Firstly, because I've seen many different definitions of the fourier transform with many different variations of the coefficients, my definition is $F(\omega)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt$
I want to find the fourier transform of $f(t)=\begin{bmatrix} 0, t\leq 1 \\e^{-at}, t>1\end{bmatrix}$
So using the definition, it's $$F(\omega)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt=\frac{1}{2\pi}(\int_{-\infty}^{1}0dt+\int_{1}^{\infty}e^{-at}e^{-i\omega t}dt)=\frac{1}{2\pi}\int_{1}^{\infty}e^{-(a+i\omega)t}dt=\frac{1}{2\pi}[\frac{e^{-(a+i\omega)t}}{-(a+i\omega)}]_{t=1}^{\infty}$$
and here is exactly my problem.
$e^{-\infty}$ approaches zero, that is obvious. But what is $\lim_{t \to \infty}e^{it}$? doesnt the imaginary unit ruin everything? that's the same as $\lim_{t \to \infty} \cos(t)+i\sin(t)$ and the trigonometric functions dont have a defined limit at infinity