Show that$$(1)\lim\limits_{R \uparrow \infty}\frac{1}{2\pi}\int_{-R}^{R}e^{-i\mu x}\hat{f}_{ab}(\mu)d\mu = f_{ab}(x)$$
where $f_{ab}(x)$ and $\hat{f}_{ab}(\mu)$ are:
More clarifications:
Let the Fourier transform $\hat{f}(\mu)$ of a function $f(x)$ specified on $\mathbb R$ is defined (often) by the formula:
$\hat{f}(\mu) = \int_{-\infty}^{\infty}e^{i\mu x}f(x)dx$ for $\mu \in \mathbb C$ whenever the integral makes sense.
Let $f_{ab}(x) = 1$ for $a \le x \le b$ and $f_{ab}(x) = 0$ for $x \neq [a,b]$.
the points $x$ are $\in \mathbb R$ except for $a, b$.
My intuition:
I have the proof of the following equation (so I can use it), and I'd like to know how I can use that equation $(2)$ to show $(1)$.
Let $a_1 \lt b_1 \le a_2 \lt b_2 \le ... \le a_{n-1} \lt b_{n-1} \le a_n \lt b_n$ and let $$f(x) = \sum_{j=1}^nc_jf_{a_jb_j}(x).$$
Then this holds, $$(2)\int_{-\infty}^{\infty}|f(x)|^2dx = \frac{1}{2\pi}\int_{-\infty}^{\infty}|\hat{f}(\mu)|^2d\mu$$
In other words, It's not required, but I think this is the way to go and show that $f_{ab}(x)$ can be recovered from its transform $\hat{f}_{ab}(\mu)$ by means of the inverse Fourier transform, for all points $x\in \mathbb R$ except for $a$ and $b$.
You want to show that the following converges to $\chi_{[a,b](x)}$ as $R\rightarrow\infty$ for $x\ne a$ and $x\ne b$: $$ \frac{1}{\sqrt{2\pi}}\int_{-R}^{R}e^{isx}\widehat{\chi_{[a,b]}}(s)ds \\ = \frac{1}{2\pi}\int_{-R}^{R} e^{isx}\int_a^b e^{-isu}duds \\ = \frac{1}{2\pi}\int_{a}^{b}\int_{-R}^{R}e^{is(x-u)}ds du \\ = \frac{1}{\pi}\int_a^b\frac{\sin(R(u-x))}{u-x} du \\ = \frac{1}{\pi}\int_{a-x}^{b-x}\frac{\sin(Rv)}{v}dv \\ = \frac{1}{\pi}\int_{R(a-x)}^{R(b-x)}\frac{\sin(w)}{w}dw $$ If $x < a < b$ or if $a < b < x$, then the above clearly tends to $0$ as $R\rightarrow\infty$ because the upper and lower limits both tend to $\infty$ or both tend to $-\infty$. If $a < x < b$, then the above tends to the improper integral $$ \frac{2}{\pi}\int_{0}^{\infty}\frac{\sin(w)}{w}dw = 1. $$ (You can look up this improper integral in any number of sources.)