Why is Fourier transform of exp(ix) a delta?

jasonc65
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Why is it that the Fourier transform of e^{2\pi ikx} is equal to \delta(k) ? The delta function is supposed to be zero except at one point. But the integral doesn't converge for k \ne 0. Yet I see a lot of books on QFT use this identity.
 
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delta isn't a function (what is it at the point where it is not zero?)

It is the delta 'function' because it behaves as the delta function.
 
Suppose you were to ask for the Fourier Series for f(x)= cos(x)?

Since the Fourier Series is, by definition, a sum of sines and cosines that add to f(x).
Since f(x)= cos(x), its Fourier series coefficients are just a1= 1, all other coefficients are 0. The delta "function" (it's really a "distribution" or "generalized function") is the functional version of that.
 
HallsofIvy said:
Suppose you were to ask for the Fourier Series for f(x)= cos(x)?

Since the Fourier Series is, by definition, a sum of sines and cosines that add to f(x).
Since f(x)= cos(x), its Fourier series coefficients are just a1= 1, all other coefficients are 0. The delta "function" (it's really a "distribution" or "generalized function") is the functional version of that.
Very interesting. The integral \int^\infty_{-\infty}e^{2\pi ikx} dx does in some ways behave like a delta function. And the delta function is an ideal function. However it's own Fourier transform is an exponential, which is a real funtion. The Fourier transform as an operator on Hilbert space is unitary, and squares to -1. Neither the delta function nor the exponential function are in Hilbert space, the latter because it doesn't satisfy boundary conditions, and the former because it isn't even a funtion. The idea is very informal and lacks rigour. I have never seen it given a rigorous basis.
 
Then get a book on "distributions" or "generalized functions" everything is done with complete rigor.
 
Thanks for the suggestion. :)
 
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