- #1

- 8

- 0

Let

[itex]f=f(x)[/itex] and let f be fourier transformable (whatever that implies)

Let

[itex]\tilde{f}(k)=∫^{\infty}_{-\infty}dx e^{-ikx}f(x)[/itex] (1)

then we should have:

[itex]f(x)=∫^{\infty}_{-\infty}dk e^{ikx}f(k)[/itex] (2)

This implies:

[itex]f(x)=∫^{\infty}_{-\infty}dk e^{ikx}∫^{\infty}_{-\infty}dx' e^{-ikx'}f(x')[/itex] (3)

Note that x'≠x

My solution is as follows:

[itex]f(x)=∫^{\infty}_{-\infty}dk e^{ikx}∫^{\infty}_{-\infty}dx' e^{-ikx'}f(x')[/itex] (4)

[itex]f(x)=∫^{\infty}_{-\infty}dk ∫^{\infty}_{-\infty}dx' e^{-ik(x'-x)}f(x')[/itex] (5)

[itex]f(x)=∫^{\infty}_{-\infty}dx' ∫^{\infty}_{-\infty}dk e^{-ik(x'-x)}f(x')[/itex] (6)

[itex]f(x)=∫^{\infty}_{-\infty}dx' δ(x'-x)f(x')[/itex] (7)

[itex]f(x)=f(x)[/itex] (8)

Is this correct? Step (6) to (7) bothers me. And what about the change in integration variables? I guess that is correct as well?