I have a problem which surely someone else had on here, but in reading similar questions, I haven't found an answer.
The problem is really trivial:
Integrate[Sin[k*x]^2, {x, -Pi, Pi},
Assumptions -> {k ∈ Integers}]
That shall simply give me
$$\pi$$
Because the whole result would be
$$\pi -\frac{\sin (2 \pi k)}{2 k}$$
But since $k\in\mathbb{Z}$ we have $\sin(2\pi k) = 0$ for every $k$.
Why doesn't mathematica print $\pi$ in the output, simply?
P.s. Of couse "Elements" denotes the $\in$ symbol. Here something went wrong during the copy of the input text!
k == 0
? $\endgroup$