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Convolution And deconvolution of functions

All this text is just verbalizing that the pdf of the sum of two independent random variates is the convolution of their respective pdf-s. That is if you have two rv-s, say $\mathbf {x,y}$ and their ...
hyportnex's user avatar
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1 vote
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Resolution and delta function

A 'delta function' is actually a distribution (a generalized function) that only can be computed in combination with an integration-over-a-range. The delta distribution is an ideal one to represent a ...
Whit3rd's user avatar
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Does the exponential representation of Dirac delta function depend on choice of Fourier convention?

The equation you wrote is a mathematical statement and as such it should not (and it doesn't) depend on any convention. Its proper meaning is intended in the sense of distribution. The distribution on ...
lcv's user avatar
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4 votes

Does the exponential representation of Dirac delta function depend on choice of Fourier convention?

The identity $$ \delta(\omega) = \frac{1}{2\pi} \int dt\, e^{i\omega t} $$ does not depend on the Fourier convention (but of course requires regularization of the integral). When you use a different ...
Sebastian Riese's user avatar
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Closed form expression of 2D CFT integral

A very common integral which is almost enough to prove this is \begin{equation} \int \frac{d^2 z}{|z - w_1|^{2\beta_1} |z - w_2|^{2\beta_2}} = \frac{\Gamma(\beta_1 + \beta_2 - 1)}{\Gamma(2 - \beta_1 - ...
Connor Behan's user avatar
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Closed form expression of 2D CFT integral

Disclaimer: The following is only an incomplete calculation/answer, but it may still serve as a first step. By translation and rotational symmetry, OP's integral can only depend on the distance $|w_{...
Qmechanic's user avatar
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