I have a Laplace transform function of the following form and I'm trying to evaluate it. From my research I think I need to take the Inverse Laplace Transform and then integrate, but I'm having trouble working that out, or if that's even what I have to do.
$$\frac{s^2}{s^2+10s+25} $$ and $$\frac{s}{s+5}$$
Any pointers on where to go from here is much appreciated.
Also, this is not homework, merely my own side project that I'm attempting to work on, but this has got my stymied.
On
The definition of a Laplace transform leads to the following expression for the inverse Laplace transform of a function $F(s)$:
$$f(t) = \frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, F(s) \, e^{s t}$$
where $c$ is a real number larger than the real parts of all poles of $F$ in the complex $s$ plane. That is, the line $\Re{s}=c$ is to the right of all poles of $F$.
When $F$ consists of poles only in the complex plane, the above integral may be shown to be equal to the sum of the residues of $F(s) e^{s t}$ evaluated at the poles of $F(s)$.
For $F(s)=s/(s+5)$, there is a single, simple pole at $s=-5$. The ILT is then
$$f(t)=\lim_{s\to-5} (s+5) F(s) e^{s t} = -5 e^{-5 t}$$
For the case $F(s) = s^2/(s+5)^2$, there is a double pole at $s=-5$, so that the residue calculation involves a derivative:
$$f(t) = \lim_{s\to-5} \frac{d}{ds} (s+5)^2 F(s) e^{s t} = (25 t-10) e^{-5 t} $$
$$ \frac{s^2}{s^2+10s+25} = \left(\frac{s}{s+5}\right)^2 = \left(1-\frac{5}{s+5}\right)^2 = 1 - \frac{10}{s+5} + \frac{25}{(s+5)^2} $$ Since the Laplace transform is linear, you can treat each term separately.
Whenever the degree of the expression in the numerator is at least as big as that in the denominator, you can do long division and get a quotient (in this case $1$) and a remainder (in this case $-5$).