Clear Filters
Clear Filters

Why Can't int Find a Simple Integral?

4 views (last 30 days)
Paul
Paul on 26 Feb 2023
Commented: Paul on 17 Sep 2023
syms x real
E(x) = 99/50*dirac(x) + rectangularPulse(0, 300, x)/30000
E(x) = 
int on first term yields expected result
int(99/50*dirac(x),x,-inf,inf)
ans = 
int on second term yields expected result
int(rectangularPulse(0, 300, x)/30000,x,-inf,inf)
ans = 
int on sum fails
int(E(x),x,-inf,inf)
ans = 
Shouldn't that work?

Accepted Answer

Pranav
Pranav on 27 Jun 2023
I am very glad to let you know that this bug has been fixed in MATLAB R2023b.
Now, this code produces
syms x real
E(x) = 99/50*dirac(x) + rectangularPulse(0, 300, x)/30000
int(99/50*dirac(x),x,-inf,inf)
int(rectangularPulse(0, 300, x)/30000,x,-inf,inf)
int(E(x),x,-inf,inf)
this output
E(x) =
(99*dirac(x))/50 + rectangularPulse(0, 300, x)/30000
ans =
99/50
ans =
1/100
ans =
int((99*dirac(x))/50 + rectangularPulse(0, 300, x)/30000, x, -Inf, Inf)
Cheers
  5 Comments
Torsten
Torsten on 28 Jun 2023
Edited: Torsten on 28 Jun 2023
Once the perfect version of R2023b is shipped, you should be able to verify this answer.
I hope not :-)
Paul
Paul on 17 Sep 2023
Now that 2023b is here, let's check for its perfection :)
syms x real
E(x) = 99/50*dirac(x) + rectangularPulse(0, 300, x)/30000
E(x) = 
int on first term yields expected result
int(99/50*dirac(x),x,-inf,inf)
ans = 
int on second term yields expected result
int(rectangularPulse(0, 300, x)/30000,x,-inf,inf)
ans = 
int on sum succeeds
int(E(x),x,-inf,inf)
ans = 

Sign in to comment.

More Answers (1)

VBBV
VBBV on 26 Feb 2023
Tips
  • In contrast to differentiation, symbolic integration is a more complicated task. If int cannot compute an integral of an expression, check for these reasons:
  • The antiderivative does not exist in a closed form.
  • The antiderivative exists, but int cannot find it.
If int cannot compute a closed form of an integral, it returns an unresolved integral.
Try approximating such integrals by using one of these methods:
  • For indefinite integrals, use series expansions. Use this method to approximate an integral around a particular value of the variable.
  • For definite integrals, use numeric approximations.
Look into Tips section of page for that reason, Going by tips section, if you consider the numeric approximations
syms x real
E(x) = 99/50*dirac(x) + rectangularPulse(0, 300, x)/30000
E(x) = 
int(E(x),x,-4,4) % with a numeric approximation it works
ans = 
  1 Comment
Paul
Paul on 26 Feb 2023
Edited: Paul on 26 Feb 2023
I don't think that's what the Tips section means for "numeric approximation," which, for example, would be using vpaintegral.
In this case, the exact answer is, I think:
sym(99)/sym(50) + 1/sym(100)
ans = 
And that answer can be obtained with finite limits of integration that cover the full range of E(x) where it's non-zero
syms x real
E(x) = 99/50*dirac(x) + rectangularPulse(0, 300, x)/30000;
int(E(x),x,-1e6,1e6)
ans = 
Seems like this problem with +- inf as the limits of integration is easy enough that int should return a result. Unless there's some sublety that I'm missing.
This works
int(E(x),x,-inf,1e6)
ans = 
But this doesn't?
int(E(x),x,-1e6,inf)
ans = 
I wonder if this is related to the issues raised in this Question, which I thought were fixed in R2022a.
But this does?
expand(int(E(x),x,-inf,inf))
ans = 
This sure looks like a bug to me.

Sign in to comment.

Tags

Products


Release

R2022b

Community Treasure Hunt

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

Start Hunting!