We are talking about a non-compact Riemannian manifold, right? Then Weyl's law may be incorrect, at least out of the box. Let us consider $H = -\Delta$ (so $V(x)=0$) in the domain $X \subset \mathbb{R}^d$ with the Dirichlet or Neumann boundary condition (the case of the closed manifold is very similar to the Neumann case).

Let non-compact part of $X$ be $\{x\colon x_1 > c, \ x'\in \rho(x_1) \Omega\}$ with $x'=(x_2,\ldots, x_d)$, $\Omega$ a compact domain with a smooth boundary in $\mathbb{R}^{d-1}$ and $\rho \to 0$ as $x_1\to \infty$.

Observe that depending on the rate of decay of $\rho$ the volume of $X$ could be finite or infinite (and the area of $\partial X$ also could be finite or infinite).

The literal Weyl's law says: if $X$ has a finite volume then the spectrum is discrete and Weyl's law holds and if $X$ has an infinite volume then there is an essential spectrum.

However the reality is more nuanced:

**Dirichlet Laplacian**
In this case spectrum is discrete for sure; simply in the case of the infinite volume the correct asymptotic expansion is different. Let $\rho (t)= t^{-\mu}$, $\mu >0$. Volume is finite for $\mu (d-1)> 1$ and infinite otherwise.

- Literal Weyl's law holds for $\mu (d-1)> 1$.
- For $\mu (d-1)=1$ (logarithmic divergence) $N(\lambda)\asymp \lambda ^{(d-1)/2}\ln (\lambda)$ and is given by the same formula but with integration over $x_1\colon \rho(x_1)>\lambda^{-1/2}$.
- For $\mu (d-1)<1$ (power divergence)
$$
N(\lambda)\sim \sum_{j} n_j(\lambda)
\tag{D}
$$
where $n_j(\lambda)$ is a Weyl's expression for $1$-dimensional Schrödinger operator $\mathsf{h}_j = -\partial_1^2 +\nu_j \rho(x_1)^{-2}$ and $\nu_j >0$ are eigenvalues of $(d-1)$-dimensional Dirichlet Laplacian in $\Omega$.

**Neumann Laplacian**
Then things change drastically. There is an essential spectrum unless $\rho\to 0$ really fast. To understand why look at (D) but instead of a Dirichlet Laplacian in $\Omega$ one should consider a Neumann Laplacian, and $\nu_1=0$. Does it mean *essential spectrum*?

- If $\rho =x_1^{-\mu}$ or even $\rho =\exp (-c x_1)$ then yes.
- But if $\rho = \exp(-x_1^{k+1})$ with $k>0$ then no, and
$$
N(\lambda) \sim N^W(\lambda) + N^c(\lambda)
\tag{N}
$$
where $N^W(\lambda)$ is a Weyl expression for an original operator, and $N^c(\lambda)$ is a Weyl expression for $1$-dimensional Schrödinger
$\mathsf{h}_0= -\partial_1^2 + W(x_1)$ with a potential $W =\frac{1}{4} (\partial_{x_1} \log \rho (x_1))^2$ and depending on $k$ either the first or the second term in (N) may prevail.

**For Dirichlet/Neumann case in more general setting see subsections 3.2.2/3.2.3 of ***100 years of Weyl's law* either in arXiv or in Bull Math Sci