Frequency shift theorem

Remarks

The proof is one line: $e^{a t} f(t) e^{-s t} = f(t) e^{-(s - a) t}$, so the two integrals are the same integral. In particular, the half-plane of convergence moves to the right by $\operatorname{Re} a$. Example: from $\mathcal{L}\{\sin \omega t\}(s) = \omega / (s^2 + \omega^2)$ one gets $\mathcal{L}\{e^{a t} \sin \omega t\}(s) = \omega / ((s - a)^2 + \omega^2)$. Likewise $\mathcal{L}\{t^n e^{a t}\}(s) = n! / (s - a)^{n + 1}$ for $\operatorname{Re} s > \operatorname{Re} a$.

Going deeper

For a student

This rule is also called the first shifting theorem or the $s$-shift rule. It is the main tool for inverting terms such as $1/((s + 2)^2 + 9)$: complete the square, then read off $\tfrac{1}{3} e^{-2t} \sin 3t$. It is the mirror image of the time shift theorem, where a shift in $t$ becomes a factor $e^{-a s}$.