Inverse Laplace transform for $\frac{1-e^{-\pi s}}{s(s^2 + 16)}$

Question

I want to find the inverse Laplace transform for the following:$$\frac{1-e^{-\pi s}}{s(s^2 + 16)}$$

This was obtained from a piecewise function and required the heaviside step function to simplify.

I have two questions:

Question 1: Must this be solved using the heaviside step function in reverse.

Question 2: If so, is this usually/always the case?

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I can see it may be a good idea to separate the top two terms, but I am unsure where to go from there. It would seem that the numerator can't have any terms divided out by a derivative, nor is it a standard expression since we have $e$ on the top, which makes me think that partial fraction decomposition won't work here.

Thanks for any ideas.

Answer 1

Let be $$ G(s)=(1-e^{\pi s})F(s) $$ where $$ F(s)=\frac{1}{s(s^2+16)}=\frac{1}{16}\left(\frac{1}{s}-\frac{s}{s^2+16}\right) $$ so that the $\mathcal{L}^{-1}(F(s))=f(t)$ is $$ f(t)=\frac{1}{16}\left(u(t)-\cos(4t)u(t)\right)=\frac{1}{8}\cdot\underbrace{\frac{1}{2}\left(1-\cos(2\cdot 2t)\right)}_{\sin^2(2t)}u(t)=\frac{1}{8}\sin^2(2t)u(t) $$ where $u(t)$ is the Heaviside step function. Thus $$ g(t)=\mathcal{L}^{-1}(G(s))=f(t)u(t)-f(t-\pi)u(t-\pi)=\frac{1}{8}[u(t)-u(t-\pi)]\sin^2(2t) $$