# Fourier Transform Proof

## 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.

Last edited:

MathematicalPhysicist
Gold Member
Use the fact that: ##\int_{-\infty}^{\infty}\exp(ik(x-x'))dk/2\pi = \delta(x-x')##.

Use the fact that: ##\int_{-\infty}^{\infty}\exp(ik(x-x'))dk/2\pi = \delta(x-x')##.
Is there an issue with the the bounds of integration in this case? I wouldn't think so but I'm not positive.

MathematicalPhysicist
Gold Member
Is there an issue with the the bounds of integration in this case? I wouldn't think so but I'm not positive.
No there's no issue.

I think but not sure that that is one of the many definitions of Dirac Delta Distribution (DDD).
Anyway, you can search google or wiki for further explanations.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
No there's no issue.

I think but not sure that that is one of the many definitions of Dirac Delta Distribution (DDD).
Anyway, you can search google or wiki for further explanations.
If you take that as the definition of the delta distribution then you must show that it has the other properties that are more common to use for its definition. If you define it (more normally) by its integral property, then you must show that what is given is actually a representation of the delta distribution. This is often done in a hand-waving manner in introductory courses.

Ray Vickson
Homework Helper
Dearly Missed

## 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.

Last edited: