Let $$f(x)=\begin{cases}e^{kx},& (x<0)\\ 0, & (x>0).\end{cases}$$ where $k>0$
Fourier transform for $f(x)$ is $f_T(w)=\frac{\sqrt{2}}{\sqrt{\pi}} \int_{-\infty}^{\infty}f(x)e^{-\iota wx}dx$
substituting :
$f_T(w)= \frac{\sqrt{2}}{\sqrt{\pi}} \int_{-\infty}^{0}e^{kx}e^{-\iota wx}dx=\frac{\sqrt{2}}{\sqrt{\pi}}\int_{-\infty}^{0}e^{(k-\iota w)x}dx$
Now how do I decide the behaviour of $e^{(k-\iota w)x}$ as $x\to -\infty$?
$e^{(k-i\omega)x}|_{x=-\infty} = e^{kx}e^{-i\omega x}|_{x=-\infty} \leq e^{kx}\|e^{-i\omega x}\| |_{x=-\infty}=0$.