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Can I do this in parallel programming??

I want to integrate this function where all parameters are constant variable is z and summation over n but i am not be able to do it can anyone help to figure out this. it would be a great pleasure,

aMie[n_, x_] := (m psi[n, m x] psidev[n, x] - psi[n, x] psidev[n, m x])/ (m psi[n, m x] xidev[n, x] - xi[n, x] psidev[n, m x]); bMie[n_, x_] := (psi[n, m x] psidev[n, x] - m psi[n, x] psidev[n, m ,x])/ (psi[n, m x] xidev[n, x] - m xi[n, x] psidev[n, m x]); pn[θ_, n_] := (LegendreP[n - 1, Cos[θ]] - LegendreP[n + 1, Cos[θ]])/(1 - (Cos[θ])^2); tn[θ_, n_] := -Cos[θ]*pn[θ, n] + (2*n + 1)*LegendreP[n, Cos[θ]]; dn[θ_, n_] := (pn[θ, n] + tn[θ, n])/(2*n + 1); Gn[n_] := NIntegrate[NIntegrate[Sin[θ]*Sqrt[Cos[θ]]* Exp[-γ^2*(Sin[θ])^2]*Exp[I*δ*Cos[θ]]* dn[θ, n], {θ, 0, ϕmax}],{δ,0,Infinity};
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  • $\begingroup$ Do you want an analytical solution? $\endgroup$ Commented Jun 2, 2017 at 14:00
  • $\begingroup$ Taking a quick inspection in the code one can see that the integration will do not proceed unless you define the following variables m, gama, phimax. Also the evaluation in the Table[] command needs the definition of alpha, k and function A[phimax]. $\endgroup$ Commented Jun 3, 2017 at 1:30

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Everything is actually ok with your code and the full integration cannot proceed mainly because the function call psidev[n, m ,x]) instead psidev[n, m x]). However the analytic integration seems to be very difficult to accomplish with these complicated expressions and here I give a numerical approach to the problem. In this manner, in order to evaluate the NIntegrate[] kernel one needs to adopt a convenient integration strategy/rule to circumvent the function subtleties. In fact this integration takes place in a complex domain and an additional care must to be taken in the calculation. Here, I do not worry about this, but it may have important consequences if there exists poles in this domain. The functions are highly oscillatory and possibly a tunning procedure in the integration method can speed up the elapsed computation time. The calculation presented below took several hours in my computer. Here I've changed the Table[] command by a For[] loop to better see what is happening during the entire evaluation. 

 m = 1.33/1.5; \[Phi]max = 78*\[Pi]/100; \[Gamma] = Sqrt[0.99]; A = 0.98; k = 1.38*10^7; \[Alpha] = 13.829; psi[n_?NumericQ, x_?NumericQ] := x SphericalBesselJ[n, x](*Riccarti \.06Bessel \.06Functions*) xi[n_?NumericQ, x_?NumericQ] := x SphericalHankelH2[n, x] psidev[n_?NumericQ, x_?NumericQ] := Derivative[0, 1][psi][n, x]; xidev[n_?NumericQ, x_?NumericQ] := Derivative[0, 1][xi][n, x]; aMie[n_?NumericQ, x_?NumericQ] := (m psi[n, m x] psidev[n, x] - psi[n, x] psidev[n, m x])/(m psi[n, m x] xidev[n, x] - xi[n, x] psidev[n, m x]); bMie[n_?NumericQ, x_?NumericQ] := (psi[n, m x] psidev[n, x] - m psi[n, x] psidev[n, m x])/(psi[n, m x] xidev[n, x] - m xi[n, x] psidev[n, m x]); pn[\[Theta]_?NumericQ, n_?NumericQ] := (LegendreP[n - 1, Cos[\[Theta]]] - LegendreP[n + 1, Cos[\[Theta]]])/(1 - (Cos[\[Theta]])^2); tn[\[Theta]_?NumericQ, n_?NumericQ] := -Cos[\[Theta]]*pn[\[Theta], n] + (2*n + 1)* LegendreP[n, Cos[\[Theta]]]; dn[\[Theta]_?NumericQ, n_?NumericQ] := (pn[\[Theta], n] + tn[\[Theta], n])/(2*n + 1); Gn[n_?NumericQ, \[Delta]_?NumericQ] := NIntegrate[ Sin[\[Theta]]*Sqrt[Cos[\[Theta]]]* Exp[-\[Gamma]^2*(Sin[\[Theta]])^2]*Exp[I*\[Delta]*Cos[\[Theta]]]* dn[\[Theta], n], {\[Theta], 0, \[Phi]max}, Method -> {"LocalAdaptive", "SingularityHandler" -> "IMT", "SingularityDepth" -> Infinity, "SymbolicProcessing" -> 0}, MaxRecursion -> 200, PrecisionGoal -> 10, AccuracyGoal -> 10] // FullSimplify; GNp[n_?NumericQ, \[Delta]_?NumericQ] := Derivative[0, 1][Gn][n, \[Delta]]; (*Qe=Table[{z,Re[Sum[((4*\[Gamma]^2)/A*(2*n+1)*(aMie[n,\[Alpha]]+bMie[\ n,\[Alpha]])*Gn[n,k*z]*Conjugate[GNp[n,k*z]]),{n,1,10}]]},{z,10^-5,10^\ -4,10^-6}]*) Result = List[]; For[i = 0, i <= 90, i++, z = N[10^-5 + i*10^-6]; func = Re[ Sum[((4*\[Gamma]^2)/ A*(2*n + 1)*(aMie[n, \[Alpha]] + bMie[n, \[Alpha]])*Gn[n, k*z]* Conjugate[GNp[n, k*z]]), {n, 1, 10}]]; AppendTo[Result, {z, func}]; Print[z, " ", func] ] ListLinePlot[Result, Axes -> True, ImageSize -> 800, AspectRatio -> 0.5, PlotRangePadding -> Automatic, Background -> White, PlotStyle -> Directive[Blue, Thickness[0.0045]], FrameStyle -> Directive[Black, Thickness[0.0035]], LabelStyle -> Directive[Black, FontSize -> 16, FontFamily -> "Times"], TicksStyle -> Directive[Black, Thickness[0.004]], FrameTicksStyle -> Directive[Black, 16, Opacity[1]], RotateLabel -> True, FrameLabel -> {StyleForm["z", FontSize -> 20], StyleForm["Sum", FontSize -> 20]}, Frame -> True]

