Understanding what happens to exp((6-s)t) when t goes to inf

[oso0690]

Homework Statement

I want to understand what happens to exp((6-s)t) when t goes to infinity. Btw, this is part of a Laplace Transform so that's what the 's' is.

Homework Equations

f(t) = exp(6t)u(t)

Find Laplace Transform.

The Attempt at a Solution

I get to:

exp((6-s)t) / (6-s) in which i must integrate from 0 to infinity. I'm pretty sure the answer is 1 / (6-s) but I don't understand why. Breaking apart exp(6t) goes to infinity. And breaking apart exp(-st) goes to 0 right? *when t=inf

 

[hunt_mat]

It's okay if (6-s)<0.

 

[oso0690]

For a Laplace Transform is that always true?

 

[hunt_mat]

No, I just figured something out. Not all functions have Laplace transforms, the example of the exponential increase is one example of that.

 

[Ray Vickson]

Many functions f(t) have Laplace transforms F(s) that are valid only for certain regions of s. For example, u(t) has L.T. = 1/s, but this is valid only if s > 0 because it comes from integrating exp(-s*t) for t from 0 to +infinity.

RGV

 

[oso0690]

So when applying t=infinity for the function exp((6-s)t) , it doesn't exist? Or does exp(6t) just get neglected?

EDIT: just saw ray's post. So must it be specified that (6-s)<0 or is it just something known with Laplace Transforms?

 

[Ray Vickson]

So when applying t=infinity for the function exp((6-s)t) , it doesn't exist? Or does exp(6t) just get neglected?

EDIT: just saw ray's post. So must it be specified that (6-s)<0 or is it just something known with Laplace Transforms?

No, you cannot neglect it; that is why the L.T. makes no sense for s < 6. And, of course, we never let t = infinity; we take the *limit* as t --> infinity. (That is how integrals are _defined_ over infinite intervals: as limits.) Anyway, when s > 6 the limit = 0, so the term goes away. However, we did not "neglect" it; we examined it carefully.

RGV

 

[oso0690]

Thank you for correcting me.

So here's what I have:

1/(6-s) * (exp((6-s)t) where t is evaluated from 0 to infinity

1/(6-s) * ( exp((6-s)*inf) - exp((6-s)*0) )

For s > 6

1/(6-s) * (0 - 1)

-1/(6-s)

1/(s-6)

For s < or equal to 6, L.T. does not exist.

Is this correct?

Btw thanks to both of you for helping me.

 

[vela]

Remember s is complex, so it doesn't make sense to say s>6. What you should say is Re(s)>6, where Re(s) denotes the real part of s. The half plane Re(s)>6 is called the region of convergence.