Taking the Inverse Fourier Transform of the Fourier Transform

In this video I show that you really do get the identity back when you take the Inverse Fourier transform of the Fourier transform.

To see how you derive the delta function integral representation, visit:

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[rafaeljc]

you just used the inverse fourier transform (from the integral that returns the delta function) to proof the inverse fourier transform

[electronic]

Thank for your video. It was difficult to understand since 3:50 but with a variable change with t=x-x' I get dt/dx'=-1 and when x'=-inf => t=+inf.
So - integral from +inf to -inf = +integral from -inf to +inf (doesn't change) and I get

Integral (t, -inf, +inf) delta(t) f(x-t) dt
= f(x-0) Integral (t, -inf, +inf) delta(t)
= f(x) 1
= f(x)
and finally with your explanation I understand dx' => x' (Yes! the x') must be = x to get delta(0) => f(x') becomes f(x) !

So I understand the magic way to express
1/(2 pi) Integral ( 1 e^( i k (x-x'), k, -inf, inf )
= ( F^-1{ 1 } ) (x-x') = Dirac (x-x')
Thanks

[shahriarahmadfahim]

In the first line (in green), why do you call that x' but not just x?

[supremex]

Except I dont get the original function back. If I use something like exp(-a|x|) I will get an extra factor of pi/a times f(x).

[pawewanat]

lim_{k->inf} e^{ikx} is not 0, for x != 0.

The same for -inf.

Therefore, this integral does not converge, does it?

[notgoodatmathmmm]

why can you interchange the integrals?