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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