Deriving some Laplace transforms

[Jamin2112]

Homework Statement

Find the Laplace transform of each of the following functions:

...

2. f(t)=e^atcos(bt)

3. f(t)=tⁿ, were n is a positive integer

Homework Equations

As you well know, taking the Laplace of f(t) means ∫f(t)e^-stdt from 0 to ∞

The Attempt at a Solution

These problems are tripping me up, since integration by parts goes on forever.

∫e^atcos(bt)e^-stdt =
∫e^t(a-s)cos(bt)dt =

I suppose I should call e^t(a-s) "dv" (?)
----> v = ^t(a-s)/(a-s)
----> u = cos(bt)
----> du = -bsin(bt) dt

Still, I don't see where this gets me. Help, please!

 

[hgfalling]

Do the integration by parts again, and then collect the integral you are interested in on one side.

This is like the following simpler integral:

[tex] \int e^x \sin x dx [/tex]
[tex] \int e^x \sin x dx = -e^x \cos x - \int -e^x \cos x dx [/tex] (one IBP)
[tex] \int e^x \sin x dx = -e^x \cos x + e^x \sin x - \int e^x \sin x dx [/tex] (second IBP)
[tex] 2 \int e^x \sin x dx = e^x \sin x - e^x \cos x [/tex] (collect like terms)
etc...

 

[Jamin2112]

Do the integration by parts again, and then collect the integral you are interested in on one side.

This is like the following simpler integral:

[tex] \int e^x \sin x dx [/tex]
[tex] \int e^x \sin x dx = -e^x \cos x - \int -e^x \cos x dx [/tex] (one IBP)
[tex] \int e^x \sin x dx = -e^x \cos x + e^x \sin x - \int e^x \sin x dx [/tex] (second IBP)
[tex] 2 \int e^x \sin x dx = e^x \sin x - e^x \cos x [/tex] (collect like terms)
etc...

In the case of this particular problem, what integral am I interested in?

 

[hgfalling]

Well, you want

[tex] \int_{0}^{\infty} e^{t(a-s)} \cos (bt) dt [/tex]

which is related, but with, you know, constants and things.