Intuition behind the proof of the Inverse Fourier Transform? I am interested in the proof of the Inverse Fourier Transform for absolutely integrable real valued functions.
The proof I have read asks you to consider an auxiliary function $g_{a}(x)$ defined as
$$g_{a}(x) = \int_{-\infty}^{\infty} e^{-a \lvert t\rvert}e^{2\pi i t x}\operatorname{d}\!t = \frac{2a}{4\pi^{2}x^{2}+a^{2}}.$$
It then proceeds to prove that $\lim_{a\rightarrow 0 }f*g_{a}(x) = f(x)$ and therefore shows that $\hat{\hat{f}}(x) = f(-x)$ by the dominated convergence theorem.
My question is as follows: Where did $g$ come from?
So far, I've noticed that $g(x)$ is the form of a polynomial of degree two higher than the numberator, so perhaps there is a relationship between this expression and hyperbolic cotangent or the digamma function? 
I've also noticed that $g_{a}(x)$ is a probability distribution function which looks a lot like the Cauchy Distribution. As a consequence to this, it feels like computing the Fourier transform of $g$ is going to be somewhat related to the Cauchy Distribution's characteristic function?
That being said, I have a feeling that I am just looking for similarities which are not useful at all. Can I gain any useful insights from those two observations? If not, how can I reasonably arrive at this integral equation for $g_{a}(t)$ "naturally"?
As for my meaning of "naturally", consider a standard epsilon-delta proof. When you let $\epsilon>0$ be given and consider $\delta = f(\epsilon)$, most of the intuition behind the proof itself was completely hidden by considering such a $\delta$ if you never show how you arrived at that $\delta$. In a similar vein, how do I reasonably arrive at $g_{a}(x)$?
 A: If you consider the operator $A=\frac{1}{i}\frac{d}{dx}$ on the domain consisting of periodic and differentiable functions on $[-\pi,\pi]$, then you can consider, as Cauchy did, $(\lambda I-A)^{-1}f$. The led to a expression with poles at $\lambda=0,\pm 1,\pm 2,\cdots$ and residue at $N$ equal to $\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)e^{-int}dt e^{inx}$. Cauchy noted that the sum of all such residues is the Fourier series sfor $f$. Cauchy then looked at trading all of the residues in the finite plane for a single residue at $\infty$, which would appear to be
$$
       \lim_{\lambda\rightarrow\infty}\lambda(\lambda I-A)^{-1}f=\lim_{\lambda\rightarrow\infty}\frac{\lambda}{\lambda I-A}f = f.
$$
This thought was never fully developed in Cauchy's lifetime, but it did pave the way for general and workable proofs about a century later for all kinds of classical expansions in orthogonal functions, not just the exponentials.
The same analysis works for the operator $A=\frac{1}{i}\frac{d}{dx}$ on $(-\infty,\infty)$ as well, and this leads to equating the above with an integral surrounding the real axis:
$$
          \frac{1}{2\pi i}\int_{-\infty}^{\infty}\{ ((t+i\epsilon)I-A)^{-1}f-((t-i\epsilon)I-A)^{-1}f \}dt
$$
This integral gives your integral with $a=\epsilon$ convolved with $f$, and it boils down to the Poisson integral representation of a harmonic function. The details are little tedious, but natural in this context.
