Counting prime points in a bounded region

Question

Let $R$ be a semi-algebraic, compact region in $\mathbb{R}^n$ with positive Lebesgue measure. Let $N(R) = \# (R \cap \mathbb{Z}^n)$. Davenport's lemma asserts that we have

$$\displaystyle N(R) = \operatorname{Vol}(R) + O(\max\{\operatorname{Vol} \overline{R}\}),$$

where the maximum is taken over all projections $\overline{R}$ of $R$ onto coordinate subspaces obtained by setting some coordinates to zero. Is there an analogous statement for

$$\displaystyle N'(R) = \# (R \cap \mathfrak{P}^n),$$

where $\mathfrak{P}$ denotes the set of primes and their negatives (so that $N'(R)$ counts the number of lattice points in $R$ whose coordinates are all primes or the negative of a prime)?

Answer 1

The best known version of the prime number theorem in short intervals is $\pi(x+y)-\pi(x)\sim\frac{y}{\log x}$, provided that $x^{7/12}<y=o(x)$. So you can cut a connected set $C\subseteq[0,x]^k$ into cubes with side length $x^{7/12}$, and discard all cubes intersecting the boundary. In this way you obtain that the number of lattice points with prime coordinates equals $(1+o(1))\int_C \frac{dx_1 \dots dx_k}{\log x_1\dots\log x_k} + \mathcal{O}(x^{7/12}|\partial C|)$. You now need some bound on the surface of your set, which was already done by Davenport.