Let $F(z)$ be a Laplace transform with singularities $z_1,z_2,\ldots,z_n$.

  • In order to find the inverse Laplace transform, we can evaluate $$ f(t)=\frac{1}{2\pi i}\int_{-\infty}^{+\infty} F(\gamma +is)e^{(\gamma + is)t} \, ds $$ which is equal to $\sum_{k\geqslant1} \operatorname{res}(F(z)e^{zt},z_k)$ if $t\geqslant0$, and is equal to $0$ if $t<0$ .
  • However for the inversion theorem to work, we need to check that: for all $\gamma>\max \operatorname{Re} z_k$, $|F(\gamma + is)| = \mathcal{O}(s^{-2})$ as |s| goes to infinity. Why does $\gamma$ need to be greater than the real parts of all the singularities ?.
  • Is it for the integral $f(t)=\frac{1}{2\pi i}\int_{-\infty}^{+\infty} F(\gamma +is)e^{(\gamma + is)t} \, ds$ to converge using the residues ?. And for what reasons will it converge only if this condition is satisfied ?.
  • Also, if we want to find the inverse laplace transform of $\frac{s}{s^2-1}$ using the residues (ie. the laplace transform of $\cosh(t)$), what should be done, as the function does not satisfy the criteria $|F(\gamma + is)|=\mathcal{O}(s^{-2})$. Thank You

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