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

I need to get this curve in my 3D graph:

My code which I was working is this:

Thank you dear friends!

[Torsten]

Correction to your f:

f = @(p) 1/(pi*sqrt(2*pi)) ...

instead of

f = @(p) 1/pi*sqrt(2*pi) ...

Correction for your tspan:

The integral does not exist for t<=0

And what is plotted against what in your graph ?

[Lewis HC]

Dear Torsten

I need to obtain this graph dear friend:

[Paul]

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.

[Torsten]

The imaginary part of the function is odd in p - so the integral is 0.

Answer 1

[Voss]

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)');

[Torsten]

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]

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!

Answer 2

[Lewis HC]

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