Question

Integrating challenge

Prove

Answer 1

\[\frac{1}{\pi }\int _{-\infty }^{\infty }e^{i 2 t} e^{-i t \omega }dt=\] =\[2 \delta (\omega -2)\]

Answer 2

Using w for omega here.
e^(-itw) = e^(-it(2pi/t)) = e^(-i2pi) = 1

So the problem reduces to
(1/pi) int e^(2it) dt
= (1/pi) 2pi delta(w-2)
= 2 delta(w-2).