Choosing a z interval starting at 10^-5 and ending at 10^-4, we have the table output

{{0.00001, -0.0144983}, {0.000011, -0.0119027}, {0.000012, -0.00979571}, {0.000013, -0.00814835}, {0.000014, -0.00692311}, {0.000015, -0.00604408}, {0.000016, -0.00540571}, {0.000017, -0.00490035}, {0.000018, -0.00444597}, {0.000019, -0.0040016}, {0.00002, -0.003566}, {0.000021, -0.0031629}, {0.000022, -0.00282066}, {0.000023, -0.00255521}, {0.000024, -0.00236225}, {0.000025, -0.0022199}, {0.000026, -0.00209895}, {0.000027, -0.00197487}, {0.000028, -0.00183635}, {0.000029, -0.00168711}, {0.00003, -0.0015412}, {0.000031, -0.00141466}, {0.000032, -0.00131755}, {0.000033, -0.00124969}, {0.000034, -0.00120152}, {0.000035, -0.00115903}, {0.000036, -0.00111008}, {0.000037, -0.00104928}, {0.000038, -0.000979359}, {0.000039, -0.000908869}, {0.00004, -0.000847562}, {0.000041, -0.000801744}, {0.000042, -0.000771639}, {0.000043, -0.000751732}, {0.000044, -0.000733649}, {0.000045, -0.000710058}, {0.000046, -0.000677772}, {0.000047, -0.000638726}, {0.000048, -0.000598594}, {0.000049, -0.000563853}, {0.00005, -0.000538736}, {0.000051, -0.000523447}, {0.000052, -0.000514335}, {0.000053, -0.00050579}, {0.000054, -0.000492889}, {0.000055, -0.000473527}, {0.000056, -0.000449134}, {0.000057, -0.000423741}, {0.000058, -0.000401943}, {0.000059, -0.000386746}, {0.00006, -0.000378291}, {0.000061, -0.000373955}, {0.000062, -0.000369706}, {0.000063, -0.000362005}, {0.000064, -0.000349369}, {0.000065, -0.0003329}, {0.000066, -0.000315611}, {0.000067, -0.000300922}, {0.000068, -0.000291065}, {0.000069, -0.000286126}, {0.00007, -0.000284117}, {0.000071, -0.000281992}, {0.000072, -0.000277088}, {0.000073, -0.000268312}, {0.000074, -0.000256544}, {0.000075, -0.000244122}, {0.000076, -0.000233689}, {0.000077, -0.000226959}, {0.000078, -0.000223977}, {0.000079, -0.000223175}, {0.00008, -0.00022217}, {0.000081, -0.000218888}, {0.000082, -0.000212494}, {0.000083, -0.000203714}, {0.000084, -0.000194416}, {0.000085, -0.000186701}, {0.000086, -0.000181922}, {0.000087, -0.000180093}, {0.000088, -0.000179944}, {0.000089, -0.000179555}, {0.00009, -0.000177268}, {0.000091, -0.000172431}, {0.000092, -0.000165653}, {0.000093, -0.000158465}, {0.000094, -0.000152576}, {0.000095, -0.000149076}, {0.000096, -0.000147959}, {0.000097, -0.000148169}, {0.000098, -0.000148128}, {0.000099, -0.00014648}, {0.0001, -0.000142707}}

This list gives us the following plot

enter image description here

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  • $\begingroup$ Thank you so much for your great effort. Can I use a loop for n summation because I have to sum it over n=1 to infinity. $\endgroup$ Commented Jun 9, 2017 at 18:29
  • $\begingroup$ How did you got this plot?? $\endgroup$ Commented Jun 9, 2017 at 21:10
  • $\begingroup$ @AnitaMaheshwari I've edited the answer above and inserted the ListLinePlot part. $\endgroup$ Commented Jun 11, 2017 at 0:16
  • $\begingroup$ is there any way to reduce timing in summing over 1 to infinity??Here we add it for 1 to 10 but extending it to 20 is also taking long timing. $\endgroup$ Commented Jun 12, 2017 at 3:47
  • $\begingroup$ Probably yes, the evaluation time can be reduced by adjusting the integration strategy. The LocalAdaptive strategy takes a long time in the process since secctionates the interval in a certain number of partitions. Maybe another approach can work better and faster than this one. You can take a look at the documentation. johnboccio.com/MathematicaTutorials/…. I'll try some more adjustments and post the results too soon as possible. $\endgroup$ Commented Jun 12, 2017 at 10:58

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