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I want to understand this and the step that I'm struggling with is that apparently

$$ \frac{1}{2\pi} \int_{-\infty}^\infty \exp \left (i k x \right) \text{d}k = \delta\left(x\right) $$

with $\delta$ delta distribution. I haven't done much with the delta distribution, so I don't understand how to arrive at the identity above. WolframAlpha tells me that the Cauchy principal value is 0. Which I guess makes sense, right...? Thanks for your help in advance.

integralette
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    So, the integral on the LHS doesn't actually exist as an ordinary integral. It needs to be interpreted with care – FShrike Apr 23 '23 at 19:50
  • @FShrike I'm sorry, but what is the conclusion of "LHS not an ordinary integral"? Is that a clue or did you just want to give a warning? I'm sorry if I don't understand. :) – integralette Apr 23 '23 at 20:00
  • @Gonçalo Aha! Thank you! That's interesting. So it's a Fourier transformation of 1. But I don't understand where your link explains my question from here? In the answer, in equation (2), my question just shows up again. Basically I don't understand how equation (2) of your link arrives at $\delta$. – integralette Apr 23 '23 at 20:02
  • The integral on the left-hand side does not converge as an improper integral. This means that it cannot be studied solely using the usual rules of calculus. However, the left-hand side can be rigorously interpreted as the "Fourier transform of a tempered distribution". While the concept of tempered distribution requires a certain level of mathematical maturity to understand, this particular example can be explained using a simpler tool called Gaussian regularization. – Sangchul Lee Apr 23 '23 at 20:10
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    Indeed, according to Gaussian regularization, the left-hand side can be seen as the "limit" of $$\frac{1}{2\pi}\int_{-\infty}^{\infty}\exp(ikx)\exp(-\varepsilon k^2),\mathrm{d}k = \frac{1}{2\sqrt{\pi\epsilon}}\exp(-x^2/4 \epsilon)$$ as $\varepsilon \to 0^+$. It can be shown that under the same "limit", the right-hand side also tends to $\delta(x)$. (If we regard the right-hand side as the density of some mass distribution over the real line, then the total mass is always $1$ and the mass concentrates to the origin as $\varepsilon \to 0^+$.) – Sangchul Lee Apr 23 '23 at 20:14

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