# Laplace transform of te^t

Pi Face

## Homework Statement

f(t)=te^t, find laplace

## The Attempt at a Solution

I started doing integration by parts and after doing it three times I wasn't sure if I was going in the right direction/making any progress. I'm not supposed to use a table to solve this (I have to do the integral out) so could anyone give me a hint as to how to start?

I can manipulate the original equation to
(0 to inf for all integrals)
∫ te^(t(1-s)) dt and then i set u=t, du=dt, dv=e^(t(1-s)), v=e^(t(1-s))/(1-s)
and then I end up with a longer expression and I need to integrate by parts again. is this is right direction?

## Answers and Replies

susskind_leon
It's the right direction. After all, your t-factor in the integrand goes away, so you just have to integrate e^(t*something).

Pi Face
I think I'm missing a step.

after doing interation by parts the first time, i get

te^[t(1-s)] / (1-s) - ∫ e^[t(1-s)] / (1-s) dt

=te^[t(1-s)] / (1-s) - e^[t(1-s)] / (1-s)^2

i then need to evaluate fro 1 to inf but what does that do to te^[t(1-s)], assuming s>1? t is inf and e^[t(1-s)] becomes 0?

EDIT: nevermind, I had a brainfart, I got the answer

Last edited:
sunjin09
I think I'm missing a step.

after doing interation by parts the first time, i get

te^[t(1-s)] / (1-s) - ∫ e^[t(1-s)] / (1-s) dt

=te^[t(1-s)] / (1-s) - e^[t(1-s)] / (1-s)^2

i then need to evaluate fro 1 to inf but what does that do to te^[t(1-s)], assuming s>1? t is inf and e^[t(1-s)] becomes 0?

Laplace transform generally has a region of convergence to the right of some axis parallel to the y-axis on the complex s plane, in this case this axis happens to be x=1.