How do I show that the synthesis equation of the Fourier Transform equals the original function?
I want to expand equation 2 using equation 1 and show that the integral indeed equals the original function. I am fine with some handwavy math.
I have seen this derivation of the Fourier Transform:
where the Fourier Transform is the limit of Fourier series. But that notation uses $\omega$ and also doesn't directly expand the integral. I am looking for something along these lines:
Only the component that was at frequency $\xi$ can produce a non-zero value of the infinite integral, because all the other shifted components are oscillatory and integrate to zero.
Inserting (1) into (2) and formally swapping the order of integration yields $$\int_{-\infty}^\infty \left( \int_{-\infty}^\infty f(y) e^{-i2\pi\xi y} dy \right) e^{i2\pi\xi x}d\xi =\int_{-\infty}^\infty \int_{-\infty}^\infty e^{i2\pi\xi(x-y)} f(y) dy d\xi = \int_{-\infty}^\infty f(y)\left(\int_{-\infty}^\infty e^{i2\pi\xi(x-y)} d\xi\right) dy$$
Formally, the inner integral on the right is the Dirac delta $\delta(y-x)$ (since the Fourier transform of $1$ is a Dirac delta, due to $1$ only containing one frequency), so then we get
$$ \int_{-\infty}^\infty f(y)\delta(y-x)dy = f(x)$$
Of course, this is all hand-wavy and you need to appeal to the theory of distributions to make this argument rigorous. The standard proof of the inversion formula uses an approximation argument: approximate a Dirac delta by a sequence of Gaussians, whose Fourier transforms are also Gaussians, and pass to the limit.