Skip to main content

calculus and analysis - How to find a more precise value of integral?


Mathematica 11 produces



NIntegrate[ArcCot[x]*Sin[x]/(5/4 + Cos[x]), {x, 0, Infinity}, 
 Method -> "ExtrapolatingOscillatory", AccuracyGoal -> 10, 
 WorkingPrecision -> 20]
0.746733605129940959

This is 4.573.3 from Gradshteyn & Ryzhik. Up to G&R the result is


 N[Pi*Log[3/2/(1 + 1/2*Exp[-1])], 10]
   0.7433557514

The question arises: how to obtain it in Mathematica?




Answer



Here's an extrapolatory approach based on Richardson extrapolation of the integrals over each "period." (Richardson[] code from Anton Antonov's answer here):


Clear[Richardson]
Richardson[A_, n_, N_] := 
 Total@Table[(A[n + k]*(n + k)^N*If[OddQ[k + N], -1, 1])/(k! (N - k)!), {k, 0, N}]

(* the integral from 2 n Pi to 2 (n+1) Pi *)
Clear[aa];
aa[n_Integer /; n >= 0] := 
  aa[n] = NIntegrate[ArcCot[x]*Sin[x]/(5/4 + Cos[x]), {x, 2 n Pi, 2 (n + 1) Pi}, 

    WorkingPrecision -> 50];

We get around machine precision with 100 terms plus 6 extrapolation terms:


sf[n_] := Sum[aa[i], {i, 0, n}]; (* partial sum = integral over {0, 2 (n+1) Pi} *)
Richardson[sf, 100, 6]
% - Pi*Log[3/2/(1 + 1/2*Exp[-1])]
(*
  0.743355751361227474780881457479018423835    
  -2.97535713733265290609713*10^-16
*)


Without extrapolation:


sf[100]
% - Pi*Log[3/2/(1 + 1/2*Exp[-1])]
(*
  0.74207790286093203962531860751097243182349263171074
  -0.00127784850029573269127658323333660172377741318898
*)



Update


More extrapolatory terms yields an approximation accurate to machine precision with many fewer terms overall:


Richardson[sf, 1, 16]
% - Pi*Log[3/2/(1 + 1/2*Exp[-1])]
(*
  0.74335575136122767678505611095516483469029
  -9.553153907978914419885698*10^-17
*)

Comments

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...
Post a Comment
Read more

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...
Post a Comment
Read more