10
$\begingroup$

I have the following code:

 Simplify[Integrate[f[x] + g[x], x] == Integrate[f[x], x] + Integrate[g[x], x]]

To test:

$$\int{\left(f(x) + g(x)\right)dx}=\int{f(x)dx}+\int{g(x)dx}$$

Why doesn't this simplify to True? More importantly, how can I get this to evaluate to True? I'm going through some manipulations of equations, and I'd like to use Mathematica to doublecheck my math and help with it.

I run into the same problem with Sum, where Integrate in the previous equation can be replaced by Sum. I'm using Mathematica 8.0.4.0.

| improve this question | |
$\endgroup$
  • $\begingroup$ I found that adding d[expression ,x] with Simplify or FullSimplify will evaluate to true. In other words, I am taking the derivative of the expression, and that seems to eliminate the integrals. I'd like to know if there are other workarounds, though. The new code is: FullSimplify[ D[Integrate[a[x] + b[x], x] == Integrate[a[x], x] + Integrate[b[x], x], x]] $\endgroup$ – Matt Groff Jul 15 '12 at 22:56
  • $\begingroup$ Why do you need any workarounds ? Your last issue with FullSimplify is just an inconsequence of M. Look at Integrate[a[x] + b[x], x], Integrate[a[x], x] and, all they return the results as the integrals would actually exist, moreover as they would be differentiable. While that is not correct in general. $\endgroup$ – Artes Jul 16 '12 at 0:40
  • $\begingroup$ @Artes: I was just wondering if there are any related features available. What did you mean by M? $\endgroup$ – Matt Groff Jul 16 '12 at 2:03
  • $\begingroup$ M like Mathematica. I meant it wasn't a good feature of M when it returned True in that case. Look that you need not FullSimplify to get True, evaluate that expression without FullSimplify. It shouldn't return True, unless you assume a[x] to be integrable/ differentiable e.g. a[x_]:=x^2. There are many imperfect issues. $\endgroup$ – Artes Jul 16 '12 at 2:14
14
$\begingroup$

There is nothing wrong with the issue in the question.

Mathematica shouldn't evaluate 

Simplify[ Integrate[f[x] + g[x], x] == Integrate[f[x], x] + Integrate[g[x], x]]

to True, because in general such a rule would be mathematically simply wrong.

Consider e.g. $ \forall_{x } f(x) = - g(x)$, while $f$ is not e.g. Lesbegue integrable. Of course Mathematica does not distinguish integrability subtleties, nevertheless there should be 

Integrate[f[x] + g[x], x] == 0

while $\int{f(x)dx}$ and $\int{g(x)dx}$ don't not exist. Q.E.D

One should emphasize that the rule works well if we define appropriate integrable functions, e.g. 

f[x_] := x^3 + 2 x + 1
g[x_] := Hypergeometric1F1[3, 7, x]
Simplify[ Integrate[ f[x] + g[x], x] == Integrate[ f[x], x] + Integrate[ g[x], x]]

True

Edit

If there is a need for a rule distributing integrals over a sum of functions (knowing that we can do this in an appropriate class of functions) we could define an adequate rule, e.g. 

intRule = Integrate[a_ + b_, c_] :> Integrate[a, c] + Integrate[b, c];

and use it with ReplaceRepeated (//.)

( Integrate[ f[x] + g[x] + h[x], x] //. intRule ) == 
  Integrate[f[x], x] + Integrate[g[x], x] + Integrate[h[x], x]

True

in the traditional form it yields :

Integrate[ f[x] + g[x] + h[x], x] //. intRule // TraditionalForm

enter image description here

| improve this answer | |
$\endgroup$
  • $\begingroup$ @MattGroff Thanks. $\endgroup$ – Artes Jul 29 '12 at 17:23
8
$\begingroup$

To get the behaviour you want, you can tell Simplify to try distributing Integrate over Plus as one of its transformations:

Simplify[Integrate[f[x] + g[x], x] == Integrate[f[x], x] + Integrate[g[x], x], 
 TransformationFunctions -> {Automatic, # /. i_Integrate :> Distribute[i] &}]

True

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.