- #1
SwordSmith
- 8
- 0
I am having problem with the inverse transformation of a Fourier transformed function which should give the function itself.
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?
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?