According to another solution here the solution should sound like $2\,\pi\,\delta(k)$
I ask myself: what's the matter of including that factor $2\pi$ ? The reason being $\delta(k)$ is infinite for $0$ and $0$ else, that $2\,\pi$ actually has no effect at all?
Don't think of the local value at a point. Here we are dealing with distributions.
The Dirac delta distribution $\delta$ satisfies $\int \delta(k) \, f(k) \, dk = f(0)$ and so, multiplication by $2\pi$ gives another value: $$\int 2 \pi\delta(k) \, f(k) \, dk = 2\pi f(0).$$