Double integral with one integral having limits as a function of other variable

Question

Shan Chu 2017-8-3

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/351395-double-integral-with-one-integral-having-limits-as-a-function-of-other-variable

评论： Shan Chu 2017-8-3

Dear all, I have the problem below:

l1=3e-3;
l2=4e-3;
r1=11e-3;
r2=12e-3;
  fun_Z0=@(x) x^(-5)* A^2 *((l2-l1)+(exp(-x*(l2-l1))-1)*x^(-1));
where 
A=integral(y*besselj(1,y), x*r1, x*r2)

I would like to calculate

Z=integral(fun_Z0, 0, Inf)

Could you please suggest a function that could handle this problem?

Thanks

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Answer 1

Teja Muppirala 2017-8-3

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/351395-double-integral-with-one-integral-having-limits-as-a-function-of-other-variable#answer_276618

编辑：Teja Muppirala 2017-8-3

在 MATLAB Online 中打开

You can put functions inside of functions

%%Embed A into fun_Z0, and tell INTEGRAL to only accept scalar inputs
l1=3e-3;
l2=4e-3;
r1=11e-3;
r2=12e-3;
A=@(z) integral(@(y) y.*besselj(1,y), z*r1, z*r2)
fun_Z0=@(x) x^(-5)* A(x)^2 *((l2-l1)+(exp(-x*(l2-l1))-1)*x^(-1));
% This seems to be a hard integral. You'll get some warnings here, but it works
Z=integral(fun_Z0, 0, Inf, 'ArrayValued', true,'AbsTol',0)

Z =

6.0922e-15

%%If you really want to make sure, break it up into pieces
intValue = 0;
tol = 1e-6;
for n = 0:100 
  prevValue = intValue;
  Z=integral(fun_Z0, 5000*n, 5000*(n+1), 'ArrayValued', true,'AbsTol',0);
  intValue = intValue + Z;
  if abs(1 - intValue/prevValue) < tol
      break 
  end
end
intValue