How is the Inverse Fourier Transform derived from the Fourier Transform? Where does the $\frac{1}{2 \pi}$ come from in this pair?
Please try to explain the Plancherel's theorem and the Parseval's theorem!
$ X(j \omega)=\int_{-\infty}^\infty x(t) e^{-j \omega t}d t$
$ x(t)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} X(j \omega)e^{j \omega t}d \omega $
 A: I am going to "derive" this heuristically, because the question is concerned with the origin of the $1/(2 \pi)$ factor.  Questions about integrability, order of integration, limts, etc., are to be smoothed over here (unless, of course, I have erred somewhere in the derivation - then all bets are of course off).
Consider the FT
$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \, f(x) \, e^{i k x}$$
Now let's assume that the inverse FT may be written as
$$f(x) = A \int_{-\infty}^{\infty} dk \, \hat{f}(k) \, e^{-i k x}$$
where we are to show, again heuristically, that $A = 1/(2 \pi)$.  To do this, I am going to rewrite the IFT as follows:
$$f(x) = A \lim_{N \to \infty} \int_{-N}^{N} dk \, \hat{f}(k) \, e^{-i k x}$$
Now, substitute the above definition of the FT into the integral, and reverse the order of integration (to repeat, I am assuming that $f$ is such this step and all the others are OK).  In doing this, we get
$$f(x) = A \lim_{N \to \infty}  \int_{-\infty}^{\infty} dx' \, f(x') \, \int_{-N}^{N} dk \, e^{i k (x'-x)}$$
Evaluating the inner integral, we get a single integral back:
$$f(x) = A \lim_{N \to \infty}  \int_{-\infty}^{\infty} dx' \, f(x') \frac{e^{i N (x'-x)}-e^{-i N (x'-x)}}{i (x'-x)} = 2 A \lim_{N \to \infty}  \int_{-\infty}^{\infty} dx' \, f(x') \frac{\sin{N (x-x')}}{x-x'}$$
Now, as $N \to \infty$, the sinc kernel behaves as a distribution which is characterized by a sifting property. Thus, we may, in this limit, take $f$ out of the integral and replace it with the value at $x'=x$.  Thus we have
$$f(x) = 2 A f(x) \lim_{N \to \infty}  \int_{-\infty}^{\infty} dx'  \frac{\sin{N (x-x')}}{x-x'}$$
or, simplifying things a bit and incorporating $N$ into the integral, we get that
$$1 = 2 A \int_{-\infty}^{\infty} dy \frac{\sin{y}}{y} = 2 A \pi$$
Thus, $A = 1/(2 \pi)$ if the above steps are valid.
