Finding the inverse laplace transform using complex analysis.

Question

I've been able to prove simple laplace transforms like $\dfrac {1}{(s+a)} $ quite easily but what about $\dfrac {1}{(s+a)^3+b^2} $ this does not seem easy to do since you cannot easily compute the residues of the denominator. How can this be done???

Answer 1

I've been able to prove simple inverse laplace transforms of $\dfrac {1}{(s+a)} $ quite easily,

but what about $\dfrac {1}{(s+a)^3+b^2} $?

Hint. One may just use a partial fraction decomposition over $\mathbb{C}(X)$, giving $$ \frac {1}{(s+a)^3+b^3}=\frac {A}{s+a+b}+\frac {B}{s+a-b\tau}+\frac {C}{s+a-b\bar{\tau}} $$ where $\tau=\frac12+i\frac{\sqrt{3}}2$, $\tau^3=-1$.

Then one may conclude with the known inverse Laplace transform of $\displaystyle \frac1{s+\alpha}$.