.
dz
=
2πi
e
−
e
z4 + 4
8i
2
= eiaR(cos θ+i sin θ) = e−aR sin θ · eiaR cos θ ,
so that
Re−aR sin θ
zeiaz
≤
4
z +4
R4 − 4
on CR , and from Jordan’s inequality, we have
Z
CR
zeiaz
2R2
dz ≤ 4
4
z +4
R −4
Z
π/2
e−aR sin θ dθ <
0
as R → ∞. Letting R → ∞ in (∗ ∗ ∗∗), we have
Z ∞
xeiax
πi
dx = e−a sin a,
4+4
x
2
−∞
and therefore,
Z
∞
−∞
π
x sin ax
dx = e−a sin a.
4
x +4
2
πR
→0
R4 − 4
(∗ ∗ ∗∗)
Question 5. [p 276, #9]
Use residues to find the Cauchy principal value of the improper integral
Z ∞
sin x dx
.
2
−∞ x + 4x + 5
Ans: −
π
sin 2.
e
Solution: We write
f (z) =
z2
1
1
=
,
+ 4z + 5
(z − z1 )(z − z1 )
where z1 = −2 + i and note that z1 is a simple pole of f (z) · eiz which lies above the real axis, with residue
B1 =
Therefore, when R >
√
1
1 e−2i
eiz1
= e−2i−1 = ·
.
z1 − z 1
2i
e 2i
5, and CR denotes the upper half of the positively oriented circle |z| = R, we have
Z
R
−R
eix
dx +
x2 + 4x + 4
Z
f (z)eiz dz = 2πiB1 ,
CR
and therefore
Z
R
−R
sin x
dx = Im(2πiB1 ) − Im
x2 + 4x + 4
Z
f (z)e−z dz.
CR
(∗ ∗ ∗ ∗ ∗)
Now, on CR , we have z = Reiθ , 0 ≤ θ ≤ π, so that z − z1 = Reiθ + 2 − i, and from the back end of the
triangle inequality, we have
√
√
|z − z1 | ≥ R − 5
and
|z − z1 | ≥ R − 5
√
for R > 5, and therefore
|f (z)| =
on CR , and MR =
1
1
1
√ ,
≤
≤
(z − z1 )(z − z1 )
|z − z1 |2
(R − 5)2
1
√
→ 0 as R → ∞.
(R − 5)2
From Jordan’s Lemma, we have
Im
Z
f (z)e
−z
CR
dz ≤
Z
CR
f (z)e−z dz → 0
as R → ∞. Letting R → ∞ in (∗ ∗ ∗ ∗ ∗) we have
P.V.
Z
∞
−∞
sin x
dx = lim
2
R→∞
x + 4x + 5
Z
R
−R
x2
sin x
π
dx = Im(2πiB1 ) = − sin 2.
+ 4x + 4
e
Question 6. [p 286, #2]
Evaluate the improper integral
Z ∞
xa
dx,
2
(x + 1)2
0
Ans:
where
− 1 < a < 3 and xa = exp(a ln x).
(1 − a)π
.
4 cos(aπ/2)
za
, and note that if a is not an integer, then this function is multiple valued,
(z 2 + 1)2
and we will choose the branch given by
z a = ea log z ,
Solution: Let f (z) =
where log z = ln r + iθ, where r > 0, and 0 ≤ θ < 2π.
In order to apply the residue theorem, the contour of integration can only enclose isolated singular points of
f, and so it cannot enclose the branch cut {z ∈ C : z = Re z ≥ 0}, so for 0 < < 1 < R, we integrate over
the contour below.
y
CR ε
,
i
Cε
U
R ,ε
0
R
x
L
R, ε
−i
Since 0 < < 1 < R, then the isolated singular points z = ±i are inside the contour, but the branch cut is
outside the contour, and applying the residue theorem we have
I
za
dz
=
2πi
Res
(f
(z))
+
Res
(f
(z))
(†)
2
2
z=i
z=−i
C (z + 1)
where the contour C = CR, + LR, + C + UR, is traversed in the counterclockwise direction.
