I'm working a homework problem whose first step is determining
$$\oint e^{-im\theta}d \theta$$
Where I already know that $m$ is an integer. I know that, in general, $\delta(m) = \oint e^{-im\theta}d \theta$. But I'm wondering what $\delta(m)$ is in this context since the m-space here is discrete. Is it the Kronecker delta?
The reason I'm thinking this is that the formula I need in the end is $\phi(r, \theta, z) = \sum_m \phi_m(r,z) e^{i m \theta} = \phi_0 (r, z)$ and if the delta is in fact the Kronecker delta it would pick out the $m=0$ from the sum like I need it to.
You are correct, $$\frac{1}{2\pi}\int_0^{2\pi} e^{-im\theta} \, \mathrm{d}\theta = \frac{1}{-2\pi im}e^{-im\theta}\Big|_0^{2\pi} = 0$$ for $m \neq 0$ and $$\frac{1}{2\pi}\int_0^{2\pi} e^{-im\theta} \, \mathrm{d}\theta = \frac{1}{2\pi}\int_0^{2\pi} 1 \, \mathrm{d}\theta = 1$$ for $m = 0$.
Another way of seeing this is that you are trying to expand the constant function $1$ as a Fourier series: $$1 = \sum\limits_{m = -\infty}^{\infty} c_me^{im\theta} \tag{1}$$ and the Fourier coefficients $c_m$ are given by $$c_m = \frac{1}{2\pi}\int_0^{2\pi} 1 \cdot e^{-im\theta} \, \mathrm{d}\theta.$$ Now it is clear that you must have $c_0 = 1$ and $c_m = 0$ for all $m \neq 0$ to satisfy $(1)$.