Comparing the Fourier transform and the Laplace transform of $ x(t) = e^{-at}u(t) $
Let the signal $ x(t) = e^{-at}u(t) $. The Fourier transform $ X(\omega) $ converges for $ a > 0 $ and is given by:
$ X(\omega) = \int_{-\infty}^{\infty} e^{-at}u(t)e^{-j\omega t} \, dt = \int_{0}^{\infty} e^{-at}e^{-j\omega t} \, dt = \frac{1}{j\omega + a}, a > 0 $ (1)
the Laplace transform is:
$ X(s) = \int_{-\infty}^{\infty} e^{-at}u(t)e^{-st} \, dt = \int_{0}^{\infty} e^{-(s + a)t} \, dt $
with $ s = \sigma + j\omega $
$ X(\sigma + j\omega) = \int_{0}^{\infty} e^{-(\sigma + a)t} e^{-j\omega t} \, dt $
By comparing with (1) above, this last equation is the Fourier transform of $ e^{-(\sigma + a)t}u(t) $
$ \Rightarrow X(\sigma + j\omega) = \frac{1}{(\sigma + a) + j\omega}, \sigma + a > 0 $
since $ ^{s = \sigma + j\omega} $ and $ \sigma = \mathbb{R} (s) $
$ X(s) = frac{1}{s + a}, \mathbb{R} (S) > -a $
that is,
$ e^{-at}u(t) \xrightarrow{\mathcal{L}} \frac{1}{s + a}, \mathbb{R}(s) > -a $
for example, for $ a = 0, x(t) $ is the unit step with Laplace transform
$ X(s) = \frac{1}{s}, \mathbb{R} (s) > 0 $
