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## Homework Statement

I (came up with)/(heard about) a way of using Laplace transforms that I didn't think about before. The problem is that it doesn't work for some reason.

Look at following integral:

[tex]I = \int_{0}^{\infty }sin(t)dt[/tex]

Say that you had no idea how to integrate something. Then what you could conceivably do is consider the following function:

[tex]F(s)=\int_{0}^{\infty }e^{-st}sin(t)dt[/tex]

This converges for [itex]s > 0[/itex]. But then this is just the LT of [itex]sin(t)[/itex] (which you know), so:

[tex]F(s)=\frac{1}{s^2+1}[/tex]

Now, if [itex]s = 0[/itex], then [itex]F(0) = I[/itex], which would also mean that [itex]I = 1[/itex]. But you can't just sub in 0 because you need [itex]s[/itex] to be greater than 0. So instead take the limit as [itex]s[/itex] goes to 0, which gives the same thing. But we know that [itex]I[/itex] doesn't converge because [itex]sin(t)[/itex] oscillates...so what's going on? On a related note, this method works for some other functions like [itex]e^{-t}[/itex] ; the LT gives [itex]\frac{1}{s+1}[/itex] and then [itex]F(0) = 1[/itex].

## Homework Equations

Laplace Transform

## The Attempt at a Solution

Head-scratching, mainly.