Inverse Laplace transform of a function with removable singularities

Question

While trying to figure out the inverse Laplace transform of $f(z)=\frac{exp(-z)(z+2) +z-2}{z^{3}}$ by using the sum of residues of the function $f(z)exp(zt)$ (from the inversion theorem) I have obtained an inverse function that is $0$. This result is coherent with the fact that the only singularity ($0$) is a removable singularity, and thus the function is holomorphic. However, by spliting the integral into two parts:$f(z)=f_1(z)+f_2(z)=\frac{exp(-z)(z+2)}{z^{3}}+\frac{z-2}{z^{3}}$ and applying the residue theorem on both functions which are not holomorphic, we get the right result (not zero). Why is this the case? Why is the inversion theorem not directly applicable here? Thank you

Answer 1

Let $f$ being piecewise continuous and $$F(s) = \int_0^\infty f(t) e^{-st}dt$$ If it converges for $\Re(s) = a$ then it is analytic for $\Re(s) > a$ and for $ \sigma > a$ : $$f(t) = \frac{1}{2i \pi} \int_{\sigma-i\infty}^{\sigma + i \infty} F(s) e^{st}ds$$ (if $f$ is continuous at $t$)

This is the Fourier/Laplace inversion theorem. The residue theorem tells us that if $F(s)$ is meromorphic on $\Re(s) \ge \sigma'$ and if $F(s) e^{st} \to 0$ as $|\Im(s)| \to \infty$ then $$\frac{1}{2i \pi} \int_{\sigma-i\infty}^{\sigma + i \infty} F(s) e^{st}ds=\frac{1}{2i \pi} \int_{\sigma'-i\infty}^{\sigma' + i \infty} F(s) e^{st}ds+ \sum_{\Re(\rho) \in (\sigma,\sigma')} Res(F(s) e^{st},\rho)$$ If $F(s)$ is meromorphic everywhere and $F(s) e^{st} \to 0$ fast enough when $\Re(s) \to -\infty$ then $$\lim_{\sigma' \to -\infty} \frac{1}{2i \pi} \int_{\sigma'-i\infty}^{\sigma' + i \infty} F(s) e^{st}ds = 0$$ and hence $$f(t) = \frac{1}{2i \pi} \int_{\sigma-i\infty}^{\sigma + i \infty} F(s) e^{st}ds =\sum_{\Re(\rho) < \sigma}Res(F(s) e^{st},\rho)$$

As you see, there are quite a lot of if and that's what you need to check with your function.

(also note it is even more complicated then $F(s) = \int_{-\infty}^\infty f(t) e^{-st}dt$ is a bilateral Laplace transform, you can recognize it if $F(s)$ doesn't $\to 0$ as $\Re(s) \to \infty$)