When considering the inverse Fourier transform of a (1D) signal $x(t)$ :
$x(t) = \int{X(f)e^{j2\pi ft}df}$
considering that the forward transform to get the Fourier transform coefficients, denoted with capital X, and I use $f$ for frequency, is:
$ X(f) = \int{x(t)e^{-j2\pi f t}dt} $
(I didn't put any bounds on the integrals since it's not really important here I guess.)
It seems to make sense (sort of intuitively) that one can fall back to the original function because the basis vectors (complex exponentials) cancel each other with the exponent rule of multiplication $a^na^m=a^{n+m}$, - (that is the intuition that I have, even though I'm not 100% sure, but would be surprised otherwise) - more precisely I am tempted to write:
$x(t) = \int{ (\int{x(t)e^{-j2\pi f t}dt})e^{j2\pi f t} df}$
It seems very nice to think that there would be some way of rewriting this (last) integral in order to have the complex exponents cancel each other, i.e.: $e^{-j2\pi f t}e^{j2\pi f t} = e^{-j2\pi f t + j2\pi f t} = e^{0} = 1$ , therefore leaving the integral with just $\int{x(t)dt}$ and since there always is a normalization factor (perhaps 1/length(x)) then x is recovered.
Can someone derive this step by step please, if it is possible to derive it from that last formula? That would be very interesting to gain a better understanding of the inverse Fourier transform.
As it has been mentioned in the comments, it might be too complex to demonstrate here in fact. However, could someone confirm if the intuition that the $e^{-j...}$ and $e^{j...}$ in fact cancel each other resulting in $e^0 = 1$ at some point and therefore allows the recovery of the function ? (I mean can someone confirm that with possibly a mathematical argument but not with a full demo since it's too complex most likely)