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

[jasonc65]

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.

[matt grime]

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.

[HallsofIvy]

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.

[jasonc65]

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.

[HallsofIvy]

Then get a book on "distributions" or "generalized functions" everything is done with complete rigor.

[jasonc65]

Thanks for the suggestion. :)