I need to find the Fourier transform of
$$ f(x)= \begin{cases} e^{-ax}, & x \geq0\\ \ 0, & x < 0 \end{cases} $$
where $a>0$.
This is what I've done:
$$ F(w) = \frac{1}{2\pi}\int_{-\infty}^\infty f(x)e^{iwx}dx = \frac{1}{2\pi}\int_{0}^\infty e^{-ax}e^{iwx}dx $$ but I don't know what to use here to continue.
Let $z:=a-iw$ so$$|e^{-zx}|=e^{-ax}\implies\lim_{x\to\infty}|e^{-zx}|=0\implies\lim_{x\to\infty}e^{-zx}=0.$$So$$\int_0^\infty e^{-zx}dx=\left[\frac{-e^{-zx}}{z}\right]_0^\infty=\frac1z=\frac{a+iw}{a^2+w^2}.$$