Inverse Laplace transform of complicated function shift

Question

Evaluate

$$\mathcal{L}^{-1}\frac{e^{-as}}{(s + 3)^2 +2}$$

I am not sure how to proceed because of the $e$.

If it was just $s^2$ in the bottom, then the second shift would have applied. Any hints?

Answer 1

Hint: Apply the inverse transform rule. If $L^{-1}\{F(s)\}=f(t)$ then $L^{-1}\{e^{-as}\}F(s)\}=H(t-a)f(t-a)$ where $H(t)$ is the Heaviside step function.

Answer 2

By using \begin{align} \delta(t - a) &\Doteq e^{-as} \\ e^{at} \, \sin(b t) &\Doteq \frac{b}{(s-a)^2 + b^2} \\ \int_{0}^{t} f(t-u) \, g(u) \, du &\Doteq \overline{f}(s) \, \overline{g}(s) \end{align} then \begin{align} \mathcal{L}^{-1}\left\{ \frac{e^{-as}}{(s-a)^2 + 2} \right\} &= \frac{1}{\sqrt{2}} \, \mathcal{L}^{-1}\left\{ e^{-as} \cdot \frac{\sqrt{2}}{(s-a)^2 + (\sqrt{2})^2} \right\} \\ &= \frac{1}{\sqrt{2}} \, \int_{0}^{t} \delta(u-a) \, e^{-3(t-u)} \, \sin(\sqrt{2}(t-u)) \, du \\ &= \frac{1}{\sqrt{2}} \, e^{-3(t-a)} \, \sin(\sqrt{2} \, (t-a)). \end{align}