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Greetings dear friends, I am trying to graph this integral so that I can obtain a graph x =f(t), thanks for your help!

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Lewis HC
Lewis HC 2022-2-25
回答： Lewis HC 2022-3-1
I need to get this curve in my 3D graph:
My code which I was working is this:
Thank you dear friends!
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Paul
Paul 2022-2-26
How does exp(i*p*x) in the equation become just cos(p*x) in the code? Unless of course only the real part of the integral is goal.

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Voss
Voss 2022-2-25
编辑：Voss 2022-2-25
Ran in:
Note that t > 0 and the grids on the surface in the desired image are more widely spaced than the actual points where the surface has been calculated (i.e., there is curvature in between grid lines).
t = linspace(0.0001,2,50);
x = linspace(-2,2,50);
[T,X] = meshgrid(t,x);
for i = 1:numel(x)
for j = 1:numel(t)
% f = @(p)1/pi*sqrt(2*pi)*sin(p)./p.*exp(-T(i,j).*p.^2).*cos(p.*X(i,j));
f = @(p)1/(pi*sqrt(2*pi))*sin(p)./p.*exp(-T(i,j).*p.^2).*cos(p.*X(i,j));
F(i,j) = integral(f,-Inf,Inf);
end
end
surf(T,X,F);
colormap(flip(autumn()));
xlabel('t');
ylabel('x');
zlabel('u(x,t)');
  2 个评论
Torsten
Torsten 2022-2-25
f = @(p)1/(pi*sqrt(2*pi))*sin(p)./p.*exp(-T(i,j).*p.^2).*cos(p.*X(i,j));
instead of
f = @(p)1/pi*sqrt(2*pi)*sin(p)./p.*exp(-T(i,j).*p.^2).*cos(p.*X(i,j));
Voss
Voss 2022-2-25
Oh yeah! Thanks!
I saw your comment before, but then I just typed in the code from the screenshot in the question anyway. D'oh!

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Lewis HC
Lewis HC 2022-3-1
I really don't know what I would do without your great help, thank you very much dear friends, you are the best!

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