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how can wirte this integration in matlab.

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Pawan Kumar
Pawan Kumar on 4 Jun 2018
Commented: Walter Roberson on 5 Jun 2018
file is attached
  2 Comments
Walter Roberson
Walter Roberson on 5 Jun 2018
The solution depends upon the relationship between k and q, and can be further resolved by adding assumptions. Some of the solutions are infinite.
I was certain I had already posted this information.
Walter Roberson
Walter Roberson on 5 Jun 2018
I was right, I did post it. You asked the question in a different topic as well.. leading to duplicated effort. :(
The solution depends on whether k is positive, 0, or negative, and on the relative values of qs and 2*k . In some combinations of circumstances it is undefined. MATLAB is able to resolve some of the combinations if you add appropriate assumptions to the variables, but it is not able to tell you the full conditional resolution under other assumptions.

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Answers (2)

Paridhi Yadav
Paridhi Yadav on 4 Jun 2018
fun = @(q) q^4/(2*pi*(k^3)*((q + qs)^2)*sqrt(1-(q/2*k)^2));
jdp(k) = integral(fun,0,2*k);

Ameer Hamza
Ameer Hamza on 4 Jun 2018
If you try to find a closed-form solution, then MATLAB is unable to solve for it for the given integral,
syms k q qs
integrand = 1/(2*pi*k^3*(q+qs)^2*sqrt(1-(q/(2*k))^2))*q^4;
J = int(integrand, q, 0, 2*k)
J =
int(q^4/(2*k^3*pi*(q + qs)^2*(1 - q^2/(4*k^2))^(1/2)), q, 0, 2*k)
The result is same as the input statement. But if you try to solve it numerically then you can do it as follow
integrand = @(q,qs,k) 1./(2*pi.*k.^3.*(q+qs).^2.*sqrt(1-(q./(2*k)).^2)).*q.^4;
J = @(k, qs) integral(@(q) integrand(q, qs, k), 0, 2*k);
qs = 1;
k = 10;
J(k, qs)
ans =
0.8855
  1 Comment
Pawan Kumar
Pawan Kumar on 5 Jun 2018
from 1st ans i got result like 1*1 sym and from 2nd is same as code replicate

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