Well, see the thing is that there's this formula in my book which I have absolutely no idea as to how they got... here it is... could someone show a proof of this?
[tex]\int_{-\infty}^{\infty} e^{ix(k'-k)} dx = 2\pi \cdot \delta(k' - k)[/tex]
It makes absolutely no sense to me... if we have that [tex]k' \neq k[/tex], then
[tex]\int_{-\infty}^{\infty} e^{ix(k'-k)} dx = \int_{-\infty}^{\infty} \cos [(k'-k) x] dx[/tex]
since sin is an odd function. But even this is a large step to claim... it's like saying that 1 - 1 + 1 - 1 + 1 - 1 + ... = 0. But can we just say that cos is a sideways shift of sin, so we can just say that it evaluates to zero as well? Or do the limits mean that we take a value that approaches infinity and it always equals zero? I think I understand why the k'=k case works, however...delta(0) = infinity, which is also the lhs. Lastly, why does the 2 pi have to be on the RHS?