There is a system of N non-interacting particles (Ideal Gas). The Hamiltonian of a system of free particles is given by:
$$H = \sum_{i=1}^{N}\frac{p_{i}^2}{2m} + \sum_{i=1}^{N} \psi(q_i)$$
where to the kinetic term we have added a confining potential:
$$\psi(q_i) = \begin{cases} 0 & q_i \in V\\ \infty & otherwise\end{cases}$$
which keeps the particles inside the volume $V$.
First of all, some definitions:
$$\Omega (E, V, N) = \int \frac{d\Gamma}{N!\hbar^3N} \Theta(E - H(\Gamma))$$
Where $d\Gamma = dp_1...dp_Ndq_1...dq_N$ (p is momentum and q position) and $\Theta$ is the step function.
I want to compute the microcanonical phase space volume for an ideal gas. To do so I have to solve the following integral:
$$\Omega (E, V, N) = \int \frac{d\Gamma}{N!\hbar^3N} \Theta(E - H(\Gamma)) = \int q_1\int q_2\int q_N\int \frac{dp_1...dp_N}{N!\hbar^3N}\Theta(E - \sum(\frac{p_{i}^2}{2m}))$$
I know the solutions follows as:
$$\int q_1\int q_2\int q_N\int \frac{dp_1...dp_N}{N!\hbar^3N}\Theta(E - \sum(\frac{p_{i}^2}{2m})) = \frac{V^N}{N!\hbar^3N}\int_{\sum \frac{p_{i}^2}{2m} \leq E} dp_1...dp_N = \frac{V^N}{N!\hbar^3N} V'_{3N}\sqrt{2mE}$$
But I do not really understand how can we go from having the step function to just the integral over $dp_1...dp_N$.
Consider the definition of the step function:$$\Theta(x)=\cases{1,x>0\\0,x<0}$$The step function is equal to $1$ if all particles are inside the volume $V$ and if $\sum \frac{p_{i}^2}{2m} \leq E$. If one particle is outside, $\psi_i(q_i)=\infty$, so if everything else is finite you get $\Theta(-\infty)=0$. That's why the integral over each of $dq_i$ yields a factor of $V$. If now the sum over the kinetic energies is grater than $E$, the $\Theta$ function is also $0$, so if you write $$\int dp_1...dp_n=\int_{\sum \frac{p_{i}^2}{2m} \leq E} dp_1...dp_n\Theta(E-\sum \frac{p_{i}^2}{2m})+\int_{\sum \frac{p_{i}^2}{2m} \gt E} dp_1...dp_n\Theta(E-\sum \frac{p_{i}^2}{2m})$$ then the second integral is $0$, while the first integral is just $$\int_{\sum \frac{p_{i}^2}{2m} \leq E} dp_1...dp_n$$