I'd like to show this formula that I found without proof in a book: $$\sum\limits_{z\in\mathbb{Z}^3,|z|\leq\omega}\frac{1}{|z|^2-k^2}=4\pi\omega + O(1) \qquad\text{ as }\quad\omega\rightarrow\infty$$ where $k$ is a fixed complex number with positive imaginary part, $\omega$ is real and $|.|$ denotes the Euclidean norm in $\mathbb{R}^3$.
But I really don't known by which angle to attack this problem. Does anyone have an idea/suggestion on the method to use?
Without making any claims as to the degree of validity of the approximation, it looks like they approximated the sum with an integral over the ball defined by the summation limits. That is, they replaced the sum by
$$\int_0^{\omega} dr \, \frac{r^2}{r^2-k^2} \, \int_0^{\pi} d\theta \, \sin{\theta} \, \int_0^{2 \pi} d\phi = 4 \pi \int_0^{\omega} dr \, \frac{r^2}{r^2-k^2}$$
Note that the integral over $r$ has no singularities because $k$ is complex with a positive imaginary part. The integral over $r$ may be done easily by observing that
$$\frac{r^2}{r^2-k^2} = 1 + \frac{k^2}{r^2-k^2}$$
so that the integral evaluates to
$$4 \pi \omega + 2 \pi k \int_0^{\omega} dr \left (\frac{1}{r-k} - \frac{1}{r+k} \right ) = 4 \pi \omega + 2 \pi k \left [\log{\left ( \frac{\omega-k}{\omega+k}\right)} - \log{\left ( \frac{-k}{k}\right)} \right ] $$
or
$$4 \pi \omega + 2 \pi k\left [\log{\left ( \frac{\omega-k}{\omega+k}\right)} -i \pi\right]$$
As $\omega \to \infty$, the log goes to zero and we get that the integral approaches
$$4 \pi \omega - i 2 \pi^2 k = 4 \pi \omega + O(1)$$
as asserted.
(Note: I chose $\log{(-1)}=i \pi$. One may choose a different branch; it does not affect the result so long as the choice is consistent throughout.)
