1. The problem statement, all variables and given/known data If laplace of [e^-(at)] u(t) is 1/(s+a) and ROC is s > -a Find laplace and ROC of -e^(-at) u(-t) 2. Relevant equations Laplace is integral over minus infinity to plus infinity of f(t) e^(-st) dt 3. The attempt at a solution Well i integrated f(t) over the limits with e(-st)u(-t) dt the u(-t) changes limits to from (-inf to +inf) to (-inf to zero.) So i integrated it and got Laplace as 1/(s+a) Now ROC is region for which laplace goes positive that is more than zero So in case of (1/(s+a)) s+a must be > 0 for F(s) to be positive. For that s > (-a) The book says that laplace is right but ROC is s < (-a) Book didnt integrate. It went through other formula. like L(f(t)) is F(s) then laplace of f(-t) is F(-s) so Laplace of e^(at)u(-t) becomes 1/(-s + a) ROC is -s + a > 0 so -s > -a Now a becomes -a so Laplace of e ^ (-at) u(-t) becomes 1/(-s -a) = -1 /(s + a) ROC is s + a > 0 so s > -a Multiply both side f(t) and F(s) by -1 Laplace of -1 * e ^ (at) u(-t) becomes 1 / (s + a) Here it said that ROC remains same as power remains same. And ROC is -s > a But shouldnt ROC for (1/(s+a) ) be s + a > 0, so s > -a I'm confused as to why is ROC different. What does underlined part mean?