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

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I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.

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