清空筛选条件
清空筛选条件

Numerical integration with a variable integration limit

1 次查看（过去 30 天）
JohnM
JohnM 2016-9-28
回答： JohnM 2016-9-29
Is there a way to implement the following:
f2=@(z) integral (@(x) f1(x,z) ,0 ,z);
P2=@(z) integral(f2,z,Inf);
With f2 and P2 as two separate anonymous functions? It seems z in f2 must be a scalar (not a vector), so a for loop is required to make multiple evaluations at z I can deal with that, but Matlab will not evaluate P2, giving the error: Error using integral (line 85) A and B must be floating-point scalars.
It appears that
P2Alt=@(z) integral2(@(x,z) f1(x,z), 0, z, z, Inf);
accomplishes the desired end result (z must be a scalar), but I would prefer the function f2, separate.

请先登录，再回答此问题。

回答（2 个）

Ujjal Dey
Ujjal Dey 2016-9-29
At first, try with "int" command and form a final function of your desired variable. Ex: f2=@(x) int (f1(x,z),z);
  1 个评论
JohnM
JohnM 2016-9-29
Hi Ujjal. I should have mentioned in my post that I don't have the symbolic math toolbox, so invoking int is not an option for me.

请先登录，再进行评论。

JohnM
JohnM 2016-9-29
After giving this some thought, the following appears to be one way to solve the problem:
f2=@(z) integral(@(x) f1(x,z), 0, z);
f2Vect=@(z) arrayfun(@(z)f2(z),z); %vectorize f2
zval=0:0.1:3;
P2=zeros(1,length(zval)); %Preallocate
for i=1:length(zval)
P2(i)=integral(f2Vect,zval(i),Inf); %Evaluate the integral of f2
end

类别

在 Help Center 和 File Exchange 中查找有关 Calculus 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by