Problem Set 4: Solutions
Physics 330
M. Seifert
Due Date: November 9, 2007
1.
(MW 310) Evaluate
I
=
∞
0
d
x
1 +
x
4
.
Viewed as a complex function, the integrand has four simple poles, at the
roots of the polynomial
z
4
+ 1 = 0; we denote these values by
z
n
=
e
(2
n

1)
iπ/
4
,
where
n
runs from 1 to 4. We choose the contour shown in Figure 1:
˜
I
=
C
d
z
1 +
z
4
This can be decomposed into three parts:
•
A line running from 0 to
R
, along which
z
=
x
(
x
∈
[0
, R
]);
•
A quartercircle running from
R
to
iR
, along which
z
=
Re
iθ
(
θ
∈
[0
,
π
2
]);
and
•
A line running from
iR
to 0, along which
z
=
ix
(
x
∈
[
R,
0]).
Splitting this up into three parts, then, we have
˜
I
=
R
0
d
x
1 +
x
4
+
π
2
0
iRe
iθ
d
θ
1 +
R
4
e
4
iθ
+
0
R
i
d
x
1 + (
ix
)
4
= (1

i
)
R
0
d
x
1 +
x
4
+
π
2
0
iRe
iθ
d
θ
1 +
R
4
e
4
iθ
In the limit where
R
→ ∞
, the second integral will scale as
R/R
4
, and so can
be neglected:
˜
I
= (1

i
)
∞
0
d
x
1 +
x
4
= (1

i
)
I
Now we simply apply the residue theorem. We have one simple pole enclosed
1
z
1
z
4
z
2
z
3
Figure 1: Integration contour for Problem 1.
in the contour, at
z
=
e
iπ/
4
; so the residue is
Res
1
1 +
z
4
z
=
e
iπ/
4
=
z

e
iπ/
4
1 +
z
4
z
=
e
iπ/
4
=
1
(
e
iπ/
4

e

iπ/
4
)(
e
iπ/
4

e
3
iπ/
4
)(
e
iπ/
4

e

3
iπ/
4
)
=
1
(2
i
)
3
(sin
π
4
)(

e
iπ/
2
sin
π
4
)(
e

iπ/
4
sin
π
2
)
=

1 +
i
4
√
2
Thus,
˜
I
= (1

i
)
I
= 2
πi

(1 +
i
)
4
√
2
I
=
√
2
π
4
2
!
ia
ia
Figure 2: Integration contour for Problem 2.
2.
(MW 313) Evaluate
I
=
d
3
x
(
a
2
+
r
2
)
3
.
Assume that
a >
0; the
a <
0 case can be obtained by substituting
a
→ 
a
in the following derivation. In spherical polar coordinates, this is simply
I
=
r
2
d
r
d
θ
d
φ
(
a
2
+
r
2
)
3
= 4
π
∞
0
r
2
d
r
(
a
2
+
r
2
)
3
= 2
π
∞
∞
r
2
d
r
(
a
2
+
r
2
)
3
where we’ve used the evenness of the integrand in the last step.
We can now use contour integration to evaluate this integral. Viewed as an
analytic function, the integrand has poles of order three at
z
=
±
ia
. We choose
the contour indicated in Figure 2. This consists of two portions, one due to the
integration along the real axis and the other due to the upper halfcircle:
C
z
2
d
z
(
a
2
+
z
2
)
3
=
R

R
r
2
d
r
(
a
2
+
r
2
)
3
+
π
0
R
2
e
2
iθ
iRe
iθ
d
θ
(
a
2
+
R
2
e
2
iθ
)
3
3
The integral along the upper halfcircle will go as
R
3
/R
6
for
R
1, and thus
in the limit
R
→ ∞
we have
C
z
2
d
z
(
a
2
+
z
2
)
3
=
∞
∞
r
2
d
r
(
a
2
+
r
2
)
3
=
I
2
π
All that remains is to calculate the residues to obtain the contour integral.
We have one pole of order three inside the contour, at
z
=
ia
; the residue there
is
Res
z
2
(
a
2
+
z
2
)
3
z
=
ia
=
1
2
d
2
d
z
2
z
2
(
z
+
ia
)
3
z
=
ia
=
1
2
2
(
z
+
ia
)
3

12
z
(
z
+
ia
)
4
+
12
z
2
(
z
+
ia
)
5
z
=
ia
=

i
16
a
3
Thus,
C
z
2
d
z
(
a
2
+
z
2
)
3
= 2
πi

i
16
a
3
I
=
π
2
4
a
3
3.
(MW 315) Evaluate
I
=
∞
0
x
d
x
1 +
x
5
.
As a complex function, the integrand has five simple poles; we will denote
them as
z
n
=
e
(2
n

1)
iπ/
5
, where
n
runs from 1 to 5.
The symmetry of the
denominator of the integrand leads us to choose the contour shown in Figure 3.
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 Physics, Methods of contour integration, cot πz dz