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

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

Answer 1

[Teja Muppirala]

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

intValue =

6.0922e-15

[Shan Chu]

Thank you so much sir, It is what I am looking for. I didn't know it is possible to put functions inside functions.

Answer 2

[Torsten]

Directly evaluate "A" and insert the expression in "fun_Z0":

http://www.wolframalpha.com/input/?i=integration&rawformassumption=%7B%22F%22,+%22Integral%22,+%22integrand%22%7D+-%3E%22x*besselj(1,x)%22&rawformassumption=%7B%22F%22,+%22Integral%22,+%22variable%22%7D+-%3E%22x%22&rawformassumption=%7B%22F%22,+%22Integral%22,+%22rangestart%22%7D+-%3E%22a*y%22&rawformassumption=%7B%22F%22,+%22Integral%22,+%22rangeend%22%7D+-%3E%22b*y%22&rawformassumption=%7B%22C%22,+%22integration%22%7D+-%3E+%7B%22Calculator%22%7D

There are several implementations of the Struve function, e.g.

https://de.mathworks.com/matlabcentral/fileexchange/37302-struve-functions

Best wishes

Torsten.

[Shan Chu]

Thank you so much for your time. But it is rather complicated. Thanks