Now f (z) =
za
has a pole of order m = 2 at z = ±i, and
(z 2 + 1)2
a log z d
d
e
za
2
=
Res(f (z)) =
(z − i) 2
2
z=i
dz
(z + 1)
dz (z + i)2
z=i
and
Res (f (z)) =
z=−i
d
za
(z + i)2 2
dz
(z + 1)2
=
z=−i
a log z d
e
dz (z − i)2
=
z=i
z=−i
i(a − 1) iπa/2
e
4
=−
i(a − 1) i3πa/2
e
,
4
and from (†) we have
I
π
za
i(a − 1) iπa/2
i3πa/2
iπa/2
i3πa/2
e
−
e
dz
=
2πi
=
(1
−
a)
e
−
e
.
2
2
4
2
C (z + 1)
Now, on C , we have
as → 0 if and only if a > −1.
Z
C
π1+a
za
dz
≤
→0
(z 2 + 1)2
(1 − 2 )2
On CR, , we have
Z
(z 2
CR,
2πRa+1
2πRa−3
za
dz ≤
=
→0
2
2
2
+ 1)
(R − 1)
(1 − 1/R2 )2
as R → ∞ if and only if a < 3.
Letting R → ∞ and → 0 in (†), we have
"Z
Z 0 a(ln x+i2π) #
R a(ln x+i0)
e
e
π
iπa/2
i3πa/2
,
lim
dx
+
dx
=
(1
−
a)
e
−
e
2
2
R→∞
(x2 + 1)2
2
0
R (x + 1)
so that
1 − e2πia
and since e
2πia
6= 1, then
Z ∞
0
(x2
Z
∞
0
xa
π
iπa/2
i3πa/2
,
−
e
dx
=
(1
−
a)
e
(x2 + 1)2
2
xa
π
eiπa/2 − ei3πa/2
π
e−iπa/2 − eiπa/2
dx = (1 − a)
= (1 − a) −iπa
,
2
2πia
+ 1)
2
1−e
2
e
− eiπa
that is,
Z
∞
0
(x2
π
sin(πa/2)
(1 − a)π
xa
dx = (1 − a)
=
+ 1)2
2
sin(πa)
4 cos(πa/2)
from the double angle formula.
Question 7. [p 290, #2]
Use residues to evaluate the definite integral
Z
Ans:
√
π
−π
dθ
.
1 + sin2 θ
2 π.
Solution: On the circle z = eiθ , 0 ≤ θ ≤ 2π, we have
dz = ieiθ dθ = iz dθ,
and
z + z −1
dz
z − z −1
,
cos θ =
,
dθ =
2i
2
iz
so we can rewrite the integral as a contour integral
Z π
I
dθ
dz
1
=
2 ! iz
2
−1
−π 1 + sin θ
|z|=1
z−z
1+
2i
sin θ =
that is,
Z
π
−π
1
=
i
I
=−
4
i
|z|=1
I
1−
|z|=1
4
dθ
=−
2
i
1 + sin θ
1 2
4 (z
1
dz
− 2 + 1/z 2) · z
z dz
,
z 4 − 6z 2 + 1
I
|z|=1
z4
z dz
.
− 6z 2 + 1
√
Note that z 4 − 6z 2 + 1 = 0 if and only if (z 2 − 3)2 = 8, that is, if and only if z 2 = 3 ± 2 2, so that the
integrand has four simple poles
q
q
q
q
√
√
√
√
z1 = 3 + 2 2,
z2 = − 3 + 2 2,
z3 = 3 − 2 2,
z4 = − 3 − 2 2,
but |z1 | > 1 and |z2 | > 1, while |z3 | < 1 and |z4 | < 1, and only z3 and z4 lie inside the contour C : |z| = 1.
The residues of the integrand at these poles are
z3
1
√ =− √
z32 − (3 + 2 2) (2z3 )
8 2
Res (f (z)) =
z=z3
and
Res (f (z)) =
z42
z=z4
so that
Z
π
−π
z4
1
√ =− √ ,
− (3 + 2 2) (2z4 )
8 2
√
4
1
1
1
dθ
= 2πi −
− √ − √
= 8π · √ = 2 π.
