# PHY7110. Solutions to HW 4

## PHY7110. Solutions to HW 4

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.

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