how can wirte this integration in matlab.
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file is attached
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Walter Roberson
2018년 6월 5일
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
2018년 6월 5일
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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Paridhi Yadav
2018년 6월 4일
fun = @(q) q^4/(2*pi*(k^3)*((q + qs)^2)*sqrt(1-(q/2*k)^2));
jdp(k) = integral(fun,0,2*k);
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Ameer Hamza
2018년 6월 4일
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
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