I already figured out FT of e^(-at^2), but this one seems to be too hard for me. I found that it should be 1/(a+i*p), but why?
Edit: Sorry, new to the page and didnt undestand the rules that I need to show effort. Apologizes. I tried to calculate it for $t>0$ and $a>0$. I got $$F = \int_{0}^{\infty} e^{-at}e^{-ipt}dt = \int_{0}^{\infty} e^{t(-a-ip)}dt = \left[\frac{e^{t(-a-ip)}}{-a-ip}\right]_0 = \frac{1}{a+ip}$$ Thanks, Andrew
Actually the function $e^{-at}$ does not have a Fourier transform - it's not integrable, not even a tempered distribution. What you've calculated here is the Fourier transform of the function $f$ defined by $$f(t)=\begin{cases}e^{-at},&(t\ge0),\\0,&(t<0).\end{cases}$$