# Inverse Fourier Transformation of a Fourier Transformation

1. Jan 14, 2012

### SwordSmith

I am having problem with the inverse transformation of a fourier transformed function which should give the function itself.

Let

$f=f(x)$ and let f be fourier transformable (whatever that implies)

Let

$\tilde{f}(k)=∫^{\infty}_{-\infty}dx e^{-ikx}f(x)$ (1)

then we should have:

$f(x)=∫^{\infty}_{-\infty}dk e^{ikx}f(k)$ (2)

This implies:

$f(x)=∫^{\infty}_{-\infty}dk e^{ikx}∫^{\infty}_{-\infty}dx' e^{-ikx'}f(x')$ (3)

Note that x'≠x

My solution is as follows:
$f(x)=∫^{\infty}_{-\infty}dk e^{ikx}∫^{\infty}_{-\infty}dx' e^{-ikx'}f(x')$ (4)

$f(x)=∫^{\infty}_{-\infty}dk ∫^{\infty}_{-\infty}dx' e^{-ik(x'-x)}f(x')$ (5)

$f(x)=∫^{\infty}_{-\infty}dx' ∫^{\infty}_{-\infty}dk e^{-ik(x'-x)}f(x')$ (6)

$f(x)=∫^{\infty}_{-\infty}dx' δ(x'-x)f(x')$ (7)

$f(x)=f(x)$ (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?

2. Jan 14, 2012

### marcusl

This looks correct to me. In step 6 you can move f(x') from the right integral into the left one, leaving int{ dk exp(-ik(x'-x)) }. This becomes the delta function in 7 as you state, through application of the principle of stationary phase.

3. Jan 14, 2012

### mathman

The analysis is essentially correct, except that you need to include 2π in the exponents. That is 2πikx or -2πikx'.

4. Jan 16, 2012

### marcusl

Oh, right! Mathman points out that I left off the normalization--sorry! You can solve the problem using the variables he mentions, or you can continue to use k but then the inverse transform has a normalizing constant 1/2π in front like this:

$$f(x)=\frac{1}{2\pi}∫^{\infty}_{-\infty} e^{ikx}\tilde{f}(k)dk.$$
The key piece of eq. (6) becomes

$$\frac{1}{2\pi}∫^{\infty}_{-\infty} e^{ik(x'-x)}dk$$
which is evaluated most easily as a limit

$$\frac{1}{2\pi}\lim_{r\rightarrow\infty}∫^{r}_{-r} e^{ik(x'-x)}dk.$$
The integral becomes

$$\lim_{r\rightarrow\infty}r\frac{\sin r(x'-x)}{r(x'-x)}$$
which is single-valued only if x'=x. At that point the ratio equals one by L'Hospital's rule, the overall expression becomes infinite, and this expression is a Dirac delta function.