- #1
Microzero
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Complex Variable---definite integral
Show that
f(k) = [tex]\frac{1}{2i\pi}[/tex] [tex]\int^{\infty}_{-\infty}[/tex] [tex]\frac{e^{ikx}}{x-i\epsilon}dx[/tex] =
1, if k>0
0, if k<0
where [tex]\epsilon[/tex] > 0
The attempt at a solution
Consider : [tex]\oint[/tex] [tex]\frac{e^{ikz}}{z-i\epsilon}dz[/tex]
the residule is : e^(-k[tex]\epsilon[/tex])
so [tex]\int^{\infty}_{-\infty}[/tex] [tex]\frac{e^{ikx}}{x-i\epsilon}dx[/tex]
= [tex]{2i\pi}[/tex] e^(-k[tex]\epsilon[/tex])
f(k) = [tex]\frac{1}{2i\pi}[/tex] [tex]\int^{\infty}_{-\infty}[/tex] [tex]\frac{e^{ikx}}{x-i\epsilon}dx[/tex] = e^(-k[tex]\epsilon[/tex])
I can't prove it equal to 0 or 1 and the required condition.
Show that
f(k) = [tex]\frac{1}{2i\pi}[/tex] [tex]\int^{\infty}_{-\infty}[/tex] [tex]\frac{e^{ikx}}{x-i\epsilon}dx[/tex] =
1, if k>0
0, if k<0
where [tex]\epsilon[/tex] > 0
The attempt at a solution
Consider : [tex]\oint[/tex] [tex]\frac{e^{ikz}}{z-i\epsilon}dz[/tex]
the residule is : e^(-k[tex]\epsilon[/tex])
so [tex]\int^{\infty}_{-\infty}[/tex] [tex]\frac{e^{ikx}}{x-i\epsilon}dx[/tex]
= [tex]{2i\pi}[/tex] e^(-k[tex]\epsilon[/tex])
f(k) = [tex]\frac{1}{2i\pi}[/tex] [tex]\int^{\infty}_{-\infty}[/tex] [tex]\frac{e^{ikx}}{x-i\epsilon}dx[/tex] = e^(-k[tex]\epsilon[/tex])
I can't prove it equal to 0 or 1 and the required condition.