Matlab Numerical integral improvement

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Shan Chu
Shan Chu 2017-2-4
编辑： Karan Gill 2017-10-17
Hi, I have the integral below:
F_A_I=@(x) besselj(1,x.*3.5).*besselj(1,x.*0.5);
A=integral(F_A_I,0,Inf,'RelTol',1e-6,'AbsTol',1e-12,'ArrayValued',true);
But Matlab said:
Warning: Reached the limit on the maximum number of intervals in use. Approximate bound on error is 1.7e+00. The integral may not exist, or it may be difficult to approximate numerically to the requested accuracy.
while Mathematica can give the answer straightforward A=0.0205664
Could you please help me to improve my code. Thanks
  1 个评论
Niels
Niels 2017-2-4
probably a definition gap in your function, integral might converge to inf, in these cases matlab displays -> Reached the limit on the maximum number of intervals in use.

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Karan Gill
Karan Gill 2017-2-13
编辑：Karan Gill 2017-10-17
Updated answer for R2017b. Use int and convert the symbolic solution to floating point.
>> syms x
f = int(besselj(1, x/2)*besselj(1, (7*x)/2),x,0,inf)
f =
-(4*(100*ellipticE(1/49) - 99*ellipticK(1/49)))/(21*pi)
>> f_dbl = double(ans)
f_dbl =
0.0022
>> f_vpa = vpa(f)
f_vpa =
0.0022054352588140668793354496265733
OLD ANSWER from 13-Feb-2017
Try using "vpaintegral": https://www.mathworks.com/help/symbolic/vpaintegral.html#bvaajp9-1
The convert to double using "double".
  2 个评论
Walter Roberson
Walter Roberson 2017-2-14
编辑：Walter Roberson 2017-2-14
vpaintegral() with up to 10000 MaxFunctionCalls complains it cannot reach required precision.
The ratio oscillates a lot.
Karan Gill
Karan Gill 2017-9-25
Updated answer with solution starting R2017b.

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