Simplification of integrals depending on a parameter [duplicate]

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

-Sin[n*Pi]/(-1+n^2)


which is indetermined for n=1.

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

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marked as duplicate by Artes, rcollyer, Michael E2, Szabolcs, bobthechemistFeb 11 at 15:40

The last expression is not undetermined if you take the limit Limit[-(Sin[n \[Pi]]/(-1 + n^2)), n -> 1]. –  b.gatessucks Feb 11 at 11:23
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 . –  chris Feb 11 at 11:29
I answered a similar question on stackoverflow, and I don't have time right now to transcribe it here. –  rcollyer Feb 11 at 12:46
In document (F1) of Simplify and FullSimplify, the Possible Issues section is related to this problem. –  Yi Wang Feb 11 at 12:55
Thanks for pointing me to the Possible Issues section. –  chris Feb 11 at 13:33

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.
This is easy to see with Integrate[Cos[n x], {x, 0, Pi}] which equals $\pi$ when $n=0$, but $0$ otherwise. –  rcollyer Feb 11 at 13:40