Laplace transform of the exponential function

Example

Let $a \in \C$ and $f(t) = e^{at}$ for $t \ge 0$, an exponential function. Its Laplace transform is $$\mathcal{L}\{e^{at}\}(s) = \frac{1}{s - a} \quad (\operatorname{Re} s > \operatorname{Re} a).$$ Check: for $s \ne a$ and $T > 0$, $$\int_0^T e^{at} e^{-st}\, dt = \frac{1 - e^{-(s - a)T}}{s - a}.$$ Since $\abs{e^{-(s - a)T}} = e^{-\operatorname{Re}(s - a) T}$, this tends to $0$ as $T \to \infty$ exactly when $\operatorname{Re} s > \operatorname{Re} a$, which gives the formula. When $\operatorname{Re} s < \operatorname{Re} a$ the term $e^{-(s-a)T}$ grows without bound. When $\operatorname{Re} s = \operatorname{Re} a$ it stays on the unit circle without a limit (or the integrand is $1$ if $s = a$), so the integral diverges. So the abscissa of convergence is $\operatorname{Re} a$. The case $a = 0$ gives $\mathcal{L}\{1\}(s) = 1/s$.