Laplace transform of a sawtooth wave 
Find the Laplace transform of the periodic function such that $f(t) = t$ if $0\leq t < 2\pi$

I am having trouble setting up this question. Am I on the right path? 
$$ 
\mathcal{L}\{f(t)\} = \int_0^{2\pi} t e^{-st} dt
$$
 A: Let $w(t)$ be one tooth of this function: $w(t)=t$ when $0\le t<2\pi$ and $w(t)=0$ otherwise. Then the whole sawtooth function is an infinite sum of such teeth: 
$$f(t) = w(t)+w(t-2\pi)+w(t-4\pi)+\dots$$
The integral you've got is $\mathcal{L}[w]$, which is
$$W(s) = \int_0^{2\pi} t e^{-st} dt = \frac{e^{-4\pi s}-e^{-2\pi s}}{s^2} - \frac{2\pi e^{-2\pi s}}{s}$$
Recalling the time shift property of the Laplace transform, you'll find that 
$$
F(s) =W(s) + e^{-2\pi s}W(s) +  e^{-4\pi s}W(s) +\dots
$$ 
where you can factor out $W(s)$ and sum the geometric series.
A: You can also answer this question without integration if you are familiar with delays. The sawtooth plus the floor function is the unit ramp. So if $F(s)$ is the Laplace transform of the floor function and $G(s)$ that of the sawtooth you defined, then,
$$
G(s) + F(s) = \frac1{s^2}.
$$
On the other hand, the floor function plus the Heaviside, all delayed by one, is simply the floor function. Hence the Laplace transform of the floor function satisfies the following functional equation,
$$
\left(F(s)+\frac1s\right)e^{-s} = F(s).
$$
Therefore,
$$
F(s) = \frac1{s(e^s-1)},
$$
and finally,
$$
G(s) = \frac1{s^2} - \frac1{s(e^s-1)}.
$$
A: You are close. One way to solve the Laplace transform of a periodic function is:
$$\mathscr L\{f(t)\} = \frac{1}{1-e^{-sT}}\int_{0}^{T}e^{-st}f(t)dt$$
For your problem:
The period is: $T=2\pi$
The sawtooth function is given as: $f(t) = t$  if $0\le t < 2\pi$
All you have to do from there is plug in and solve the integral. You were just missing the  $\frac{1}{1-e^{-sT}}$ before the integral sign.
