Complex Variable-definite integral

[Microzero]

Complex Variable---definite integral

Show that
f(k) = $$\frac{1}{2i\pi}$$ $$\int^{\infty}_{-\infty}$$ $$\frac{e^{ikx}}{x-i\epsilon}dx$$ =

1, if k>0
0, if k<0

where $$\epsilon$$ > 0

The attempt at a solution

1st. step:
Consider : $$\oint^{\infty}_{-\infty}$$ $$\frac{e^{ikz}}{z-i\epsilon}dz$$

the residule is : e^(-k$$\epsilon$$)

so $$\int^{\infty}_{-\infty}$$ $$\frac{e^{ikx}}{x-i\epsilon}dx$$

= $${2i\pi}$$ e^(-k$$\epsilon$$)

$$\frac{1}{2i\pi}$$ $$\int^{\infty}_{-\infty}$$ $$\frac{e^{ikx}}{x-i\epsilon}dx$$

= e^(-k$$\epsilon$$)

 

[HallsofIvy]

Show that
f(k) = $$\frac{1}{2i\pi}$$ $$\int^{\infty}_{-\infty}$$ $$\frac{e^{ikx}}{x-i\epsilon}dx$$ =

1, if k>0
0, if k<0

where $$\epsilon$$ > 0

The attempt at a solution

1st. step:
Consider : $$\oint^{\infty}_{-\infty}$$ $$\frac{e^{ikz}}{z-i\epsilon}dz$$

This makes no sense. What contour are you integrating over?

the residule is : e^(-k$$\epsilon$$)

so $$\int^{\infty}_{-\infty}$$ $$\frac{e^{ikx}}{x-i\epsilon}dx$$

= $${2i\pi}$$ e^(-k$$\epsilon$$)

$$\frac{1}{2i\pi}$$ $$\int^{\infty}_{-\infty}$$ $$\frac{e^{ikx}}{x-i\epsilon}dx$$

= e^(-k$$\epsilon$$)

 

[Microzero]

um...the pole is at z= iε
so i find the residule at z=iε