# Inverse Laplace Transform Help

1. Jan 27, 2014

### GreenPrint

1. The problem statement, all variables and given/known data

Is there a way to evaluate $L^{-1}(\frac{F(s)}{s + a})$? I'm sure if it can be evaluate.

2. Relevant equations

3. The attempt at a solution

2. Jan 27, 2014

### pasmith

It depends on what $F$ is. If $F$ has no singularities then the inverse transform
$$\mathcal{L}^{-1}\left(\frac{F(s)}{s + a}\right) = \frac{1}{2\pi i}\int_{c-i\infty}^{c + i\infty} \frac{F(s)e^{st}}{s+a}\,ds$$
(where $c \in \mathbb{R}$ is such that there are no singularities of $F(s)/(s+a)$ to the right of the line $\mathrm{Re}(s) = c$) reduces to the residue of $\dfrac{F(s)e^{st}}{s+a}$ at $s = -a$, which is $F(-a)e^{-at}$.

3. Jan 27, 2014

### Ray Vickson

Use convolution: if
$$f(t) = L^{-1}[F(s)](t),$$
then
$$L^{-1} \left( \frac{F(s)}{s+a} \right) (t) = \int_0^t f(t-\tau) e^{-a \tau} \, d \tau.$$

4. Feb 1, 2014

### GreenPrint

So if you have no idea what one of the functions in the frequency domain is and you get something like

inverse Laplace transform( F(s)G(s) )

and you know what G(s) of is but F(s) is not given then you have no way to evaluate the expression?

5. Feb 1, 2014

### Ray Vickson

Of course. If I give you two different F(s), you will get two different answers.