Laplace transform of a exponential.

Question

we know that Laplace transform of $x^n$ is

$\mathcal{L}[x^n]$=$\frac{n!}{s^{n+1}}$

provided $n$ is a positive integer

but what is Laplace transform of

$a^x$ where $a$ is some constant number

please help i searched on google so many times and try to solve this by its original defination that is

$\mathcal{L}[x^n]=\int_{0}^{\infty}a^x e^{-sx}dx$

but how to solve further ( i am using $x$ insted of $t$ here)

please help

thank you.

Answer 1

$\int_0^{\infty} a^{x} e^{-sx}dx=\int_0^{\infty} e^{x\ln \, a} e^{-sx}dx=\int_0^{\infty} e^{-(s-\ln \, a)x}dx=\frac 1 {s-\ln \, a}$.

Answer 2

$a^xe^{-sx} = (ae^{-s})^x$

$$\mathcal L(ae^{-s})^x = \int^{\infty}_0(ae^{-s})^x = \frac{1}{\ln(ae^{-s})}(ae^{-s})^x\bigg\vert^\infty_0 = \frac{1}{\ln a -s}(0-1) = \frac{1}{s-\ln a}$$