Assuming[Element[n, Integers], Integrate[Sin[x]*Sin[n*x],{x,0,Pi}]]

returns 0, which is obviously wrong for n=1.

Assuming[n==1, Integrate[Sin[x]*Sin[n*x],{x,0,Pi}]]

does return Pi/2 (so in particular not 0).

Just evaluating the integral yields


which is indetermined for n=1.

Can someone explain to me what Mathematica is doing and how to obtain a correct result?

  • $\begingroup$ The last expression is not undetermined if you take the limit Limit[-(Sin[n \[Pi]]/(-1 + n^2)), n -> 1]. $\endgroup$ Feb 11, 2014 at 11:23
  • $\begingroup$ Yes, l'Hospital works, however Mathematica does not seem to use that for simplification. Simplify[-Sin[n Pi]/(-1+n^2), Assumptions -> Element[n,Integers]] also gives 0 . $\endgroup$
    – chris
    Feb 11, 2014 at 11:29
  • $\begingroup$ I answered a similar question on stackoverflow, and I don't have time right now to transcribe it here. $\endgroup$
    – rcollyer
    Feb 11, 2014 at 12:46
  • $\begingroup$ In document (F1) of Simplify and FullSimplify, the Possible Issues section is related to this problem. $\endgroup$
    – Yi Wang
    Feb 11, 2014 at 12:55
  • $\begingroup$ Thanks for pointing me to the Possible Issues section. $\endgroup$
    – chris
    Feb 11, 2014 at 13:33

1 Answer 1


If you look in the help file for Integrate under the section on "Possible Issues", there is an explanation. The docs comment: "Parameters like n are assumed to be generic inside indefinite integrals:" and the example is given of Integrate[x^n, x] which returns x^(1 + n)/(1 + n). As with the OPs integral, the answer is true for generic n, but not for a specific value n=-1 (or n=1 for the OPs integrand).

I had not seen it before, but rcollyer's explanation of this issue is very detailed and has some good pointers on how to handle this kind of situation. His post begins: "Sometimes, it pays to understand the integrand better before you integrate..." In the particular integral here, the integral of Sin[x]*Sin[n x], it is clear that n=1 is very special -- it is also clear from the indefinite integral answer -Sin[n*Pi]/(-1+n^2) that n=1 is special. But I know of no way to automatically tell what the generic/specific conditions are in all cases.

  • $\begingroup$ This is easy to see with Integrate[Cos[n x], {x, 0, Pi}] which equals $\pi$ when $n=0$, but $0$ otherwise. $\endgroup$
    – rcollyer
    Feb 11, 2014 at 13:40
  • 2
    $\begingroup$ I could agree with this explanation, nonetheless genericity makes sense in the reals (meaning e.g. non-zero measure) but it doesn't (or at least it does mean something very different) in the integers. Apparently the documentation should be improved. Similar problems which I find even more harmful one finds here Why the inequality does not take into account the domain?. I think you should mention explicitly this problem (+1). $\endgroup$
    – Artes
    Feb 11, 2014 at 13:46
  • 1
    $\begingroup$ @rcollyer This is very weird. Try quitting the kernel, then evaluate your example first with Assuming then with Assumptions. Both give 0. Now quit again, then evaluate first using Assumptions then with Assuming. The first one does not give 0, the second one does. The first one doesn't give zero on re-evaluation either. (Cached result?) I smell a bug. $\endgroup$
    – Szabolcs
    Feb 11, 2014 at 13:54
  • 1
    $\begingroup$ @Szabolcs I don't see that behavior on more recent sources, so if it is a bug, it is fixed. $\endgroup$
    – rcollyer
    Feb 11, 2014 at 13:56
  • 2
    $\begingroup$ @rcollyer Are you sure it's not due to the order you evaluated them in? Please see here. $\endgroup$
    – Szabolcs
    Feb 11, 2014 at 14:00

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