Matt Chu said:
Homework Statement
Given a continuous non-periodic function, its Fourier transform is defined as:
$$f(x) = \int_{-\infty}^\infty c(k) e^{ikx} dk, \ \ \ \ \ \ \ \ \ \ \ \ \ c(k) = \frac{1}{2\pi} \int_{-\infty}^\infty f(x) e^{-ikx} dx$$
The problem is proving this is true by evaluating the integral when ##c(k)## is plugged into the equation for ##f(x)##.
Homework Equations
$$f(x) = \int_{-\infty}^\infty c(k) e^{ikx} dk$$
$$c(k) = \frac{1}{2\pi} \int_{-\infty}^\infty f(x) e^{-ikx} dx$$
The Attempt at a Solution
This ends up with a long integral:
$$f(x) = \int_{-\infty}^\infty \left( \frac{1}{2\pi} \int_{-\infty}^\infty f(x') e^{-ikx'} dx' \right) e^{ikx} dk$$
I'm not sure really how to proceed from here. I moved the ##e^{ikx}## into the inner integral, which I figured was fine since it's constant relative to ##x'##.
$$f(x) = \int_{-\infty}^\infty \left( \frac{1}{2\pi} \int_{-\infty}^\infty f(x') e^{ik(x-x')} dx' \right) dk$$
I tried to kill at least one of the integrals by seeing if something evaluated to a Dirac Delta but I can't seem to get that result. I also tried integrating by parts, but that led me nowhere.
You can do it by a limiting argument; that is the way every treatment I have ever seen has done it. Define
$$c(k) = \frac{1}{2\pi} \int_{-\infty}^\infty f(x') e^{-ikx'} dx'$$
and
$$f_K(x) = \int_{-K}^K c(k) e^{ikx} dk$$
We want the value of ##\lim_{K \to \infty} f_K(x).##
Note that
$$f_K(x) = \int_{ -\infty}^{\infty} f(x') D_K(x-x') \, dx' = \int_{-\infty}^\infty f(x-y) D_K(y) \, dy$$
where
$$D_K(y) = \frac{1}{2 \pi} \int_{-K}^K e^{iky} \, dk = \frac{1}{\pi} \frac{\sin(Ky)}{y}.$$
The function ##D_K(y)## is sharply peaked at ##y = 0##, and ##\int_{-\infty}^{\infty} D_K(y) \, dy = 1##--see the link below. Intuitively, this suggests that as ##K \to \infty## the integral of ## f(x-y) D_K(y)## will just pick out the value ##f(x)## coming from ##y = 0##. That is, ##\lim_{K \to \infty} D_K(y) = \delta(y).##
In fact, this is not quite true. What can be proven is that for "reasonably nice functions" ##f(x)##---absolutely integrable, of bounded variation on finite intervals and having a finite number of jump discontinuities on finite intervals---we will have
$$\lim_{K \to \infty} f_K(x) = \frac{1}{2} [ f(x+) + f(x-) ],$$
where ##f(x+)## and ##f(x-)## are the right-hand and left-hand limits of ##f(y)## as ##y \to x##; that is,
$$ f(x+) = \lim_{y \downarrow x} f(y), \;\; f(x-) = \lim_{y \uparrow x} f(y).$$ Of course we have ##f_K(x) \to f(x)## at any point where ##f(x)## is continuous.
The function ##D_K(y)## is called the Dirichlet Kernel, and is well-studied in Fourier series treatments; the convergence theorems proved for Fourier series go through for Fourier transforms, because they both involve limits of integrals like ##f_K(x)## above.
See, eg.,
http://www.sosmath.com/fourier/fourier3/fourier31.html#proof3 . Be sure to click on the word "proof" that appears a few times in that article. Better still, go to the library and take out a book about Fourier series.