I want to find the inverse fourier transform of
$$
\frac 1 {j \omega – 1}
$$
The fourier transform of
$$
e^{-at} u(t)
$$
is
$$
\frac {1}{j \omega + a}
$$
This result if true ONLY if a > 0. If a < 0, the demonstration shows there is a limit that does not exist.
So I was wondering what can we do if a is < 0.
I have the idea of using u(-t) instead of u(t). Which mean that if a is < 0, the inverse fourier transform would be :
$$
-e^{-at} u(-t)
$$
This can be demonstrate using the definition of the inverse fourier transform, but I am not sure. Can this result be used (in the case it is right) ?
Best Answer
In general, we have ${\cal F}(t \mapsto f(-t))(\omega) = ({\cal F}f) (-\omega)$.
If $f(t) = e^{-at} u(t)$, with $a>0$, we have $({\cal F}f)(\omega) = { 1 \over i \omega +a}$.
If we let $g(t) = -f(-t)$, we have $({\cal F}g)(\omega) = -({\cal F}f)(-\omega) = {1 \over i \omega -a}$.
Hence the inverse is $t \mapsto - e^{at} u(-t)$.