# Laplace Transforms for improper integrals?

1. Jul 25, 2011

### Screwdriver

1. The problem statement, all variables and given/known data

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:

$$I = \int_{0}^{\infty }sin(t)dt$$

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

$$F(s)=\int_{0}^{\infty }e^{-st}sin(t)dt$$

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

$$F(s)=\frac{1}{s^2+1}$$

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

2. Relevant equations

Laplace Transform

3. The attempt at a solution

2. Jul 26, 2011

### Dick

I'm not sure what you are expecting here. Sure, the integral of sin(x) does not converge. The integral of sin(t)*e^(-s*t) does for s>0. And it does have a limit, again sure. What you've done is 'regularize' an integral by multiplying by e^(-s*t) to force it to converge. But the limit of that as s->0 doesn't make the original integral converge magically.

3. Jul 26, 2011

### Screwdriver

So does that mean this (?):

$$\lim_{s\to0}\int_{0}^{\infty }e^{-st}f(t)dt \neq \int_{0}^{\infty }f(t)dt$$

But how can that be true? Certainly it must be true for proper integrals.

4. Jul 26, 2011

### Dick

e^(-s*t) is less than 1 for s=>0. So yes, for any convergent integral, I think it works. But applying that to a divergent integral is a different story.

5. Jul 26, 2011

### Screwdriver

Okay, thanks! By the way, I realized what I was expecting from the LT. The limit exists - I would expect it to not exist. Why should the limit exist if the integral diverges?

If the LT of $sin(t)$ was something like $\frac{1}{s}$, then the limit as $s$ goes to zero doesn't exist. That would make sense, because the integral also essentially doesn't exist.

6. Jul 26, 2011

### Dick

Ok, take that example. The LT of f(t)=1 is 1/s. That does diverge like the integral of 1 as s->0 because e^(-st) approaches 1 as s->0. So the limit of the LT would have to diverge as s->0. On the other hand sin(t) oscillates and the LT can average out the oscillation. Take cos(t). The limit there is 0 rather than 1. But there's not that much difference between the behavior of sin(t) and cos(t) as integrals.

7. Jul 26, 2011

### Screwdriver

Okay, I can live with that. Thanks again