Why doesn't Mathematica cancel this constant out of the integral?

Question

I have the mathematical expression

Integrate[f[r], {r, τ, t}] == a*Integrate[f[r]/a, {r, τ, t}]

which is obviously true for all functions f[r] if a is a constant. However, Mathematica does not simplify the above expression to "true". Even Simplify and FullSimplify don't work. What am I doing wrong?

It's only true, if Integrate[f[r], {r, \[tau], t}] exists, but Mathematica doesn't know about the unknown function f[r]! Substitute f by Sin for example and you get True as expected. — Ulrich Neumann, Aug 06 '20 at 11:11
@UlrichNeumann You're right. Is there a way to tell Mathematica that the integral exists for all upper and lower bounds for the general function f? — HerpDerpington, Aug 06 '20 at 11:13
Sorry no idea. Surprisingly Simplify[Integrate[f[r], r] == a*Integrate[f[r]/a, r] , Assumptions -> a > 0] works! — Ulrich Neumann, Aug 06 '20 at 11:18
@UlrichNeumann Maybe thats because it's not defined for a == 0... — HerpDerpington, Aug 06 '20 at 11:19
Simplify[Integrate[f[r], r] == a*Integrate[f[r]/a, r], Assumptions -> a != 0] does the job. — user64494, Aug 06 '20 at 11:51
In the case of indefinite integrals, the cancellation appears to work even without Simplify (in MA 12.0.0.0) — Hausdorff, Aug 06 '20 at 12:11
@user64494 It does not matter in this case, since the result does not depend on the value of a (the limit $a\rightarrow 0$ is defined). Integrate[f[r], r] == a*Integrate[f[r]/a, r] yields True for me — Hausdorff, Aug 06 '20 at 12:28
@Hausdorff: Thank you for your personal opinion. However, I was taught division by zero is impossible. I think the answer is generic. — user64494, Aug 06 '20 at 12:34
In V12.1.1, a*Integrate[f[r]/a, r] evaluates to Integrate[f[r], r] — Michael E2, Aug 06 '20 at 12:38
@Michel E2: That was discussed some time ago. I have no time to find a reference. — user64494, Aug 06 '20 at 12:39
@HerpDerpington Try linearExpand[ a Inactive[Integrate][f[r]/a, r]] — Michael E2, Aug 06 '20 at 12:48
@usr64494 Yes, it was, so I'm surprised you're bringing it up. You still haven't learned how Mathematica works. You might also extend your knowledge of the mathematics of rational functions — Michael E2, Aug 06 '20 at 13:00
Michael E2: I was taught traditional math where the domains are taken into account. — user64494, Aug 06 '20 at 13:20
@MichaelE2 Unfortunately that method does not work in MA 8, since Inactivate only exists from MA 10 onward. — HerpDerpington, Aug 06 '20 at 13:21
Remove the Inactive? Change to plain Integrate. That might work. You can also replace Inactive with Hold or HoldForm (expr /. Integrate -> HoldForm[Integrate]). You have to do that in the code of — Michael E2, Aug 06 '20 at 14:08

score 2 · Accepted Answer · edited Aug 06 '20 at 22:12

2

Mathematica does not like to take things out of integrals.

You can define the following rule:

repl = Integrate[x_ y__, {var_, bounds__}] /; FreeQ[x, var] :> 
  x Integrate[y, {var, bounds}];

And then Mathematica will happily factor out the "constants"

Integrate[f[r], {r, τ, t}] == 
  a*Integrate[f[r]/a, {r, τ, t}] /. repl
(* True *)

edited Aug 06 '20 at 22:12

march

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answered Aug 06 '20 at 11:19

Natas

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That breaks in this case for me: (f[t] * Integrate[g[r], {r, a, t}] == Integrate[f[t]*g[r], {r, a, t}]) /. repl evaluates the RHS integral without integrating. (But it does cancel constants...) – HerpDerpington Aug 06 '20 at 11:36
@HerpDerpington Your example works for me (MA 12.0.0.0) – Hausdorff Aug 06 '20 at 12:05
@Hausdorff I have MA 8.0.0.0 – HerpDerpington Aug 06 '20 at 12:06

Why doesn't Mathematica cancel this constant out of the integral?

1 Answers1