Questions and Answers in MSE and two other references:
Updates.
Ad 2). The Fourier transform
is a special case of the double-sided Laplace transform:
$$
F(p) = \int_{-\infty}^{+\infty} e^{-pt}\,f(t) dt \quad \Longrightarrow \quad
F(i\omega) = \int_{-\infty}^{+\infty} e^{-i\omega t} f(t)\,dt
$$
The Fourier transform, in turn, is a generalization of the complex
Fourier series:
start with equation (20) in the Wolfram reference and read until the end.
Ad 5). In this reference
it is argued on page 23 (with obviously a typo in it) that the Boltzmann probability distribution $f(E)$ must have the following form:
$$
f(E_1) \times f(E_2) = h(E_1+E_2)
$$
Let's elaborate on this a little bit:
$$
f(E_1) \times f(E_2) = h(E_1+E_2) \quad \Longrightarrow \quad h(E) = h(E+0) = f(0)f(E)
$$
Derivative:
$$
h'(E) = f(0)f'(E) = \lim_{\delta\to 0} \frac{h(E+\delta)-h(E)}{\delta} =
\lim_{\delta\to 0} \frac{f(E)f(\delta)-f(0)f(E)}{\delta} =\\
f(E)\,\lim_{\delta\to 0} \frac{f(\delta)-f(0)}{\delta} =
f(E) f'(0) \quad \Longrightarrow \quad f'(E) = \frac{f'(0)}{f(0)} f(E)
$$
Continuing in terms of the article:
$$
\frac{df(E)}{f(E)} = \frac{-dE}{E_c} \quad \Longrightarrow \quad f(E) = A e^{-E/E_c}
$$
Which is the Ansatz for the Boltzmann distribution.
Ad 3). According to Wikipedia, De Moivre's formula is:
$$
\left[cos(x)+i\sin(x)\right]^n = \cos(nx) + i\sin(nx)
$$
And this can be proved for any integer $n$ , quite independent of Euler's formula.
It's rather the other way around: because of the pattern $\;f(x)^n = f(nx)\;$ , de Moivre's formula can be considered as a heuristics for Euler's formula.
But uniquesolution is quite right: <quote> Probably the most fundamental fact about it is that it is the only measurable function for which $f(x+y)=f(x)f(y)$ for all $x,y$ </quote> Can we mimic this behavior of $\,e^x$ with the function $f(x) = \cos(x)+i\sin(x)$ ? From trigonometry we know that:
$$
\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)\\
\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)
$$
Hence:
$$
f(x+y) = \cos(x+y) + i \sin(x+y) =\\ \left[\cos(x)\cos(y) - \sin(x)\sin(y)\right] + i \left[\sin(x)\cos(y) + \cos(x)\sin(y)\right]
=\\ \left[\cos(x) + i \sin(x)\right]\left[\cos(y) + i \sin(y)\right] = f(x)f(y)
$$
So our $f(x)$ behaves like an exponential function.