Roger Griffith
Math 121A
hw # 10
Evaluation of Defenite Integrals By Use of Residue Theorem
When z0 has a simple pole than we can evaluate the residue by
R(z0) = lim (z − z0 ) f (z)
z→z0
For problems that have multiple poles we can find the residue by using
d m−1
1
f (z)
z→z0 (m − 1)! dzm−1
res(z0) = lim
where m is the order of the pole
Problem #1
Evaluate the integral
Z ∞
sin(x)
x
−∞
dx
(Note: You can look at the book on page 690)
we can write this as
Z
the residue at z0 = 0 is
!
"
sin(x)
eiz
1
residue of
dx = 2πi ·
at z0 = 0
x
2
z
Res(0) = lim z
z→0
eiz
= lim eiz = 1
z→0
z
Z ∞
sin(x)
−∞
x
dx = πi
taking the imaginary part of both sides we get
Z ∞
sin(x)
−∞
x
dx = π
Problem #2
Page 699 of M.L. Boas book. Problems 1,3,6,10,11,12,13,15,17,19,20
(1).
Z 2π
0
dθ
13 + 5 sin θ
If we let
z = eiθ
dz = ieiθ dθ = izdθ → dθ =
1
dz
iz
and also
1
eiθ − e−iθ z − z
z2 − 1
=
=
2i
2i
2zi
thus we substitute into the integral to find
sin θ =
dz
$ =2
#
2
C z(13 + 5 z2 −1
2zi
I
I
1
dz = 2
2
5z + 26iz − 5
poles
z=
%
I
1
dz = 2 · 2πi · res(−i/5)
(5i + z)(i + 5z)
−5i
−i/5
we do not consider the −5i pole because it is outside the contour integral. The residue is given by
Res(−i/5) =
lim z + i/5
z→−i/5
1
(5i + z)(i + 5z)
i + 5z = 5(z + i/5)
thus we find
=
1
1
=
24i
z→−i/5 5(5i + z)
Z 2π
π
dθ
=
13 + 5 sin θ 6
0
(3).
lim
Z 2π
0
If we let
z = eiθ
dθ
5 − 4 sin θ
dz = ieiθ dθ = izdθ → dθ =
dz
iz
and also
1
eiθ − e−iθ z − z
z2 − 1
sin θ =
=
=
2i
2i
2zi
thus we substitute into the integral to find
dz
$=
#
C z(5 − 4 z2 −1
2zi
I
I
1
dz =
2
2z + 5iz − 2
poles
z=
I
%
1
dz = 2πi · res(−i/2)
(2i + z)(i + 2z)
−5i
−i/5
we do not consider the −2i pole because it is outside the contour integral. The residue is given by
Res(−i/2) =
lim z + i/2
z→−i/2
1
(2i + z)(i + 2z)
i + 2z = 2(z + i/2)
Res(−i/2) =
1
1
=
3i
z→−i/2 2(2i + z)
lim
2
thus we find
Z 2π
2π
dθ
=
5 − 4 sin θ
3
0
(6).
Z π
0
dθ
(2 + cos θ)2
Since this function is even, it can be written as
1
2
Z 2π
0
dθ
(2 + cos θ)2
If we let
z = eiθ
dz = ieiθ dθ = izdθ → dθ =
dz
iz
and also
1
z2 + 1
eiθ + e−iθ z + z
=
=
cos θ =
2
2
2z
thus we substitute into the integral to find
1
2i
so we find
I
&
C
(2 +
zdz
1
#
$'2 =
2i
z2 +1
z
2
&
' ] ] dz =
2
i
2z + z +1
I
2 2
2
2z
I
z
dz
(1 + 4z + z2 )2
√
z
√
√
= 2πi · res(−2 + 3)
C (z + (2 + 3)2 (z + (2 − 3)2
%
√
−(2 + 3)
√
poles z =
−(2 − 3)
√
we do not consider the −(2 + 3) pole because it is outside the contour integral. To find the residue
we can use we know that
2z
√
f (z) =
i(z + (2 + 3)2
√
√
2z
1
3
√
√
Res(−2 + 3) = lim √ −
+
=
3
2
9i
(2 + 3 + z)
z→−2+ 3 (2 + 3 + z)
I
thus we find that the integral is
Z π
0
(10).
√
2 3π
dθ
=
(2 + cos θ)2
9
Z ∞
−∞
dx
x2 + 4x + 5
From M.Boas, we know that solving this integral is equivalent to solving
I
C
1
z2 + 4z + 5
dz = 2πi · res(z0)
3
which becomes
1
dz
C (z + (2 + i))(z + (2 − i))
%
−(2 + i)
poles z =
−(2 − i)
I
we do not consider the −2 − i pole because it is outside the contour integral. To find the residue we
can use we know that.
so we know that the residue is given as
res(−2 + i) = lim z − (−2 + i)
z→−2+i
thus we find
1
1
1
= lim
=
(z + (2 + i))(z + (2 − i)) z→−2+i (z + (2 + i)) 2i
Z ∞
dx
−∞
(11).
x2 + 4x + 5
Z ∞
= π
dx
(4x2 + 1)3
0
as before we can solve this integral by solving
1
2
I
1
C
which becomes
(4z2 + 1)3
1
2
I
C
dz =
1
· 2πi · res(z0)
2
1
(2z + i)3 (2z − i)3
poles
z=
%
dz
i
2
− 2i
we do not consider the − 2i pole because it is outside the contour integral. To find the residue we can
use we know that
1
f (z) =
2(2z + i)3
24
3
=
2i
z→i/2 (1 + 2z)5
res(i/2) = lim
thus
Z ∞
0
(12).
dx
2
(4x
+ 1)3
Z ∞ 2
x dx
0
1
=
4
x + 16 2
thus
1
2
=
3π
2
x2
dx
4
−∞ x + 16
Z ∞
1
x2
dx = · 2πi res(z0)
4
2
−∞ x + 16
Z ∞
4
with a little work and effort we find
1
2
this becomes
1
z2
dz
=
4
2
C z + 16
Z
z2
dz
2
2
C (z + 4i)(z − 4i)
Z
z2
√
√
√
√ dz
C (z + 2i i)(z − 2i i)(z + 2 i)(z − 2 i)
√

