How the partial fraction decomposition works for finding this Inverse Laplace Transform?

Question

I've been working to find inverse Laplace transform for the following : $$ \frac{A}{(s-a)(s-r_1)(s-r_2)} $$ However, I'm getting stuck on the partial fraction decomposition. When I run the decomposition in Wolfram Alpha, it comes back as $$-\frac{A}{(s-r_1)(a - r_1)(r_1 - r_2)} -\frac{A}{(s - r_2)(a - r_2)(r_2 - r_1)} + \frac{A}{(a - r_1)(a - r_2)(s - a)}$$ Any thoughts on how this decomposition works? I can solve the inverse Laplace easily from this point but for the life of me I can't figure out how this partial fraction is working. Thanks for the help!

Answer 1

For $a,r_1,r_2$ not equals $$\frac{A}{(s-a)(s-r_1)(s-r_2)}=A\left (\frac 1{(a-r_1)(a-r_2)(s-a)} \\+\frac 1 {(r_1-a)(r_1-r_2)(s-r_1)} \\+ \frac 1 {(r_2-a)(r_2-r_1)(s-r_2)} \right)$$ In case of equality ( for example $a=r_1$ or $r_1=r_2$) the decomposition is different.

Take a simple example: $$f(s)=\dfrac 1 {(s-2)(s-3)}$$ It is decomposed as $$f(s)=\dfrac {A}{(s-3)}+\dfrac B{(s-2)}$$ For A you calculate $h(s)=(s-3) f(s) \implies A=h(3)=1$ . For B you calculate $h(s)=(s-2) f(s) \implies B=h(2)=-1$ Therefore we have: $$f(s)=\dfrac {1}{(s-3)}-\dfrac 1{(s-2)}$$

Answer 2

Calling

$$ f(s) =\frac{A}{(s-a)(s-r_1)(s-r_2)} $$

supposing that $a \ne r_1\ne r_2$ we have by residues

$$ f(s) = \left(\lim_{s\to a}(s-a)f(s)\right)\frac{1}{s-a}+\left(\lim_{s\to r_1}(s-r_1)f(s)\right)\frac{1}{s-r_1}+\left(\lim_{s\to r_2}(s-r_2)f(s)\right)\frac{1}{s-r_2} $$