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Please, help to find a mistake in the code for double integral

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Hexe
Hexe 2023-2-24
回答： Torsten 2023-2-24
Ran in:
I have a double integral (a picture of formula is attached) and a code. But this integral should be equal 1 at s = 0 at any parameters n, t, k. Thus, all curves should start from the point (0,1). But, I obtain at vpa the ans = 0.35331 at s = 0, and curves start from different points at different parameters. I suspect that there is an error somewhere in the code entry itself, but I can't see it now. Please help me to find it.
n = 1 ;
t = 1;
k = 1; %(k is ksi in formula)
s = 0:0.5:3;
fun = @(x,q,p) ((x.*exp(2*n*t.*(x.^2)))./sqrt(1-x.^2)).*(exp(-2*t.*(q.^2)).*besselj(0,q.*p.*x).*(1+((sqrt(-pi)./(q*k)).*(1+((q*k).^2)/2).*erfi(q*k*sqrt(-1)/2).*exp(-((q*k).^2)/4))));
f3 = arrayfun(@(p)integral2(@(x,q)fun(x,q,p),0,1,0,Inf),s);
Cor = -((2*k*sqrt(2*n*t)*exp(-2*n*t))/(pi*erf(sqrt(2*n*t))*(atan(k/(2*sqrt(2*t)))-sqrt(2*t)/(2*k*(0.25+(2*t)/(k^2)))))).*f3;
plot(Cor,'b-')
%vpa(Cor,5)
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显示 2更早的评论
Torsten
Torsten 2023-2-24
And if "erf" does not work, you take another function that sounds similarily and hope you get the correct result ?
Google
error function for complex argument & matlab
and you will find some functions from the file exchange to compute the error function with complex argument (if this is really what is meant in the formula).
Hexe
Hexe 2023-2-24
I thought that erfi is the error function for the imaginary argument which is in the formula.

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回答（2 个）

Torsten
Torsten 2023-2-24
移动：Image Analyst 2023-2-24
You can use
erf(i*z) = i*erfi(z)
Thus in your case
erf(i*q*k/2) = i*erfi(q*k/2)
But you will come into trouble here since you integrate from 0 to Inf with respect to q.
And erfi(x) -> Inf very fast as x-> Inf.
You will have to evaluate
i*erfi(q*k/2)*exp(-(q*k/2).^2)
as one expression to avoid Inf * 0 here.
  1 个评论
Hexe
Hexe 2023-2-24
移动：Image Analyst 2023-2-24
Thank you for your last comment. You are right. In MathCad the integral was calculated only after cutting it from 0 to 50. Now it works as it should be.

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Torsten
Torsten 2023-2-24
Ran in:
n = 1 ;
t = 1;
k = 1; %(k is ksi in formula)
s = 0:0.5:20;
fun = @(x,q,p) ((x.*exp(2*n*t.*(x.^2)))./sqrt(1-x.^2)).*(exp(-2*t.*(q.^2)).*besselj(0,q.*p.*x).*(1+((sqrt(-pi)./(q*k)).*(1+((q*k).^2)/2).*func(q,k))));
f3 = arrayfun(@(p)integral2(@(x,q)fun(x,q,p),0,1,0,Inf),s);
Cor = -((2*k*sqrt(2*n*t)*exp(-2*n*t))/(pi*erf(sqrt(2*n*t))*(atan(k/(2*sqrt(2*t)))-sqrt(2*t)/(2*k*(0.25+(2*t)/(k^2)))))).*f3;
plot(s,Cor,'b-')
grid on
function values = func(q,k)
values = arrayfun(@(q)1i*2/sqrt(pi)*integral(@(t)exp(t.^2-((q*k)/2)^2),0,q*k/2),q);
end

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