The following expression $$\int_{0}^{t} f(t_0) \,dt_0 - \int_{0}^{t} f(t_1) \,dt_1 = 0$$ is zero because it is just a change of integration variable.

Why doesn't Mathematica give zero in this case?

  • $\begingroup$ Maple gives zero simplify(int(f(t1),t1=0..t)-int(f(t2),t2=0..t)) gives zero. So I think you are right, Mathematica should give zero. I tried both Simplify and FullSimplify. !Mathematica graphics may be there a trick to make it simplify it to zero $\endgroup$
    – Nasser
    Oct 31, 2023 at 11:21
  • $\begingroup$ Mathematically it is zero, I believe the logic just hasn't been implemented in mathematica $\endgroup$
    – Luca
    Oct 31, 2023 at 11:34
  • $\begingroup$ You have to define what your f[x] is. Not all functions can be integrated, for example, Tan[x] is infinite when x=pi/2 $\endgroup$
    – I. C. Lin
    Oct 31, 2023 at 11:36
  • 1
    $\begingroup$ I have an application with many nested integrals, and this behaviour is just a prerequisite so that I can calculate some results in high order perturbation theory. $\endgroup$
    – Luca
    Oct 31, 2023 at 11:58
  • 3
    $\begingroup$ I agree. Mathematica understands many symbolic relations. The fact that an integral shouldn't depend on the name of the dummy variable should be fairly easy to spot. $\endgroup$ Oct 31, 2023 at 12:16

3 Answers 3


You have to tell what your function is. Mathematica cannot output undefined values, for example, the integral result is infinite in some interval.

f[x_] := Tan[x];
Integrate[f[t1], {t1, 0, \[Pi]/2}] - Integrate[f[t2], {t2, 0, \[Pi]/2}]


f[x_] := Tan[x];
Integrate[f[t1], {t1, 0, \[Pi]/3}] - Integrate[f[t2], {t2, 0, \[Pi]/3}]

You can compare the 2 results.

  • 5
    $\begingroup$ "You have to tell what your function is. Mathematica cannot output undefined values"... this doesn't sound right? It can clearly turn f(x) - f(x) into 0 despite everything being undefined in that expression. $\endgroup$
    – user541686
    Nov 1, 2023 at 7:23
  • 1
    $\begingroup$ I believe here Mathematica is correct and Maple is wrong - if f were a function such that the integral is infinite, the difference should be indeterminate, not zero. So one cannot just simplify it to zero without assuming the integral is finite. $\endgroup$
    – arkeet
    Nov 5, 2023 at 19:20
FullSimplify[Integrate[f[t1], {t1, 0, t}] - Integrate[f[t2], {t2, 0, t}], 
 Assumptions -> t1 == t2]

(* 0 *)

Or if you encounter such situation mentioned in @Domen's comment:

expr = Integrate[f[t1], {t1, 0, t}] - Integrate[f[t2], {t2, 0, t}] + 
   t1 - t2;

expr /. Integrate[x_, {y_, z_, q_}] :> 
  Integrate[x /. y -> uniquename, {uniquename, z, q}]

(* t1 - t2 *)
  • $\begingroup$ Just a note, one has to be quite careful here because this doesn't take into account the correct local scoping of integration variables. Namely, FullSimplify[Integrate[f[t1], {t1, 0, t}] - Integrate[f[t2], {t2, 0, t}] + t1 - t2, Assumptions -> t1 == t2] will give 0 instead of t1 - t2. $\endgroup$
    – Domen
    Oct 31, 2023 at 13:57

This can be grounded in such a way (of course, assuming the existence of the integral Integrate][f[t2], {t2, 0, t}]):

Activate[IntegrateChangeVariables[Inactive[Integrate][f[t2], {t2, 0, t}], t1, t1 == t2]]

Integrate[f[t1], {t1, 0, t}]


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