Is the Fourier Transform of a Constant a Dirac Delta Function?

[Kolahal Bhattacharya]

Homework Statement

Fourier transform of a constant

Homework Equations

The Attempt at a Solution

I am trying to prove that Fourier Transform of a constant is a Dirac delta function.I have fed f(x)=1 in the formula of Forward Fourier transform and got F(k)=int{exp[-ik*pi*x]}dx
I know that this is delta function with argument k.But cannot prove it.It's sure to give infinity at k=0.But,for other values of k I am not getting the integral=0.So,an attempt through the way of definition fails.Can anyone help me?

 

[Meir Achuz]

Start with \delta(x) and Fourier transform it. You get a constant.

 

[Kolahal Bhattacharya]

Here is a method how to do it.It took some time,however.

First, you find the F.T. of delta(x) and it turns out as a constant F(k)=1
Then,you did inverse transform of 1 which yields a delta function.
Like this: int{exp[ik'x']dk'}=2*pi*delta(x')
Putting k'=x and x'=-k,we have
int{exp[-ikx]dx}=2*pi*delta(-k)

Now consider the F.T. of a constant function g(x)=c
G(k)=int{c exp[-ikx]dx}=c[2*pi*delta(-k)]=2*pi*c delta(k)

 

[Andrea B DG]

If you compute Int( exp(-ikx), x=-Infinity..Infinity) you get Infinity for k=0 and 2/k sin(k Infinity) that is undefined for k=/=0. To remove this ambiguity use the Residue theorem giving Int( exp(-ikz), z on C ) = 0 where C is the contour of a half-disk of radius R. Then using the parametrisation z = R exp (i alpha) and letting R->Infinity you will find Int( exp(-ikx), x=-Infinity..Infinity) = 0 for k =/=0. Using this reasoning you get that Int( exp(-ikx), x=-Infinity..Infinity) is "proportional" to delta(k).

To understand the factor 2*pi you have to work in the context of the definitions of the Fourier transform and of his inverse. These definitions results from the Fourier serie as T->Infinity, see http://en.wikipedia.org/wiki/Fourier_Transform.

 

[paranormal]

The Fourier transform of a constant function is indeed a Dirac delta function. This can be proved by using the definition of the Fourier transform and the properties of the Dirac delta function. Let's start with the definition of the Fourier transform:

F(k) = ∫f(x)e^(-ikx)dx

Since f(x) is a constant function, it can be written as f(x) = c, where c is a constant. Substituting this in the formula, we get:

F(k) = ∫c*e^(-ikx)dx

Using the property of the Dirac delta function, we can write this as:

F(k) = c*∫δ(x)e^(-ikx)dx

Now, according to the definition of the Dirac delta function, ∫δ(x)dx = 1. Therefore, the above equation becomes:

F(k) = c

This shows that for all values of k, the Fourier transform of a constant function is a constant value, which is equal to the original constant function. In other words, the Fourier transform of a constant function is a Dirac delta function centered at zero with a magnitude equal to the constant value. This can also be seen from the graph of the Fourier transform of a constant function, which is a spike at k=0 with a height equal to the constant value.

Therefore, it can be concluded that the Fourier transform of a constant function is a Dirac delta function.