Let $S_d=\{\mathbf{\omega}=(\omega_0,...,\omega_d) \in \mathbb{R}^{d+1}_+:\sum_{i=0}^d \omega_i=1 \}$ be the unit simplex. I am reading a text which has defined a so-called spectral density $f$ on $S_d$ given by $$f(\omega)=\frac{1}{w_0^2w_1...w_d (2\pi)^{d/2}\left| \text{det}(\Sigma)\right|^{1/2}}\exp\left( -1/2 \tilde{\mathbf{\omega}}^T\Sigma^{-1} \tilde{\mathbf{\omega}} \right) $$ where $\Sigma=$ is some invertible positive definite matrix and $\tilde{\mathbf{\omega}}= (\ln(\omega_i/\omega_0)+2\lambda^2_{i}:1 \leq i \leq d)^{T}$ with $0<\lambda_{i} \leq 2$.
The text then mentions that certain quantities can be calculated as
$$\int_{S_d}g(\mathbf{\omega})f(\mathbf{\omega})d\mathbf{\omega},$$ where $g$ is another function on $S_d$.
I am wondering how the intepret/calculate the latter integral: isnt its value $0$ because we are (Riemann?) integrating over a lower dimensional region of $\mathbb{R}^{d+1}_+$? I am tempted to work with the function $f(1-\sum_{i=1}^d\omega_i, \omega_1,...,\omega_d)$ on $\{ \mathbf{\omega}=(\omega_1,...,\omega_d) \in \mathbb{R}^{d}_+:\sum_{i=1}^d \omega_i \leq1 \}$ but how can I "get" there?