For a function $f(t)$ to have a Laplace transform, it must be piece-wise continuous of exponential order. But what about the inverse Laplace ?
Is there expression $F(s)$ which has no inverse Laplace ? If so, what are the necessary conditions to have inverse Laplace ?
I'm asking because I was working on this problem to find its inverse Laplace:
$$F(s)=\ln\left(\frac{s+4}{(s-1)(s^2+1)}\right)$$
And here what I got:
$$F(s)=\ln\left(\frac{s+4}{(s-1)(s^2+1)}\right)=\ln(s+4)-\ln(s-1)-\ln(s^2+1)$$ $$F'(s)=\frac{1}{s+4}-\frac{1}{s-1}-\frac{2s}{s^2+1}$$ $$\mathcal{L}^{-1}(F'(s))=e^{-4t}-e^{t}-2\cos{t}$$
Knowing that $\mathcal{L}^{-1}(F(s))=\frac{-1}{t}\mathcal{L}^{-1}(F'(s))$:
$$f(t)=\mathcal{L}^{-1}(F(s))=\frac{-1}{t}(e^{-4t}-e^{t}-2\cos{t})$$
But, I tried to check this result and found that: $$\lim_{t\rightarrow 0^+}\frac{-1}{t}(e^{-4t}-e^{t}-2\cos{t})=D.N.E$$
Which means that I can't find Laplace transform for $f(t)$
So, if my procedure is wrong or there are some expressions which have no inverse Laplace ?
Thanks for help.
There are expressions which are not Laplace transforms of functions of exponential order. A necessary condition for $F(s)$ to be the Laplace transform of a function of exponential order is that $|s F(s)|$ is bounded as $s \to +\infty$. Your function has $|F(s)| \sim -2 \ln(s)$ as $s \to +\infty$.
However, your $F$ is the Laplace transform of a distribution that is not a function.