Since I’m away this week, the solutions would be brief. Please e-mail me if you have a question and would like a more detailed answer.
Suggested reading:
G. Arfken and H. Weber, Mathematical Methods , Chapter 7.
Problem 1: Integrals (A&W, 7.1.8-9)
(a) Show that
Z
π
0 dθ
( a + cos θ ) 2
=
πa
( a 2 − 1) 3 / 2
, a > 1 .
(1)
Solution: this problem can be done with a direct application of residue theorem with a variable substitution z = exp( iθ ). Then dθ = ( − i/z ) dz and cos θ = ( z 2 + 1) / (2 z ). The contour integral is over the unit circle in the complex z − plane. Note that the integrand is even, so se can calculate the same integral from 0 to 2 π and divide it by two.
Then, the contour integral is
I
C
= − 2 i
I zdz
C
( z 2 + 2 az + 1) 2
= − 2 i
I
C zdz
( z − z
1
) 2 ( z − z
2
) 2
= 2 πi
X
Res [ f ( z i
)] , i where z
1 , 2
= − a ± p
( a + 1)( a − 1) .
and f ( z ) in the integrand function. Note that | z
2
| = − a − p
( a + 1)( a − 1) >
1 for a > 1, so it lies outside of the integration contour (unit circle). Thus, we only need to compute a residue of the order-two pole at z = z
1
. Calculating it gives the result that we were supposed to prove.
(b) Show that
Z
2 π
0 dθ
1 − 2 a cos θ + a 2
=
2 π
1 − a 2
, | a | < 1 .
1
(2)

What happens if | a | > 1? What happens if | a | = 1?
Solution: do the substitutions 1 − a 2 to reduce the integral to the integral
= α and 2 a = β in the denominator
I =
Z
2 π
0 dθ a + b cos θ done in class. The result follows. If | a | > 1 the integral equals to 2 π/ ( α 2 − 1).
For a = − 1 there are singularities for θ = 0 and 2 π . For a = 1 there is singularity at θ = π . In both cases the integral does not exist.
Problem 2: More integrals (A&W, 7.1.14)
(a) Show that ( a > 0)
Z
−∞ cos xdx
∞ x 2 + a 2
=
π e
− a
.
a
How is the right side modified if cos x is replaced with cos kx ?
(b) Show that ( a > 0)
Z
−∞ x sin xdx
∞ x 2 + a 2
= πe
− a
.
How is the right side modified if sin x is replaced with sin kx
(3)
(4)
Solution: problem 2 was done in class.
2