What do Fourier Transforms actually do? I will illustrate this with a simple example:

Consider the exponential decay function $$f(t)=\begin{cases}
 0  &
 \ t\lt 0 , \\ A e^{-\lambda t} & \ t\ge 0 \end{cases}$$
  Where $\lambda \gt 0$ and $A$ is a constant.

Using the formula for the Fourier transform of $f(t)$: 
$$\mathcal{F}(\omega)=\frac{1}{\sqrt{2\pi}}\int_{t=-\infty}^{\infty}f(t)e^{-i\omega t}\mathrm{d}t=
\frac{1}{\sqrt{2\pi}}\int_{t=-\infty}^{0}0e^{-i\omega t}\mathrm{d}t+\frac{A}{\sqrt{2\pi}}\int_{t=0}^{\infty}e^{-\left(\lambda+i\omega \right)t}\mathrm{d}t$$ 
                                                                                                                                $$=0 + \frac{A}{\sqrt{2\pi}}\left[\frac{e^{-\left(\lambda+i\omega \right)t}}{\lambda + i \omega}\right]_{t=0}^{\infty}=\color{purple}{\frac{A}{\sqrt{2\pi}(\lambda+i\omega )}}$$
So, I started with an exponential decay function $f(t)=A e^{-\lambda t}$ and now I have $$\color{purple}{\frac{A}{\sqrt{2\pi}(\lambda+i\omega )}}$$ What exactly does this expression marked purple mean? Or put slightly differently, what does it represent? What did the Fourier transform do to $f(t)$? Basically, what I'm saying is that I have absolutely no idea what the Fourier transform has achieved.
Could someone please spell this out to me in simple English so that I can get a 'feel' for what's going on?
Thank you kindly.   
 A: You are expanding a function $f$ in a basis for which differentiation $\frac{d}{dx}$ is diagonalized. That's why the Fourier transform is so useful for studying differential equations. That normalized basis is $e_{s}(x)=\frac{1}{\sqrt{2\pi}}e^{isx}$, and $e_s$ is an eigenfunction of differentiation with eigenvalue $is$, which gives you a diagonalization:
$$
                   \frac{d}{dx}e_{s} = ise_{s},\;\;\; -\infty < s < \infty.
$$
The eigenvalues are real if you look at $\frac{1}{i}\frac{d}{dx}$ instead as the fundamental operator; this operator is selfadjoint, and $\{ e_{s}(x)\}_{s=-\infty}^{\infty}$ is a type of basis that diagonalizes $\frac{1}{i}\frac{d}{dx}$ and has real eigenvalues.
If you look at the standard basis vectors $\{ e_n \}_{n=1}^{N}$ for $\mathbb{C}^{N}$ (same as $\mathbb{R}^{N}$ but you allow complex entries,) then you can expand $f \in \mathbb{C}^{N}$ along this basis with dot products
$$
                 f = \sum_{n=1}^{N}(f,e_n)e_n
$$
The coefficient of $e_n$ is the complex dot product $(f,e_n)$. The Fourier transform looks like a "continuous" (i.e., integral sum) of this type:
$$
                 f = \int_{-\infty}^{\infty}(f,e_s)e_s ds.
$$
The coefficient of $e_s$ is
$$
       (f,e_s) = \int_{-\infty}^{\infty}f(t)\overline{e_s(t)}dt = \hat{f}(s).
$$
So,
$$
                   f = \int_{-\infty}^{\infty}\hat{f}(s)e_{s}(x)ds.
$$
This basis diagonalizes $\frac{1}{i}\frac{d}{dx}$, which makes it appear that differentiation is transformed to multiplication:
$$
               \frac{1}{i}\frac{d}{dx}f = \int_{-\infty}^{\infty}\hat{f}(s)se_{s}(x)ds.
$$
This is the historical view of the Fourier transform and is basically how Fourier saw it.
