# Easy Laplace Transform

1. Nov 18, 2013

### iRaid

1. The problem statement, all variables and given/known data
Apply the definition in (1) to find directly the Laplace transforms of the functions described (by formula or graph).
1) f(t)=t

2. Relevant equations

3. The attempt at a solution
Seems pretty easy... Question is, I don't understand the directions exactly.. Am I allowed to use the tables or do I have to solve the transform? Anyway this is how I solved, but i got stuck..
$$\mathcal{L}(f(t))=\int_0^{\infty}e^{-st}f(t)dt=\int_0^{\infty}te^{-st}dt$$
Then using an integration by parts, u=t and dv=e-stdt, I get:
$$\frac{-te^{st}}{s}\bigg|_0^{\infty}-\int_0^{\infty}\frac{-e^{st}}{s}dt=\left(\frac{-te^{st}}{s}-\frac{e^{s}t^{s}}{s^{2}}\right)\bigg|_0^{\infty}$$
Is that right up to there and what do I do after?

2. Nov 18, 2013

### Dick

Yes, you solve it by integrating by parts and you are starting with right parts. But it goes badly after that. How did $e^{-st}$ turn into $e^{st}$? And I have no idea how you got some of the other stuff.

3. Nov 18, 2013

### iRaid

Yes I messed up, it should be u= t, du=dt; $dv=e^{-st}dt$, $v=\frac{-e^-st}{s}$ Which leads to:
$$\frac{-te^{-st}}{s}-\int\frac{-e^{-st}}{s}dt=\frac{-te^{-st}}{s}-\frac{e^{-st}}{s^{2}}$$

Right?

4. Nov 18, 2013

### LCKurtz

On your last line you have dropped the minus sign in the $e^{-st}$. And check that antiderivative in red.

 Took too long to type it, others beat me to it.

5. Nov 18, 2013

### Dick

That looks much better. Now you evaluate it between 0 and infinity. It's going to involve thinking about limits. Take s>0.

6. Nov 18, 2013

### iRaid

$$\lim_{t \to \infty} \frac{-te^{-st}}{s}-\frac{e^{-st}}{s^{2}}$$
It seems like both would go to 0... Maybe I'm not seeing something correctly

7. Nov 18, 2013

### Dick

Yes, they both go to 0. That $te^{-st}$ goes to zero might take a little justification, but if you know why, that's fine. Now you have to take the limit as t->0 and take the difference.

8. Nov 18, 2013

### iRaid

$te^{-st}$ can be written as $\frac{t}{e^{st}}$ and then you can use L'Hopitals since you get $\frac{\infty}{\infty}$ and end up with $\frac{1}{se^{st}}=0$

AHHH now I see, I forgot you have to evaluate at 0 also...
So the first fraction goes to 0 and the second one will be e-s(0)=1 so you end up with: $\frac{1}{s^{2}}$

Thanks for the help!