2
i
1 + sin θ
8 2 8 2
4 2
Question 8. [p 291, #5]
Use residues to evaluate the definite integral
Z π
cos 2θ dθ
1
−
2a cos θ + a2
0
Ans:
(−1 < a < 1).
a2 π
.
1 − a2
Solution: Since
Z
π
0
cos 2θ dθ = 0, we may assume that 0 < |a| < 1, and consider the real part of
Z
π
−π
e2iθ
dθ.
1 − 2a cos θ + a2
We convert this to a contour integral around the circle |z| = 1 using dz = ie iθ dθ = iz dθ, and
sin θ =
z − z −1
,
2i
z + z −1
,
2
cos θ =
dθ =
dz
,
iz
to get
Z
π
−π
e2iθ
dθ =
1 − 2a cos θ + a2
I
=
1
i
|z|=1
I
1
=−
i
=−
that is,
Z
π
−π
1 − 2a
|z|=1
I
1
ai
z2
dz
(z − a)(az − 1)
|z|=1
1
e2iθ
dθ = −
1 − 2a cos θ + a2
ai
I
z2
dz
−1
iz
z+z
+ a2
2
z2
dz
[−az 2 + (1 + a2 )z − a]
|z|=1
I
z2
dz,
(z − a)(z − 1/a)
|z|=1
z2
dz.
(z − a)(z − 1/a)
1
The integrand has a simple pole at z1 = a and z2 = , and since |a| < 1, only z1 is inside the contour |z| = 1.
a
The residue at z1 is
2
a2
a3
z
=
=−
,
Res
z=z1
(z − a)(z − 1/a)
a − 1/a
1 − a2
so that
Z
π
−π
1
2πa2
a3
e2iθ
dθ
=
2πi
−
,
=
−
1 − 2a cos θ + a2
ai
1 − a2
1 − a2
and equating real and imaginary parts, we have
Z π
cos 2θ
2πa2
dθ
=
,
2
1 − a2
−π 1 − 2a cos θ + a
and therefore
Z
π
0
πa2
cos 2θ
dθ
=
.
1 − 2a cos θ + a2
1 − a2
Question 9. [p 291, #6]
Use residues to evaluate the definite integral
Z π
0
Ans:
√
dθ
(a + cos θ)2
(a > 1).
aπ
a2
3 .
−1
Solution: Since
Z
π
−π
dθ
=2
(a + cos θ)2
Z
0
π
dθ
,
(a + cos θ)2
we convert this to a contour integral around the circle |z| = 1 using dz = ieiθ dθ = iz dθ, and
z − z −1
,
2i
sin θ =
to get
Z
π
dθ
=
(a + cos θ)2
I
cos θ =
z + z −1
,
2
1
2
dz
4
=
iz
i
dθ =
I
(z 2
dz
,
iz
z
dz.
+ 2az + 1)2
z + 1/z
|z|=1
2
√
√
The integrand has a pole of order 2 at z1 = −a√+ a2 − 1 and a pole of order 2 at z2 = −a − a2 − 1,
however |z2 | > 1 since a > 1, while |z1 | = a − a2 − 1 < 1, and z1 is the only singular point inside the
contour C : |z| = 1.
−π
|z|=1
a+
To find the residue at z = z1 , we write
f (z) =
z
z
√
√
=
,
2
2
(z 2 + 2az + 1)2
(z + a − a − 1) (z + a + a2 − 1)2
so that
(z − z1 )2 f (z) =
and
(z + a +
z
√
a2 − 1)2
,
√
−z + a + a2 − 1
d 2
√
(z − z1 ) f (z) =
.
dz
(z + a + a2 − 1)3
Therefore,
Res f (z) = lim
z=z1
z→z1
d 2a
a
(z − z1 )2 f (z) = √
√
3 =
3 ,
dz
2
2
2 a −1
4 a −1
and
I
that is,
Z
and finally,
π
−π
|z|=1
(z 2
2πia
z
dz =
√
3 ,
2
+ 2az + 1)
4 a2 − 1
4
2πia
2πa
dθ
= · √
= √
3 ,
3
2
(a + cos θ)
i 4 a2 − 1
a2 − 1
Z
π
0
πa
dθ
= √
3 .
2
(a + cos θ)
a2 − 1