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I'm trying to calculate a certain integral that in a specific regime has the form

$ \int_{0}^{\infty} \mathrm{d}k \int_{-\infty}^{+\infty} \mathrm{d}\sigma f(k) e^{i \sigma k} $

Since $f$ doesn't depend on sigma, I integrate the exponential to get $ 2 \pi \delta(k) $. But the point where the delta blows up is at the end point of the integral and I don't know if what I'm doing is actually valid. This integral is actually a part of a calculation that, if I do it naivelly, is giving me the wrong result by a factor of 1/2. Because of this question (and answer), Delta (Dirac) function integral, I'm suspicious that the problem lies with the delta there.

Can someone help me make sense of it?

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  • $\begingroup$ How do you get $2\pi\delta(k) = \int_{-\infty}^{+\infty} e^{i \sigma k} \mathrm{d}\sigma$? I don't think that is true. $\endgroup$ – Paul Sinclair Sep 4 '15 at 0:39
  • $\begingroup$ It is the usual definition of $ \delta(x)$, like in eq. 31 here: mathworld.wolfram.com/DeltaFunction.html $\endgroup$ – Gabriel Cozzella Sep 4 '15 at 0:47
  • $\begingroup$ It's been too long since i've looked at fourier transforms when I can't even recognize one. The delta function can indeed be tricky, since it's really just a shorthand way of saying something other than it looks like. But in this case I think your problem is that you are integrating $k$ from $0$ to $\infty$ instead of $-\infty$ to $\infty$. Thus you are only picking up one side of the delta function, and getting half the value. $\endgroup$ – Paul Sinclair Sep 4 '15 at 1:45
  • $\begingroup$ I thought of the following to justify this: I can integrate from $-\infty$ to $+\infty$ and multiply the integral by $\theta(k)$. Then the $\delta$ picks up $\theta(0)$ which usually is defined as 1/2. $\endgroup$ – Gabriel Cozzella Sep 4 '15 at 12:59

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