I would like to find the Inverse Laplace transform of the following fraction:
$$\frac{1}{s^2-18s+810}$$
The problem is that the numerator has imaginary roots $9+27i$ and $9-27i$. Can someone solve this fraction step by step, because I can't find $A$ and $B$ from imaginary roots?
Instead of using partial fractions, I would start by completing the square on the denominator: $$\frac{1}{s^2-18s+810}=\frac{1}{(s-9)^2+729} \tag{*}$$ You can now easily use result 19 on the Table of Laplace Transforms, however if you wish to derive it, I provided a small proof:
Proof: By definition, we know that: $$\mathcal{L}\{\sin(bt)\}=\int_0^{\infty} e^{-st}\sin(bt)~dt$$ Consider using integration by parts twice on the indefinite integral: $$\begin{align}\int e^{-st}\sin(bt)~dt&=-\frac{1}{s}e^{-st}\sin(bt)+\frac{b}{s}\int e^{-st}\cos(bt)~dt\\&=-\frac{1}{s}e^{-st}\sin(bt)+\frac{b}{s}\left(-\frac{1}{s}e^{-st}\cos(bt)-\frac{b}{s} \int e^{-st}\sin(bt)~dt\right)\\&=-\frac{1}{s}e^{-st}\sin(bt)-\frac{b}{s^2}e^{-st}\cos(bt)-\frac{b^2}{s^2}\int e^{-st}\sin(bt)~dt \end{align}$$ This implies that: $$\frac{s^2+b^2}{s^2}\cdot \int e^{-st}\sin(bt)~dt=\displaystyle -e^{-st} \left({\frac{1}{s} \sin (bt) + \frac b {s^2} \cos (bt) }\right)$$ Therefore: $$\begin{align}\mathcal{L}\{\sin(bt)\}&=-\frac{s^2}{s^2+b^2}\left[e^{-st}\left(\frac{1}{s}\sin(bt)+\frac{b}{s^2}\cos(bt)\right)\right]_0^{\infty}\\&=-\frac{s^2}{s^2+b^2}\left(0-1\cdot \left(\frac{1}{s}\cdot 0+\frac{b}{s^2}\cdot 1\right)\right)\\&=\frac{b}{s^2+b^2} \tag*{$\blacksquare$} \end{align}$$
Proof: By definition: $$\begin{align}\mathcal{L}\{e^{at}f(t)\}&=\int_0^{\infty} e^{-st}[e^{at}f(t)]~dt\\&=\int_0^{\infty}e^{-(s-a)t}f(t)~dt\\&=F(s-a) \tag*{$\blacksquare$} \end{align}$$
Therefore, it follows from $(1)$ and $(2)$ that: $$\mathcal{L}\{e^{at}\sin(bt)\}=\frac{b}{(s-a)^2+b^2} \tag{3}$$ We can easily apply $(3)$ to evaluate the Inverse Laplace Transform of $(*)$.