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4

It's much simpler than what you're describing, if I understand you correctly. Assuming the space index comes first, you can simply call Fourier /@ data, where data is your cube of data. For example: dat = RandomReal[{0, 1}, {3, 100, 200, 300}]; recipDat = Fourier /@ A; Otherwise, Transpose the indices so that the spatial index (ie, the one with length 3) ...


5

Adding assumptions finds two of the cases. Assuming[{a, t, r, z} ∈ Reals && t > z, Integrate[ DiracDelta[ t - a - (r^2 + (z - a)^2)^(1/2)]/(4 Pi (r^2 + (z - a)^2)^(1/2)), {a, -Infinity, Infinity}] ] (* 1/(4 π t - 4 π z) *) Assuming[{a, t, r, z} ∈ Reals && t < z, Integrate[ DiracDelta[ t - a - (r^2 + (z - ...


1

Although Mathematica cannot solve Integrate[DiracDelta[g[x]] h[x], x] as written, it can be transformed to Integrate[DiracDelta[x-x0] h[x] / (Abs[D[g, x]]/.x -> x0), x] using standard formulas, where g[x0] == 0. (Note that, if g[x] has multiple real zeroes, the preceding expression must be summed over those zeroes.) This can be applied to the ...


0

Try this: 1/(4 Pi (r^2 + (z - a)^2)^(1/2)) /. (Solve[ t - a - (r^2 + (z - a)^2)^(1/2) == 0, a][[1, 1]]) // Simplify (* 1/(2 \[Pi] Sqrt[(r^2 + (t - z)^2)^2/(t - z)^2]) *) Have fun!



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