Derive the Laplace transform of the function
f(t)=te^-3t

I've started it using the integral definition of the Laplace, but haven't been able to integrate it yet.
I think my integral is from 0 to infinity of te^[(3+s)t]dt

Any help would be appreciated. Thanks!

Question

Derive the Laplace transform of the function
              f(t)=te^-3t

I've started it using the integral definition of the Laplace, but haven't been able to integrate it yet.
I think my integral is from 0 to infinity of te^[(3+s)t]dt

Any help would be appreciated.  Thanks!

anonymous · Answer

so u want to evaluate$$\int_0^\infty te^{-(3+s)t} \ \text{d}t$$

anonymous · Answer

id say use integration by parts

turingtest · Answer

I will get to this after breakfast if nobody else does :)

anonymous · Answer

Sorry for the delay, I had to get to class.  Mukushla, I think that's the integral I'm working with but I formed it myself from the definition of the Laplace.

turingtest · Answer

$$\mathcal L\{te^{-3t}\}$$correct?

anonymous · Answer

Yes

turingtest · Answer

you have your integral set up correctly, where are you stuck?

anonymous · Answer

I'm not sure how to do the integration, is it integration by parts?

turingtest · Answer

yeah but I should stop talking now as it seems I am getting the wrong answer :/
give me a moment please to see if I can find my error

anonymous · Answer

Ok, thanks for your help :)

turingtest · Answer

Oh man I was spacing... I had just typed the wrong thing into wolfram so it looked wrong
ok yes, start by integration by parts...

anonymous · Answer

Does it come out to be $$-\frac{ 1 }{ (3-s)^2 }$$

turingtest · Answer

no minus signs, but aside from that, yes

anonymous · Answer

ok, I'll check my signs again.  Thanks :)