Dirac Delta function as a Fourier transform

[andrewkirk]

It is fairly easy to demonstrate that the Dirac delta function is the Fourier transform of the plane wave function, and hence that:

$\delta(x)=∫_{-∞}^{∞}e^{ikx}dk$ (eg Tannoudji et al 'Quantum Physics' Vol 1 p101 A-39)

Hence it should be the case that $∫_{-∞}^{∞}e^{ik}dk = \delta(1) = 0$

However the integral on the LHS does not even seem to exist, let alone equal zero. We can write:

$∫_{-∞}^{∞}e^{ik}dk = ∫_{-∞}^{∞}cos(k)+i sin(k) dk$

$= ∫_{-∞}^{∞}cos(k)dk+i ∫_{-∞}^{∞}sin(k)dk$

and neither of these integrals exist.

Yet, based on the Fourier argument, they must both exist and equal zero.

I have probably done something silly to get myself into this contradiction. Can somebody please help me by showing what it is?

 

[maverick_starstrider]

There are a number of things going on here I don't really understand but chronologically why does delta(1)=0? Also there is nothing trivial about a Dirac delta function, it's always been a trouble maker in terms of rigorous mathematics.

 

[strangerep]

I have probably done something silly to get myself into this contradiction. Can somebody please help me by showing what it is?

It seems you're missing the crucial piece of information that the Dirac delta is not an ordinary function in the usual sense. Rather it's a generalized function known as a "linear functional" or "distribution".

http://en.wikipedia.org/wiki/Distribution_(mathematics)

It only makes mathematical sense when integrated with another function, e.g.,
$$
\int_{-\infty}^{\infty}\!\! dx f(x) \delta(x-a) ~=~ f(a)
$$

 

[ParticleGrl]

Hence it should be the case that $∫_{-∞}^{∞}e^{ik}dk = \delta(1) = 0$

Rigorously, the dirac delta function is actually a distribution, not a function. Even still, you can do a bit of 'physics math' by adding a term like -epsilon*k^2 to the exponential in the integral to help it converge, and then limiting it away after the calculation.

 

[lugita15]

That is indeed the Cauchy principle value of the integral, so you can at least make some sense of what you got.