What's the laplace inverse of this function?

Question

I'm completely stuck on how to do this one. Any help is appreciated.

What is the inverse Laplace transform of:

$$\mathcal{L} ^ {-1} \left\{ \frac{e^{-2s}}{s-2} \right\} = f(t)$$

Answer 1

Using the shift theorem: $$\mathscr L\{f(t-a)\,\mathcal U(t-a)\}(s)=e^{-as}F(s).$$

Here $\mathcal U(t-a)$ is the Heaviside step function that is defined as $1$ for $t\geq a$, $0$ otherwise. What you've got is $e^{-as}F(s):a=3,F(s)=\dfrac{1}{s-2}.$ Therefore, applying the theorem yields

$$e^{2(t-3)}\,\mathcal U(t-3).$$

So the proof for the shift theorem (called second shift theorem (I guess)) is the following:

$$\mathscr L\{f(t-a)\,\mathcal U(t-a)\}(s)=\int_0^\infty e^{-st}f(t-a)\,\mathcal U(t-a)\, dt =\int_a^\infty e^{-st}f(t-a)\, dt. $$

Consider the substitution $u=t-a$, then:

$$\int_0^\infty e^{-s(u+a)} f(u)\,du=e^{-sa}F(s).$$

Recall that $F(s)$ is the laplace transform of $f(t)$.

Answer 2

We could also use the inverse Mellin transfer where the pole is at $s=2$ of order one. That is, \begin{align} \mathcal{L}^{-1}\Bigl\{\frac{e^{-2s}}{s-2}\Bigr\} &= \frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{e^{-2s}}{s-2}e^{st}ds\\ &=\sum\text{Res}\\ &=\lim_{s\to 2}(s-2)\frac{e^{s(t-2)}}{s-2} \end{align} From the limit, we have an exponential term of the form $e^{s(t-2)}$. Thus $\text{Re}(s) >2$ which means we need a unit step at $t = 2$. $$ \mathcal{L}^{-1}\Bigl\{\frac{e^{-2s}}{s-2}\Bigr\} = e^{2(t-2)}\mathcal{U}(t-2) $$ From the Wolfram link, you will see that the accepted answer is incorrect. Note that $\theta(t-2)$ is the Heaviside Theta which is the unit step. We even check our solution by taking the Laplace transform. \begin{align} \int_0^{\infty}e^{2(t-2)}\mathcal{U}(t-2)e^{-st}dt &= \int_2^{\infty}e^{2(t-2)}e^{-st}dt = e^{-4}\int_2^{\infty}e^{t(2-s)}dt\\ &= e^{-4}\frac{e^{t(2-s)}}{2-s}\Bigr|_2^{\infty}\\ &= \frac{e^{-2s}}{s-2} \end{align}