−2i i



−2√i
√
poles z =

2i
i


 √
2 i
√ √
we will only consider 2i i, 2 i. So we find the residues to be given by
1
2
Z
z2
√
√
√
√
(z + 2i i)(z − 2i i)(z + 2 i)(z − 2 i)
z→2i i
!
"
i+1
z2
√
√
√ =− √
= lim√
8 2
z→2i i (z + 2i i)(z + 2 i)(z − 2 i)
2
√
√
z
√
√
√
√
res(2 i) = lim√ z − 2 i
(z + 2i i)(z − 2i i)(z + 2 i)(z − 2 i)
z→2 i
"
!
z2
1−i
√
√
√ =
√
= lim√
8 2
z→2 i (z + 2i i)(z − 2i i)(z + 2 i)
√
res(2i i) =
so we find
1
2
√
lim√ z − 2i i
√
√
π
x2
1
i)
+
res(2
i)) = √
dx
=
·
2πi(res(2i
4
2
−∞ x + 16
4 2
Z ∞
(13).
Z ∞
0
x2 dx
1
=
2
2
2
(x + 4)(x + 9)
x2 dx
2
2
−∞ (x + 4)(x + 9)
Z ∞
this can also be written as
1
2
z2
dz
C (z + 2i)(z − 2i)(z + 3i)(z − 3i)
Z
poles

−2i



+2i
z=

−3i



+3i
and we have poles at −2i, 2i, −3i, 3i, we will consider 2i, 3i. so this becomes
1
2
x2 dx
= πi(res(2i) + res(3i))
2
2
−∞ (x + 4)(x + 9)
Z ∞
5
and we know that
z2
z→2i
(z + 2i)(z − 2i)(z + 3i)(z − 3i)
z2
1
= lim
=−
z→2i (z + 2i)(z + 3i)(z − 3i)
5i
z2
res(3i) = lim z − 3i
z→3i
(z + 2i)(z − 2i)(z + 3i)(z − 3i)
z2
9
= lim
=
z→3i (z + 2i)(z − 2i)(z + 3i)
30i
res(2i) = lim z − 2i
so we find
(15).
1
2
π
x2 dx
= πi(res(2i) + res(3i)) =
2
2
10
−∞ (x + 4)(x + 9)
Z ∞
Z ∞
cos 2xdx
1 ∞ cos 2xdx 1
=
= · 2πi · res(z0)
2
2 −∞ 9x2 + 4
2
0 9x + 4
looking at the real part this can be written as
1
2
Z
e2iz
dz
C (3z + 2i)(3z − 2i)
Z
we have two poles at − 2i3 , 2i3 we will only consider the
2i
3
, so we find
e2iz
Res(2i/3) = lim z − 2i/3
9(z + 2i/3)(z − 2i/3)
z→2i/3
e2iz
e−4/3
=
8i
z→2i/3 9(z + 2i/3)
=
thus we find
lim
Z ∞
cos 2xdx
0
(17).
This can be written as
which can be simplyfied to
2
9x + 4
Z ∞
−∞
=
π
12e4/3
x sin xdx
x2 + 4x + 5
Z ∞
x sin xdx
= Im
2
−∞ x + 4x + 5
,Z
zeiz
dz
z2 + 4z + 5
zeiz
dz
C (z + (2 + i))(z + (2 − i))
Z
6
-
we find two poles at −2 + i, −2 − i, we will only consider −2 + i, so the residue is given as
Res(−2 + i) =
lim z + 2 − i
z→−2+i
zeiz
z + (2 + i))(z + (2 − i))
zeiz
z→−2+i z + 2 + i
(i − 2)ei(i−2) (cos 2 − i sin 2)(−2 + i) −2 cos 2 + 2i sin 2 + i cos 2 + sin 2
Res(−2 + i) =
=
=
2i
2ie
2ie
taking only the imaginary part we find
cos 2 + 2 sin 2
Res(−2 + i) =
2e
thus we find
Z ∞
x sin xdx
π
= (cos 2 + 2 sin 2)
2
e
−∞ x + 4x + 5
(19).
Z ∞
cos 2xdx
2
2
−∞ (4x + 9)
we can take the real part of this which becomes
=
lim
e2iz
dz
2
2
−∞ 16(z + 3i/2) (z − 3i/2)
Z ∞
we find two poles at −3i/2, 3i/2 we will only condider 3i/2, so the residual is
Res(3i/2) = lim −
z→3i/2
thus we find
e2iz
i
=−
3i 3
27e3
(z + 2 )
Z ∞
cos 2xdx
−∞
(20).
(4x2 + 9)2
=
π
27e3
Z ∞
cos xdx
2 2
−∞ (1 + 9x )
we can look at the real part and we find
eiz
1
dz
=
2
81
−∞ [(3z + i)(3z − i)]
Z ∞
eiz
dz
i
i
−∞ (z + 3 )2 (z − 3 )2
Z ∞
we find two poles at −i/3, i/3 we will only use the i/3, so the residual is given as
Res(i/3) = lim −
z→i/3
i
2eiz
= − 1/3
3
81(z + i/3)
9e
so the results is given by
Z ∞
2π
cos xdx
=
2
2
9e1/3
−∞ (1 + 9x )
7