Answer
$\dfrac{1} {s-\alpha}$
Work Step by Step
The Laplace Transform of $f(x)$ can be found as:
$L f(s)=\int_0^{\infty} e^{\alpha x}e^{-sx} \ dx\\=\int_0^{\infty} e^{-(s-\alpha) x } dx\\=\lim\limits_{R \to \infty}\dfrac{-1}{s-\alpha}[e^{-(s-\alpha) x }]_0^R\\=\lim\limits_{R \to \infty} \dfrac{-1} {s-\alpha}(e^{-(s-\alpha) R }-1)\\=\dfrac{-1} {s-\alpha}(0-1) \\=\dfrac{1} {s-\alpha}$
