Show that for the simple random walk on $\mathbb{Z}^D$ with $D\in\mathbb{N}$ one has $(P^{2n})_{0,0}\cdot D^{\frac{n}{2}}\overset{n\to\infty}{\to}const.\cdot D$ by taking the following steps.
(i) Let $\mu$ be a probability measure on $\mathbb{Z}^D$ then one has $\mu(\{x\})=(2\pi)^{-D}\int_{[-\pi,\pi]^D} e^{-i\langle{t,x}\rangle}\phi_\mu(dt)$, where $\phi_\mu$ is the characteristic function of $\mu$.
(ii) Using (i) show that $(P^{2n})_{0,0}=(2\pi)^{-D}\int_{[-\pi,\pi]^D} D^{-2n}(\cos(x_1)+...+cos(x_D))^{2n}dx,$ where $P$ is the trasition matrix for the random walk, so $P_{x,y}=\frac{1}{2^D}$, if $d_1(x,y)=1$ and $P_{x,y}=0$ else.
(iii) Deduce the claim from (ii).
Can someone please help me out here? I can show (i) and the proof of (iii) should be straight forward too, but I am stuck with (ii).
Denote by $S_n= X_1 + \cdots + X_n$ the random walk and by $\mu_n$ the distribution of $S_n$, that is $\mu_n(\{x\}) = \mathbb{P}(S_n= x)$. To apply (i), we need to compute the characteristic function of $\mu_n$. Using that the $X_i$ are i.i.d, we have $$\phi_{\mu_{n}}(t) = \mathbb{E}\left[e^{i\langle t,S_n\rangle} \right]= \mathbb{E} \left[e^{i\langle t, X_1 \rangle}\right]^{n}. $$ Now let $v_1 = (1, 0 , \ldots,0), \ldots, v_D = (0, \ldots, 0, 1)$ be the canonical base of $\mathbb{R}^D$. Since $X_1$ is uniformly distributed over $\{ \pm v_1 , \ldots , \pm v_D \}$ (by definition of the simple random walk), we have $$ \begin{align*}\mathbb{E} \left[e^{i\langle t,X_1 \rangle} \right]&= \sum_{k = 1}^D \left(\mathbb{P}(X_1 = v_k)e^{i \langle t, v_k\rangle}+ \mathbb{P}(X_1 = -v_k)e^{-i \langle t, v_k\rangle}\right) \\&= \frac{1}{2D} \sum_{k=1}^D \left(e^{it_k} + e^{-it_k} \right) = \frac{1}{D} \sum_{k=1}^D \cos(t_k).\end{align*}$$
Finally, noticing that $(P^{2n})_{0,0} = \mathbb{P}(S_{2n} = 0) = \mu_{2n}(\{0\})$, it follows from (i) that $$\begin{align*}(P^{2n})_{0,0} &= (2\pi)^{-D} \int_{[-\pi,\pi]^D}\phi_{\mu_{2n}}(t) \, \mathrm{d} t\\&= (2\pi)^{-D} \int_{[-\pi,\pi]^D} D^{-2n} \left(\sum_{k=1}^D \cos(t_k) \right)^{2n} \, \mathrm{d} t.\end{align*}$$