Question

the previous hw question and answer

1. Consider the integral from question 2 of the previous homework assignment: too sin ma dx, and assume that both m and a are positive real numbers. By using an indented contour, evaluate this integral fully. You are allowed to resubmit material submitted as part of the previous assignment if you wish.]

Answer 1

sin(ms) 1. ·s da x (x²+0²) solution Consider s feno) dry - Imaginary part of immy с e dy Where C is the semi- Cuide diameter c m&3²7a²) upper half of the A with the bounding ER, R] . s flgs day S The poles of fig) the solutions of " ( § ta2)=0 => y = .g=Iia y = 0 is a simple pola lies insiden. g=ia is a simple pole lies insider Res [ fig),o] lim 3 (3-0) fig) immy = lim g no m (% +0² Lim immy L mo geta² a² e >

Res [fig), ia] - Lim my-ia) immy my (3² +0²) z-siq - Lim iam gasia (m-ia) 2 (aria) (%-ia) img = lim ysia 3(Bria) -ma -ma é (ia) (zid) -242- theorem , Hence by Cauchy's Cauchy's residue f(g.) dry =aai [sum of the residues tai [sum of residues on the real arus] -ma Here s fly dry =25i ] **i (] aa² ma - Ti e Ti + a² a² - TT -md [-e +1] i ) fox) dx + 5 figldg TL (i-ema) -R a²

If Roto then of lip celf 2 6 58 T ima xoxza²) Cinemas az Equating the imaginary parts pu both sides, sicles we get sin (ne) dae = ses & T a² [l-ema] x(x² + a²)