# Laplace transform, book example

by hotcommodity
Tags: book, laplace, solved, transform
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 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.
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PF Gold
P: 5,532
 Quote by hotcommodity $$\frac{1}{2-s} \underbrace{lim}_{b -> \infty} e^{(2-s)b}$$
That's only a piece of the integral, the piece that goes with the upper limit of integration. You also have to write down the piece that goes with the lower limit of integration, as per the Fundamental Theorem of Calculus.
 P: 436 Oh ok, I see why they did that now. Thank you!

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