Summation formula in dimension 2

Question

One of the most common tools in analytic number theory is the summation by parts, my question is what is the similar formula when we are, for example, in dimension two and we have the sum $$ \sum_{|z|<x}a_{z}f(z) $$ where $z$ runs on the gaussian integer and f is a differenziable function on $\mathbb{C}$. Thank you in advance.

Answer 1

When $f$ is a radial function, i.e. $f(z)=f(|z|)$ (forgive the abuse of notation), then partial summation is straightforward.

Let $A(x) = \sum_{|z| < x} a_z$. Then we can adapt the usual proof for summation by parts.

$$\sum_{|z|<x} a_z f(|z|) = \int_1^x f(t)\,dA(t) = A(x)f(x) - \int_1^xA(t)f'(t)dt.$$