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edited Nov 8, 2022 at 13:13

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Analytic integration is much more stable than numerical integration, especially when infinite and/or multi-dimensional integrals are concerned:

f[v_f[u_, x_, z_] = Assuming[x \[Element] Reals && x != 0 && z \[Element] Reals && -1 < vu <> 10, Integrate[(Cos[-kx x] Cos[-kz z])/(kx^2 + ky^2 + kz^2 (1 - v^2)u*kz^2), {kz, -∞, ∞}, {kx, -∞, ∞}, {ky, -∞, ∞}]] (* (2 π^2)/Sqrt[-((-1 + v^2)Sqrt[u x^2) + z^2] *) eA[v_, t_]eA[t_] = 0.0579484*f[v0579484*f[0.000086, 3.2, 0.999957 t] (* 1.14386/Sqrt[0.999914 t^200088064 -+ 100.24 (-1 +999914 v^2)]t^2] *)

Analytic integration is much more stable than numerical integration, especially when infinite and/or multi-dimensional integrals are concerned:

f[v_, x_, z_] = Assuming[x \[Element] Reals && x != 0 && z \[Element] Reals && -1 < v < 1, Integrate[(Cos[-kx x] Cos[-kz z])/(kx^2 + ky^2 + kz^2 (1 - v^2)), {kz, -∞, ∞}, {kx, -∞, ∞}, {ky, -∞, ∞}]] (* (2 π^2)/Sqrt[-((-1 + v^2) x^2) + z^2] *) eA[v_, t_] = 0.0579484*f[v, 3.2, 0.999957 t] (* 1.14386/Sqrt[0.999914 t^2 - 10.24 (-1 + v^2)] *)

Analytic integration is much more stable than numerical integration, especially when infinite and/or multi-dimensional integrals are concerned:

f[u_, x_, z_] = Assuming[x \[Element] Reals && x != 0 && z \[Element] Reals && u > 0, Integrate[(Cos[-kx x] Cos[-kz z])/(kx^2 + ky^2 + u*kz^2), {kz, -∞, ∞}, {kx, -∞, ∞}, {ky, -∞, ∞}]] (* (2 π^2)/Sqrt[u x^2 + z^2] *) eA[t_] = 0.0579484*f[0.000086, 3.2, 0.999957 t] (* 1.14386/Sqrt[0.00088064 + 0.999914 t^2] *)

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answered Nov 8, 2022 at 12:49

Roman

51.3k
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62
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Analytic integration is much more stable than numerical integration, especially when infinite and/or multi-dimensional integrals are concerned:

f[v_, x_, z_] = Assuming[x \[Element] Reals && x != 0 && z \[Element] Reals && -1 < v < 1, Integrate[(Cos[-kx x] Cos[-kz z])/(kx^2 + ky^2 + kz^2 (1 - v^2)), {kz, -∞, ∞}, {kx, -∞, ∞}, {ky, -∞, ∞}]] (* (2 π^2)/Sqrt[-((-1 + v^2) x^2) + z^2] *) eA[v_, t_] = 0.0579484*f[v, 3.2, 0.999957 t] (* 1.14386/Sqrt[0.999914 t^2 - 10.24 (-1 + v^2)] *)