Help with Fourier transform

Question

I need to show that the Fourier transform of $$ {\rm g}\left(t\right) = {1 \over 2\pi}\,{1 \over 1 - {\rm i}t}\quad \mbox{is}\quad {\rm G}\left(w\right) = {\rm e}^{-\omega}\quad \mbox{for}\quad \omega > 0\ \mbox{and}\ 0\ \mbox{otherwise.} $$ Any ideas?

Answer 1

Unfortunately, $g$ is not an $L^1$ function, and thus Fourier Transform of $g$ can not be calculated in the traditional way.

Fortunately, $G$ is an $L^1$ function, and this allows us to calculate its inverse Fourier Transform as $$ \frac{1}{2\pi}\int_{-\infty}^\infty \mathrm{e}^{\mathrm{i} wt}G(w)\, dw= \frac{1}{2\pi}\int_{0}^\infty \mathrm{e}^{\mathrm{i} wt}\mathrm{e}^{-w} dw. $$