Verifying inverse Fourier's Transform using Cauchy's integral formula I have the following exercise: 

Suppose $f(z)$ and $g(w)$, two analytic functions that are related by
  $$f_a(z) = \frac{1}{\sqrt{2\pi}} \int_{-a}^{a} e^{iwz} g(w) dw $$
  We define $f(z) = \lim_{a \rightarrow \infty} f_a(z)$ as the Fourier transform of $g(w)$. Use Cauchy's integral formula to verify that:
  $$g(w) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-iwz} f(z) dz$$

While I find this to be very, very interesting, I have no idea on how to approach this problem. I initally thought in rewriting $f_a(z)$ as an integral (from Cauchy's integral formula, assuming that $z_0$ can be any point) but I wasn't able to make the $z_0 - z$ "go away" from it. I thought, then, as applying the same trick to $g(w)$, but it was worse because now I had the integral of an integral that I dind't know how to solve. I'm also not sure about how to relate an integral in the real line with an integral on a closed path in $\mathbb{C}$. 
I'm pretty sure that this may be rather simple (and, why not, beautiful) so I'm posting this here to see your ideas about this.
Thanks in advance.
 A: Use
$$f_a(z) = \frac{1}{\sqrt{2\pi}} \int_{-a}^{a} e^{iwz} g(w) dw $$
in the equation
$$\begin{align}g(x)& = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}dz e^{-ixz} f(z) \\
& =  \lim_{a\to\infty}  \int_{-a}^{a}dw \left(\frac{1}{2\pi}\int_{-\infty}^{\infty}dze^{i(w-x)z}\right) g(w)  \\
\end{align}$$
The integral in parentheses amounts to the Dirac delta function :
$$\frac{1}{2\pi}\int_{-\infty}^{\infty}dze^{i(w-x)z}=\delta(w-x)$$
In the limit $a\to\infty$ the integration variable $w$ is guaranteed to hit $x$ since both are real, which gives
$$\begin{align}\lim_{a\to\infty}  \int_{-a}^{a}dw \delta(w-x) g(w)  =g(x)
\end{align}$$
as expected. What the exercise is probably trying to teach you is the fact that the localization of the Dirac delta function can be represented as a Cauchy contour integral on the space of holomorphic functions :
$$\begin{align} g(x)=\int_{-\infty}^{\infty}dw \delta(w-x) g(w)  \equiv \frac{1}{2\pi i}\oint_{|z-x|=\epsilon}dz\frac{g(z)}{z-x}
\end{align}$$
which relies on the fact that $g(z)$, being analytic, is finite at $z=x$ and has no branch cuts in the vicinity.
