How to improve the calculation accuracy and speed of double numerical integration？

Question

I want to solve the following double numerical integral:

$A=\int _{0.52}^1\int _0^{\infty }\frac{\sqrt{s^2-0.52^2} \sqrt{1-s^2} \left(2 s^2-1\right) \left(\left(2 s^2-1\right) \text{T1}-2 s^2 \text{T2}\right)}{\left(2 s^2-1\right)^4+16 \left(s^2-0.52^2\right) s^4 \left(1-s^2\right)}d\xi ds$

$\text{T1}=\frac{\cos (-8.2 \xi ) \sin \left(\frac{\xi (0.936 t)}{s}\right)}{\xi }-\frac{\cos (11.8 \xi ) \sin \left(\frac{\xi (0.936 t)}{s}\right)}{\xi }$

$\text{T2}=\frac{\cos (-8.2 \xi ) \sin (0.877 \xi t)}{\xi }-\frac{\cos (11.8 \xi ) \sin (0.877 \xi t)}{\xi }$

I just used the "Table" function to output the t vector：

Table[NIntegrate[A,{ξ ,0,infinity },{s,0.52,1}],{t,0,25,0.25}]

Adopting the "GlobalAdaptive" integration strategy takes a long time, and the error is large.And I get the following warning

NIntegrate::eincr: The global error of the strategy GlobalAdaptive has increased more than 2000
 times. The global error is expected to decrease monotonically after a number of integrand
 evaluations. Suspect one of the following: the working precision is insufficient for the
 specified precision goal; the integrand is highly oscillatory or it is not a (piecewise) smooth
 function; or the true value of the integral is 0. Increasing the value of the GlobalAdaptive 
option MaxErrorIncreases might lead to a convergent numerical integration. NIntegrate obtained
 1.939973715732701`*^-6 and 0.0010362662139119024` for the integral and error estimates.

So I want to know how to improve the integration formula, or what integration strategy to use to improve the calculation speed and accuracy. Thanks!!!

Answer 1

You can do the integration over xi with Integrate .

T1 = -((Sin[(0.936 t)/s \[Xi]] Cos[
     11.8 \[Xi]])/\[Xi]) + (Sin[(0.936 t)/
     s \[Xi]] Cos[-8.2 \[Xi]])/\[Xi];
T2 = -((Sin[0.877 t \[Xi]] Cos[11.8 \[Xi]])/\[Xi]) + (Sin[
   0.877 t \[Xi]] Cos[-8.2 \[Xi]])/\[Xi];

A = ((Sqrt[s^2 - 0.52^2] Sqrt[
   1 - s^2] (2 s^2 - 1) ((2 s^2 - 1) T1 - 
    2 s^2 T2))/((2 s^2 - 1)^4 + 16 s^4 (s^2 - 0.52^2) (1 - s^2)));
AR[ \[Xi]_, s_, t_] = A // Rationalize[#, 0] & // Together

int[s_, t_] = 
  Integrate[AR[\[Xi], s, t], {\[Xi], 0, \[Infinity]}, 
    Assumptions -> {0 < t < 25, 52/100 < s < 1}]

(*   (25 \[Pi] (-1 + 2 s^2) Sqrt[-169 + 794 s^2 - 
625 s^4] ((-1 + 2 s^2) Sign[1025 s - 117 t] + (1 - 2 s^2) Sign[
   1475 s - 117 t] + 
 2 s^2 (-Sign[41/5 - (877 t)/1000] + 
    Sign[59/5 - (877 t)/1000])))/(4 (-625 + 5000 s^2 - 
 12296 s^4 + 7296 s^6))   *)

(tab = Table[{t, NIntegrate[int[s, t], {s, 52/100, 1}]}, {t, 0, 
 25, .25}]) // Timing

ListPlot[tab]

I testet it for t=625/100. (Cases with 3 equal values should be tested further.)

NIntegrate[int[s, 625/100], {s, 52/100, 1}]
(*   0.0455044   *)

NIntegrate[
  AR[\[Xi], s, 625/100], {s, 52/100, 1}, {\[Xi], 0, \[Infinity]}]
(*   0.0454959   *)

Answer 2

A=((Sqrt[s^2 - 0.52^2] Sqrt[1 - s^2] (2 s^2 - 1) ((2 s^2 - 1) T1 - 2 s^2 T2))/((2 s^2 - 1)^4 + 16 s^4 (s^2 - 0.52^2) (1 - s^2)))
T1=-((Sin[ (0.936 t)/s ξ] Cos[11.8 ξ ])/ξ) + (Sin[ (0.936 t)/s ξ] Cos[ -8.2 ξ])/ξ
T2=-((Sin[0.877 t ξ ] Cos[11.8 ξ ])/ξ) + (Sin[0.877 t ξ ] Cos[ -8.2 ξ])/ξ