Asymptotics of the Lattice Green Function Let
$$
\mu_k=\sum_{i=1}^d (2\sin (k_i/2))^2, \quad k\in \mathbb{R}^d
$$
Then it's not hard to show that if $d>2$, then
$$
\int_{(-\pi,\pi]^d} \frac{e^{ikx}}{\mu_k} dk =\frac{1}{|x|^{d-2}}C_d(x), \quad x\in \mathbb{Z}^d
$$
where $C_d(x)\to\text{const}$ as $x\to \infty$, or formally,
$$
C_d(x)\to \int \frac{e^{ip_1}}{p^2} dp,  \quad |x|\to\infty
$$
It then seems reasonable to say that $C_d(x) = \text{const} +O(|x|^{-\epsilon})$ for some $\epsilon >0$. However, how would one go about to prove this?
 A: I think I found a nice solution using the local central limit theorem. Let me provide a sketch of the proof, since it's a lot to write down.

Theorem. Let $d>2$. Then the lattice Green function $C(x)$ satisfies
$$
C(x) \equiv\int_{(-\pi,\pi]^d} \frac{dk}{(2\pi)^d} \frac{e^{ikx}}{\mu_k} =\frac{c_d}{|x|^{d-2}}+O\left( \frac{1}{|x|^d} \right)
$$
where $c_d$ is a constant such that $c_d/|x|^{d-2}$ is the Fourier transform (as tempered distribution) of $1/k^2$, i.e., the continuum Green function $G(x)$, formally written as,
$$
G(x) \equiv\int\frac{dk}{(2\pi)^d}\frac{e^{ikx}}{k^2} =\frac{c_d}{|x|^{d-2}} 
$$

To prove this, we will need the local central limit theorem, i.e.,

Lemma (LCLT). Consider a simple random walk on $\mathbb{Z}^d$ with transition matrix $p$ and let $p_n(x) =p^n(0,x)$ be the probability of starting at $0$ and ending at $x$ after $n$ steps. Define $\bar{p}_0(x)=\delta(x)$ and
$$
\bar{p}_n(x) = 2 \left( \frac{d}{2\pi n} \right)^{d/2} e^{-dx^2/2n}
$$
Then the error $E_n(x) = p_n(x) -\bar{p}_n(x)$ satisfies
$$
E_n(x)=\frac{1}{|x|^{2m}} O\left( \frac{1}{n^{(d+2-2m)/2}}\right), \quad m\in \mathbb{N}
$$

Also notice that the lattice Green function is equal to
$$
C(x) =\sum_{n=0}^\infty p_n(x)
$$
which you can show using the fact that the characteristic function $\varphi_n(k)$ of $p_n(x)$ is equal to $=\varphi(x)^n$ where $\varphi(x)$ is the characteristic function of $p(x)$.  From my answer here, you see that the continuum lattice Green function can be written as
$$
G(x)= \frac{1}{(4\pi)^{d/2}}\int_\epsilon^\infty t^{-d/2} e^{-x^2/4t} dt
$$
Notice the similarity between $G(x)$ and $\sum \bar{p}_n(x)$. Using the lemma, we can do some approximations that will ultimately lead to an error $O(|x|^{-d})$. I might try to fill in the details later on.
