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.

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

πa dθ = 2 , 2 (a + cos θ) (a − 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)/(2z). 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 2π and divide it by two. Then, the contour integral is I I X zdz zdz = −2i = 2πi Res[f (zi )], IC = −2i 2 2 2 2 C (z − z1 ) (z − z2 ) C (z + 2az + 1) i where z1,2 = −a ±

p (a + 1)(a − 1).

p and f (z) in the integrand function. Note that |z2 | = −a− (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 = z1 . Calculating it gives the result that we were supposed to prove. (b) Show that Z 0

2π

dθ 2π = , 1 − 2a cos θ + a2 1 − a2 1

|a| < 1.

(2)

What happens if |a| > 1? What happens if |a| = 1? Solution: do the substitutions 1 − a2 = α and 2a = β in the denominator to reduce the integral to the integral Z 2π dθ I= a + b cos θ 0 done in class. The result follows. If |a| > 1 the integral equals to 2π/(α2 −1). For a = −1 there are singularities for θ = 0 and 2π. 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

−∞

∞

π cos xdx = e−a . 2 2 x +a a

(3)

How is the right side modified if cos x is replaced with cos kx? (b) Show that (a > 0) Z

−∞

∞

x sin xdx = πe−a . 2 2 x +a

How is the right side modified if sin x is replaced with sin kx Solution: problem 2 was done in class.

2

(4)

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

πa dθ = 2 , 2 (a + cos θ) (a − 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)/(2z). 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 2π and divide it by two. Then, the contour integral is I I X zdz zdz = −2i = 2πi Res[f (zi )], IC = −2i 2 2 2 2 C (z − z1 ) (z − z2 ) C (z + 2az + 1) i where z1,2 = −a ±

p (a + 1)(a − 1).

p and f (z) in the integrand function. Note that |z2 | = −a− (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 = z1 . Calculating it gives the result that we were supposed to prove. (b) Show that Z 0

2π

dθ 2π = , 1 − 2a cos θ + a2 1 − a2 1

|a| < 1.

(2)

What happens if |a| > 1? What happens if |a| = 1? Solution: do the substitutions 1 − a2 = α and 2a = β in the denominator to reduce the integral to the integral Z 2π dθ I= a + b cos θ 0 done in class. The result follows. If |a| > 1 the integral equals to 2π/(α2 −1). For a = −1 there are singularities for θ = 0 and 2π. 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

−∞

∞

π cos xdx = e−a . 2 2 x +a a

(3)

How is the right side modified if cos x is replaced with cos kx? (b) Show that (a > 0) Z

−∞

∞

x sin xdx = πe−a . 2 2 x +a

How is the right side modified if sin x is replaced with sin kx Solution: problem 2 was done in class.

2

(4)