- #1
Microzero
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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)
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)
