Numerical integration with two parameters

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Dimitar
Dimitar 2022-8-22
评论： Dimitar 2022-8-22
So I want to integrate the following integral fun = @(t, x, y) (y-t*x+t^2)/(sqrt((x-2*t)^2+(y-t^2)^2)^3). I tried using the quad function (@t from -inf to inf) because the parameters are 'vectors' ie. x=-inf:inf and y=-inf:inf but I don't get an answer. Am I using the correct function or should I do something else, I really can't figure it out. Any answer would be helpful, thanks in advance.
  2 个评论
Torsten
Torsten 2022-8-22
编辑：Torsten 2022-8-22
I tried using the quad function (@t from -inf to inf) because the parameters are 'vectors' ie. x=-inf:inf and y=-inf:inf
So you are trying to calculate the 3d integral for x=-Inf:Inf and for y=-Inf:Inf and for t=-Inf:Inf of the above function ?
Or is it a one-dimensional integral with fixed values for x and y ?
Dimitar
Dimitar 2022-8-22
Not really, I want to get one more 'vector' of answer with different values of x and y, for example x=0, y=0 the next one being x=0,y=1 etc. And after I get all the answers I want to plot them on a graph Sorry, I don't know how to explain better, I hope you can understand now.

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回答（1 个）

Torsten
Torsten 2022-8-22
编辑：Torsten 2022-8-22
Ran in:
You should check whether the integral exists when y = x^2/4. In this case, the denominator of fun is 0, and I think you'll have a pole of order 1 at t = x/2. This would mean the integral does not exist.
fun = @(t,x,y)(y-t*x+t.^2)./(sqrt((x-2*t).^2+(y-t.^2).^2).^3);
% Case where y ~= x^2/4
x = 2;
y = 4;
value = integral(@(t)fun(t,x,y),-Inf,Inf)
value = 0.7966
% Case where y = x^2/4
x = 2;
y = 1;
value = integral(@(t)fun(t,x,y),-Inf,Inf)
Warning: Reached the limit on the maximum number of intervals in use. Approximate bound on error is 1.3e-02. The integral may not exist, or it may be difficult to approximate numerically to the requested accuracy.
value = 1.8403
  1 个评论
Dimitar
Dimitar 2022-8-22
I actually succeeded getting results and plotting them, by taking the values of x and y as such: [x, y] = ndgrid(-10:10,-10:10) and adding 'ArrayValued' to the integral function. I just have to figure out how to exclude ie set to 0, the values of the integral where it doesn't exist. Thank you so much!!

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