Laplace Transform of this?

Question

I'm trying to find the Laplace transform of this:

$f(t) = e^{-2t}$

my assumption was

$ \int_{0}^{\infty}f(t)e^{-st}\,dt $

but for some reason when I try integrating with parts I end up with $\frac{1}{s-1}$ which isn't right... any guide on how to break this down would be appreciated!

Answer 1

$$I=\int_{0}^{\infty}f(t)e^{-st}\,dt$$ $$I=\int_{0}^{\infty}e^{-2t}e^{-st}\,dt$$ Substitute $u=e^t \implies du=udt$ $$I=\int_{1}^{\infty}u^{-2}u^{-s}\,\dfrac {du}{u}$$ $$I=\int_{1}^{\infty}u^{-(3+s)}\,{du}$$ $$I=\dfrac {u^{-(s+2)}}{-(s+2)}\Big|_1^{\infty}$$ $$\implies I(s)=\dfrac 1 {s+2}$$

Answer 2

By the substitution $u=(2+s)t$ we get $\int_0^{\infty} e^{-(2+s)t}dt=\frac1 {2+s} \int_0^{\infty} e^{-u}du=\frac1 {2+s} $

Answer 3

There is no need of IBP.

$$\int_0^\infty f(t)e^{-st}dt =\int_0^\infty e^{-2t}e^{-st}dt = \int_0^\infty e^{-(s+2)t}dt = \frac{-1}{s+2}[0-1] = \frac{1}{s+2}$$