# Laplace transform
From now on, LT is short for Laplace Transform
What is LT? It's denoted as $\mathcal{L}$ and it's an operator defined as the following:
$$\underset{ =F(s) }{ \mathcal{L}\{f(t)\}(s) }:=\int _{0}^\infty e^{-st}f(t) dt=\lim_{ T \to \infty }\int _{0} ^T e^{-st}f(t)\, dt$$
This doesn't look like anything useful, but later on we will learn how it is.

Here are some trivial example computations:
$\mathcal{L}\{0\}=0$ Look at your bank account, integrate 0 you still get 0 :D
$\mathcal{L}\{1\}=\int_{0}^\infty e^{-st} \, dt=-\frac{1}{s}e^{-st}|_{0}^\infty=\frac{1}{s}$ if s>0
$\mathcal{L}\{e^{at}\}$=$\int_{0}^{\infty} e^{at}e^{-st}\, dt=\int_{0}^{\infty}e^{-(s-a)t}dt=\frac{1}{s-a}$ if $s-a>0$
$\mathcal{L}\{\sin bt\}=\int _{0}^{\infty}e^{-st}\sin(bt) \, dt=\frac{b}{s^2+b^2}$ by integration by parts
similarly can be done for cos, but we have run out of time.
#end of lec 14 #start of lec 15
compute the LT of this funny function:
$f(t)=\begin{pmatrix}1, & 0\leq t\leq 1 \\ 2, & 1<t\leq 2 \\0, & 2<t \end{pmatrix}$
$F(s)=\int _{0}^1 e^{-st}\, dt+2\int _{1}^2 e^{-st}\, dt+0$
$=-\frac{1}{s}e^{-st}|_{t=0}^{t=1}-\frac{2}{s}e^{-st}|_{t=1}^{t=2}=-\frac{1}{s}(e^{-s}-1)-\frac{2}{5}(e^{-2s}-e^{-s})$
We have shown how to compute the LT of a choppy function.

$\mathcal{L}\{\alpha f(t)+\beta g(t)\}=\alpha \mathcal{L}\{f\}+\beta y\mathcal{L}\{g\}$
the LT is a linear operator. as shown above. Proof:
$\mathcal{L}\{\alpha f(t)+\beta g(t)\}=\alpha \int_{0}^{\infty}f(t)e^{-st}  \, dt+\beta \int_{0}^{\infty}g(t)e^{-st}  \, dt \quad \Box$

Def: $f(t)$ is piecewise continuous on an interval $I$ if $f(t)$ is continuous on $I$, except possibly at a /finite/ number of points of /jump/ discontinuity
what is continuouity? the limit exists and equals the value at that point.
![[Drawing 2023-10-11 13.17.32.excalidraw]]
Def: f(t) is of exponential order $\alpha$ if there $\exists\ T,\ M, T\geq0$ such that $\forall\ t>T : f(t)\leq Me^{\alpha t}$
this is important so that f(t)e^{-st} doesn't go into infinity,
this can be proven, but not necessary
Theorem: If $f(t)$ is piecewise continuous and of an exponential order $\alpha$ on $[0, \infty)$, then $\mathcal{L}\{f\}$ exists $s>\alpha$
and we are guarenteed that $s>\alpha$
$e^{\alpha t}$ is this of exponential order? Of course, $M=1,\alpha=\alpha$
what about $\sin(t)$? yes, sin is bounded b/w -1 and 1 
What about $e^{t^2}$? No, t^2 always outgrows $\alpha t$ eventually. This function does not have a LT.

![[Drawing 2023-10-11 13.21.18.excalidraw]]
Theorem if $F(s)$ is a LT then $\lim_{ s \to \infty }F(s)=0$

properties:
assume LT of f(t) exists:
first property:  $\mathcal{L}\{e^{\alpha t}f(t)\}=\int _{0}^\infty e^{\alpha t}e^{-st}\, dt=\int _{0}^\infty e^{-(s-a)t}f(t)\, dt=F(s-a)$

$\mathcal{L}\{e^{\alpha t}\cdot 1\}=\frac{1}{s-\alpha}$
$\mathcal{L}\{e^{\alpha t}\sin(bt)\}=\frac{b}{(s-a)^2+b^2}$ these properties are essential for the midterm.

using integration by parts:
$\mathcal{L}\{f'(t)\}(s)=\int _{0}^\infty e^{-st}f(t)\, dt=e^{-st}f(t)|_{t=0}^{t\to \infty}+s \underbrace{ \int e^{-st}f(t) \, dt }_{ F(s) }$$=sF(s)-f(0)$
$\mathcal{L}{f''}=s^2F(s)-sf(0)-f'(0))$
using proof by induction (try at home!):
$\mathcal{L}\{f^{(m)}\}=s^mF(s)-s^{m-1}f(0)-\dots-f^{m-1}(0)$

trig stuff:
$\mathcal{L}\{\sin bt\}=\frac{b}{s^2+b^2}$
$\mathcal{L}\{\cos bt\}=\frac{s}{s^2+b^2}$
$\cos(bt)=\frac{1}{b}(\sin(bt))'$

$\frac{dF}{ds}(s)=\frac{d}{ds}\int _{0 } ^\infty e^{-st}f(t)\, dt=\int _{0} ^\infty f(t) \frac{d}{ds}(e^{-st}) \, dt$
$=-\int _{0} ^\infty e^{-st}tf(t)\, dt$
$-\frac{dF}{ds}=\mathcal{L}\{tf(t)\}$; again we can use induction to prove:
$\mathcal{L}\{t''f(t)\}=(-1)^{n} \frac{d^nF}{ds^n}$
$\mathcal{L}\{t 1\}=-(s^-1)^1=s^-2$; $\mathcal{L}\{t^n\}= \frac{n!}{s^{n+1}}$ (can be proven by induction)
today covers all midterm material. Yay!
#end of lec 15