Why Can't int Find a Simple Integral?

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?

Answer 1

[VBBV]

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 = 

[Paul]

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.