How to find the inverse laplace transform of an arbitrary function

Question

How to find $$\mathcal{L^{-1}} \left[ \frac{F(s)}{s+a} \right]$$where $F(s)$ is the Laplace transform of $f(t).$

Answer 1

If you know about convolution, this is just a piece of cake.

$\mathcal L^{-1}\left\{\dfrac{F(s)}{s+a}\right\}$

$=\mathcal L^{-1}\left\{\dfrac{1}{s+a}\right\}*\mathcal L^{-1}\{F(s)\}$

$=e^{-at}*f(t)$

$=\int_0^te^{-a(t-\tau)}f(\tau)~d\tau$

$=e^{-at}\int_0^te^{a\tau}f(\tau)~d\tau$

Answer 2

Hint:

You know that $$\mathcal{L}(f*g)=F(s)G(s)$$ so $$\mathcal{L^{-1}}\big(F(s)G(s)\big)=f*g$$ wherein $f*g=\int_0^tf(\kappa)g(t-\kappa)d\kappa$.

Answer 3

Use laplace transform properties : $$ G(s)=\frac{F(s)}{s+a} $$ $$ \mathcal{L^{-1}}[G(s)]=g(t) $$ $$ \mathcal{L} \left[ e^{at}g(t) \right] = G(s-a)=\frac{F(s-a)}{s} $$ $$ \frac{1}{s} \triangleq \int_0^t $$ $$ e^{at}g(t)=\int_0^t e^{a \tau }f( \tau ) d \tau $$ $$ g(t)=e^{-at} \int_0^t e^{a \tau }f( \tau ) d \tau $$ or we can write : $$ g(t)= \int_0^t e^{-a(t- \tau) }f( \tau ) d \tau $$