how can wirte this integration in matlab.

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Pawan Kumar
Pawan Kumar 2018-6-4
评论： Walter Roberson 2018-6-5
file is attached
  2 个评论
Walter Roberson
Walter Roberson 2018-6-5
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 2018-6-5
I was right, I did post it. You asked the question in a different topic as well.. leading to duplicated effort. :(
https://www.mathworks.com/matlabcentral/answers/69221-indefinite-integral-with-matlab#answer_323038
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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回答（2 个）

Paridhi Yadav
Paridhi Yadav 2018-6-4
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 2018-6-4
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 个评论
Pawan Kumar
Pawan Kumar 2018-6-5
from 1st ans i got result like 1*1 sym and from 2nd is same as code replicate

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