Laplace transform, book example

[SOLVED] Laplace transform, book example

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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Tom Mattson
Staff Emeritus
$$\frac{1}{2-s} \underbrace{lim}_{b -> \infty} e^{(2-s)b}$$