No plot in double integral
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Hi! I try to solve the double integral, but there is an empty plot. Could you please tell me, what is wrong? In MathCad this integral is solved well.
n=1;
t=1;
r=1;
s=0:0.1:5;
fun = @(x,z,k) (x.*exp(2*n*t.*x.^2)./sqrt(1-x.^2)).*(exp(-2*t.*z.^2).*besselj(0,z.*k.*x).*(1+(sqrt(-pi)./(z*r)).*(1+((z*r).^2)/2).*(erfi(z*r/2)).*(exp(-((z*r).^2)/4))));
f3 = arrayfun(@(k)integral2(@(x,z)fun(x,z,k),0,Inf,0,1),s)
Cor = ((-2*r*sqrt(2*n*t)*exp(-2*n*t))/(pi*erf(sqrt(2*n*t))*(atan(r/(2*sqrt(2*t)))-sqrt(2*t)/(2*r*(1/4+2*t/(r^2))))))*f3;
plot(s,Cor,'g-');
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the cyclist
2023년 1월 29일
편집: the cyclist
2023년 1월 29일
Can you upload the formula you are trying to integrate, in math notation? Maybe a mistake was made in transcribing it into MATLAB.
Are you expecting complex values?
Maybe you wanted x to be limited to 0 1 and z to be 0 to infinity? You coded with x being 0 to infinity and z being 0 to 1
n=1;
t=1;
r=1;
syms x z k s real
fun = (x.*exp(2*n*t.*x.^2)./sqrt(1-x.^2)).*(exp(-2*t.*z.^2).*besselj(0,z.*k.*x).*(1+(sqrt(-pi)./(z*r)).*(1+((z*r).^2)/2).*(erfi(z*r/2)).*(exp(-((z*r).^2)/4))));
f3 = subs(int(int(fun,x,0,inf), z, 0, 1), k, s)
fun_zhalf = limit(subs(fun, k, 1), z, 1/2)
xlimited = limit(fun_zhalf, x, 1)
case_to_plot = children(children(xlimited, 3),1)
fplot(case_to_plot, [0 2])
xlimited = limit(fun_zhalf, x, 2)
vpa(xlimited)
the cyclist
2023년 1월 29일
I was surprised to see the explicitly imaginary
sqrt(-pi)
in the expression, which is what prompted me to ask to see the math notation version.
I did not notice that. I wonder what it would look like if you used sqrt(pi) instead?
n=1;
t=1;
r=1;
syms x z k s real
Pi = sym(pi);
fun = (x.*exp(2*n*t.*x.^2)./sqrt(1-x.^2)).*(exp(-2*t.*z.^2).*besselj(0,z.*k.*x).*(1+(sqrt(Pi)./(z*r)).*(1+((z*r).^2)/2).*(erfi(z*r/2)).*(exp(-((z*r).^2)/4))));
f3 = subs(int(int(fun,x,0,inf), z, 0, 1), k, s)
fun_zhalf = limit(subs(fun, k, 1), z, 1/2)
xlimited = limit(fun_zhalf, x, 1)
case_to_plot = children(children(xlimited, 3),1)
fplot(case_to_plot, [0 2])
xlimited = limit(fun_zhalf, x, 2)
vpa(xlimited)
You get rid of the real-valued portion of the function (at least at some values), but the complex portion remains.
Hexe
2023년 1월 30일
Dear all, thank you for your replies. Yes, there is an imaginary part where sqrt(pi). And you are right, the integration is from 0 to 1 for x and 0-Inf for z. The math form is attached.
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