Answer to Question #286772 in Complex Analysis for Bon

Question

find inverse laplace transorm : se^-2s/(s^2 + pi^2)

Answer 1

$\qquad\text{By property of Laplace Transform,}\\ if\ \mathcal{L}^{-1}(G(s))=g(t), then\\ \mathcal{L}^{-1}(e^{-as}G(s))=u(t-a)\times g(t-a)\\ \quad\\ Now,\mathcal{L}(f(t))=\frac{se^{-2s}}{s^2+\pi^2}=F(s)\\ \Rightarrow f(t)=\mathcal{L}^{-1}(F(s))=\mathcal{L}^{-1}(\frac{se^{-2s}}{s^2+\pi^2})=\mathcal{L}^{-1}(e^{-2s}\times \frac{s}{s^2+\pi^2})\\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad=\mathcal{L}^{-1}(e^{-2s}\times G(s)), where\ G(s)=\frac{s}{s^2+\pi^2}\\ But, \mathcal{L}^{-1}(G(s))=\mathcal{L}^{-1}(\frac{s}{s^2+\pi^2})=\cos(\pi t)=g(t), thus\\ \Rightarrow f(t)=\mathcal{L}^{-1}(e^{-2s}\times G(s))\\ \qquad\quad =u(t-2)\times g(t-2), \text{from the above stated property where a=2}\\ \qquad\quad =u(t-2)\times \cos(\pi(t-2)), since\ g(t)=\cos(\pi t)\\ Thus, \mathcal{L}^{-1}(\frac{se^{-2s}}{s^2+\pi^2})=u(t-2)\times \cos(\pi(t-2)).$