For a system of $N$ identical particles we deal in quantum mechanics with wave functions $\langle \{\mathbf{r}_i \} \mid \Psi \rangle=\Psi(\mathbf{r}_1,\dots,\mathbf{r}_N)$ from which determine the probabilities of the particles being at positions $\{\mathbf{r}_1,\dots,\mathbf{r}_N \}$, $$\langle \Psi \mid \Psi \rangle=\langle \Psi \mid \sum_{\mathbf{r}_i}\mid \{\mathbf{r}_i \} \rangle\langle \{\mathbf{r}_i \} \mid \Psi \rangle=\sum_{\mathbf{r}_i}|\Psi(\mathbf{r}_1,\dots,\mathbf{r}_N)|^2=1$$ where $\displaystyle \sum_{\mathbf{r}_i} = \int \prod_{i=1}^Nd^3r_i$.
I cannot see why $\displaystyle \sum_{\mathbf{r}_i} = \int \prod_{i=1}^Nd^3r_i$ holds. Surely we should integrate over the momentum as well? I think this because earlier in my notes it states that:
All microstates should in principle be included in the partition sum over states irrespective of the macroscopic properties of the system
Also why dont we have a factor of $\frac{1}{N!}$ to represent the indistinguishability of particles?