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

1. Dec 10, 2004

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.

2. Dec 11, 2004

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.

3. Dec 11, 2004

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.

4. Dec 11, 2004

jasonc65

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.

5. Dec 12, 2004

HallsofIvy

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

6. Dec 13, 2004

jasonc65

Thanks for the suggestion. :)