# Laplace transform, book example

by hotcommodity
Tags: book, laplace, solved, transform
 P: 436 I'm having trouble following one of the steps in my textbook. They give a function: $$y(t) = e^{2t}$$ and plug it into $$Y(s) = \int_{0}^{\infty} y(t) e^{-st} dt$$ and compute. They end up with: $$\frac{1}{2-s} \underbrace{lim}_{b -> \infty} e^{(2-s)b}$$ (call this part 1) Which will converge to zero if s > 2. Then they state "we see that the improper integral for Y(s) does not converge if s is less than or equal to 2, and that $$Y(s) = \frac{1}{s-2}$$ if s >2. " (call this part 2) I don't get how they got from part 1 to part 2. How does the denominator go from 2-s to s-2, and why does the exponential go away? Any help is appreciated.
 Quote by hotcommodity $$\frac{1}{2-s} \underbrace{lim}_{b -> \infty} e^{(2-s)b}$$