What is the difference between $e^{it}$ and $e^{st}$ I'm currently trying to make sense of the Laplace transform, but I'm a bit stumped by the $e^{-st}$ part of the transform.
I understand Euler's formula $e^{it} = \cos(t)+i\sin(t)$, but I don't understand why in the Laplace transform it is a whole complex number in the exponent?
I assume $e^{st}$ is not equal to $\cos(t) + i\sin(t)$.
Any help would be appreciated
 A: Laplace transform $F$ of a function $f$ is given by the formula $$F(s) = \int_0^\infty f(t)e^{-st}\,dt. \tag{1}$$
First, let's discuss what $t$ is. It's a dummy variable used for integration, it takes value in $[0,\infty)$. It often represents time in applications.
Note that $t$ would disappear in a closed-formula for $F$ after integration takes place, but $s$ will remain and therefore the above integral defines a function in variable $s$ (when it converges). So, $s$ is a complex variable, not a fixed number.
The imaginary unit $i$, on the other hand, is a fixed complex number, so comparing $s$ and $i$ in this context is like apples and oranges. For example, if we used $it$ instead of $-st$, if integral $$\int_0^\infty f(t)e^{it}\,dt\tag{2}$$ converges, the result is just another complex number, not a function of a complex variable. (Well, technically, any constant can be interpreted as a function, but that wouldn't be interesting here.)
So, to recapitulate, the expression $e^{-st}$ in this context has two parameters, $s$ a complex one, and $t$ non-negative real, so $(s,t)\mapsto e^{-st}$ is a function with domain $\mathbb C\times [0,\infty)$. The expression $e^{it}$ has only one parameter $t$, while $i$ is a constant, so $t\mapsto e^{it}$ is a function with domain $[0,\infty)$.
A: This is my opinion. The Laplace transform is usually used to convert a linear ODE to an algebraic equation to make it easier to find a solution. The transformation given by
$$
F(s) = \mathcal{L}\{f(t)\}
$$
has a polynomial common denominator. (The notation $F(s)$ shows that we are taking $s$ as a variable but not a constant.) For example,
$$
F(s) = \frac{\cdots}{(s-a) \cdots (s^2 + bs+ c) \cdots}
$$
where the order of the polynomial is not larger than the order of the ODE. From this transformation, someone observed that the roots of the polynomial have important information about the solution to the linear ODE. The roots are generally complex numbers as you might know. For that reason, it is natural to set $s = \sigma + i\omega$ as a complex number.
