Author Topic: 2.6 #4  (Read 269 times)

celina q

  • Newbie
  • *
  • Posts: 1
  • Karma: 0
    • View Profile
2.6 #4
« on: December 06, 2018, 11:52:48 PM »
Can someone show me the process for this question?

Kris

  • Newbie
  • *
  • Posts: 3
  • Karma: 1
    • View Profile
Re: 2.6 #4
« Reply #1 on: December 07, 2018, 12:13:10 AM »
We consider the function $f(z)=\frac{e^{iaz}}{(z^2+1)(z^2+4)}$.
$(z^2+1)(z^2+4)=0\Rightarrow$$z=\pm i,$$\pm 2i.$ Only i, 2i are in the upper-half plane.
$Res(f,i)=\frac{e^{iai}}{2i(-1+4)}=\frac{e^{-a}}{6i}$    $Res(f,2i)=\frac{e^{ia2i}}{(-4+1)4i}=\frac{e^{-2a}}{-12i}$. $I=2\pi i(\frac{e^{-a}}{6i}-\frac{e^{-2a}}{12i})=\pi(\frac{e^{-a}}{3}-\frac{e^{-2a}}{6})$