Explicit integral could not be found.

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David Yoo
David Yoo on 26 Sep 2012
Hello,
I am trying to get a solution for below problem, which is to evaluate double integrals of function f(x,y) where the first integral boundary is in variable y. Though the explicit integral could not be found, I expect numerical answer still exists; however I don't know how to get that. I tried double and VPA functions but they didn't work.
___________________________________________________________________________
syms x y
f = x^3*exp(1/x^2)*(x^(1/2)/2)^(1/2);
F = int(f,x,0,y);
ans = int(F,y,0,100); ___________________________________________________________________________
Warning: Explicit integral could not be found.
Warning: Explicit integral could not be found.
ans = int(int(x^3*exp(1/x^2)*(x^(1/2)/2)^(1/2), x = 0..y), y = 0..100)
___________________________________________________________________________
VPA(ans)
ans =
numeric::int(numeric::int(x^3*exp(1/x^2)*(x^(1/2)/2)^(1/2), x = 0..y), y = 0..100)

Answers (2)

Babak
Babak on 26 Sep 2012
Mathematica finds the indefinite integral of f over x
Take this and evaluate it for the boundaries (i.e. x=y and x=0) and subtract the results. this will give you the definite integral from x=0 to x=y then again try to integrate the result over y.
  3 Comments
David Yoo
David Yoo on 26 Sep 2012
Thank you for the answer. However, is there any way to solve this kind of problem in Matlab?
Babak
Babak on 26 Sep 2012
Try Matt's solution.

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Matt Fig
Matt Fig on 26 Sep 2012
Edited: Matt Fig on 26 Sep 2012
Are you sure about those limits?
Here is how one would ordinarily solve such a problem numerically. Note I use limits 1-y and 2-10:
fh = @(y) quadl(@(x) x.^3.*exp(1./x.^2).*(x.^(1/2)/2).^(1/2),1,y);
quadv(fh,2,10)
.
.
.
Let's just check the method with an easy problem. Say we want to solve:
int(int(x,0,y),1/2,1)
By simple calculus this is:
int(y^2/2,1/2,1) = 7/48
So in MATLAB
fh = @(y) quadl(@(x) x,0,y);
quadv(fh,1/2,1) % Equal to 7/48 as expected...

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