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-');
6 件のコメント
the cyclist
2023 年 1 月 29 日
編集済み: the cyclist
2023 年 1 月 29 日
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
2023 年 1 月 29 日
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
2023 年 1 月 30 日
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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