Fourier Transform of a constant

I am tryung 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 arguement 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?

 

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It's been a while since I've studied this, but one thing I do recall is that rigorously you cannot speak of the values of the delta function, because it really isn't a function at all. It's a distribution, and that distribution has various representations. One representation of the delta function is, as you have discovered,

\(\delta(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ikx}dx\)

The representation on the right hand side isn't required to vanish for x not equal to zero. (Indeed, the complex exponential isn't even integrable on \((-\infty,\infty)\)!)

What we do require is that:

\(\int_{-\epsilon}^{\epsilon}f(x)\delta(x)dx=f(0)\), for any \(\epsilon>0\).

 

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Hi neelakash,

The idea here is that for k different than zero the oscillations of the integrand should cancel in some sense. Nevertheless, you should be clear that the integral doesn't really exist as Tom2 said. To see things more clearly, insert a convergence factor \( e^{-\epsilon |x| } \) into your integral. You can now easily compute the integral and you will find a well known representation of the delta function as \( \epsilon \) goes to zero.

 

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Let f(x)=delta(x)
Then,Fourier Transform: F(k)=int{delta(x)exp[-ik*pi*x]dx}=exp[0]=1
Thus,Fourier transform of a delta function is a constant.
Now,let me carry out the inverse transform:
int{(1)exp[ik*pi*x]dk
Can you please tell me why is this delta(x)?

 

Did you perform the calculation I suggested?

 

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)