Exponentials
Translation of Inverse Laplace Transforms
Given function $f(t)$ with Laplace transform $F(s)$ and constant $\alpha \in \mathbb{C}$
\[\begin{align} \mathcal{L}\{ e^{\alpha t}f(t) \} &= \int_{0}^{\infty} e^{\alpha t}f(t) \cdot e^{-st} \; dt \\[10pt] &= \int_{0}^{\infty} f(t) e^{-(s-\alpha)t} \; dt \\[10pt] &=\text{assme that } \lvert s \rvert > \lvert \alpha \rvert \text{ as otherwise the integral will not converge} \\[10pt] &= F(s-\alpha) \end{align}\]This is a really nice property that is extremely useful for inverse Laplace transforms. This property also effectively solves any function f(t) that involves an exponential. For example, applying this result to polynomials gives us
\[\mathcal{L} \{ e^{\alpha t} \} = \frac{1}{s-\alpha}\] \[\mathcal{L} \{ e^{\alpha t}t^n \} = \frac{n!}{(s- \alpha)^{n+1}}\] \[\mathcal{L} \left \{ e^{\alpha t} \sum_{k=0}^n a_k t^k \right \} = \sum_{k=0}^n \frac{k! \ a_k}{(s- \alpha)^{k+1}}\]Translation in the Exponent
Notice that \(e^{\alpha t + \beta} = e^{\beta} \cdot e^{\alpha t}\), so in general we have
\[\mathcal{L}\{ e^{\alpha t + \beta}f(t) \} = e^{\beta} \cdot F(s-\alpha)\]