Mathematica and its default assumptions

Question

I am a bit surprised that evsluating
Integrate[Exp[I n x],{x, 0, Pi}]

gives
$qquad frac{-i(e^{inpi}-1)}{n}$.
What if $n = 0$? I didn't assume anything about $n$ and yet Mathematica assumes $nneq 0$.
It is the same with the Maxima command :
integrate(exp(%i*n*x),x,0,%pi);

and the Maple command
int(exp(n*x*I), x = 0 .. PI)

How do I handle this kind of behavior which I consider undesirable?
What is the name of this feature in symbolic computation, so that I can find a solution for Maxima as well.

yawnoc · Answer

$n = 0$ is a removable singularity. If you take the limit of the result of the integral, you will get the same as if you put $n = 0$ before taking the integral:
result = Integrate[Exp[I n x], {x, 0, Pi}]

Limit[result, n -> 0]
(* Pi *)

Integrate[Exp[I 0 x], {x, 0, Pi}]
(* Pi *)

Addendum
Re @cvgmt on Integrate[x^a, x] for a = -1: that also is a removable singularity (provided the constant of integration is chosen correctly). We have
$$
  int_1^x u^a ,mathrm{d}u = frac{x^{a+1} - 1}{a+1},
$$
and
$$
  lim_{a to -1} frac{x^{a+1} - 1}{a+1} = log x.
$$
In general Mathematica does not bother with treating removable singularities separately. For example, x/x will simplify to 1 without any assumptions on x.

Mathematica and its default assumptions

One Answer

Addendum

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