If $\mathcal{L}[f(t)] = \hat f(p)$ then $\mathcal{L}[e^{at}f(t)] = \hat f(p+a)$

Question

Let $f:[0,\infty) \rightarrow \mathbb{R} $ be continuous, with the property that $f(t)e^{-pt} \rightarrow 0$ as $t \rightarrow \infty$.

If $\mathcal{L}[f(t)] = \hat f(p)$ then $\mathcal{L}[e^{at}f(t)] = \hat f(p+a)$

I have as far as $\mathcal{L}[e^{at}f(t)] = \int^{\infty}_0 e^{-(s+a)t}f(t)dt$

How do I proceed?

Answer 1

Take a look at the other side also, as $$ \hat f(p) = \int_0^\infty e^{-pt}f(t)\,dt $$ we have $$ \hat f(p+a) = \int_0^\infty e^{-(p+a)t}f(t)\,dt $$ As $$ \mathcal L[e^{-a\cdot}f](p) = \int_0^\infty e^{-pt}e^{-at}f(t)\,dt $$ we are done.