Q:

Integrating $\prod_{i=k+1}^{kN} \left(\int_{-\infty}^{\infty}dp_i\right)\times$ with conditions on $p_i$

I am trying to integrate this expression which came up in a derivation of the momentum distribution function for an ideal gas. $\Theta(x)$ is the Heaviside step function which is $1$ when $x$ is positive and $0$ if negative.

$$\prod_{i=k+1}^{kN} \left(\int_{-\infty}^{\infty}dp_i\right)\times\left[ \Theta\left(2mE - \sum_{i=1}^{kN}p_i^2\right) - \Theta\left(2m(E-\triangle E) - \sum_{i=1}^{kN}p_i^2\right)\right]$$

The integrals here are nested like $\int_{-\infty}^\infty dp_{k+1}\int_{-\infty}^\infty dp_{k+2}\cdots \int_{-\infty}^\infty dp_{kN}$
I also want to get some physical intuition as to what this integral means. I have done the integration where the $i$ in the integral goes from $1$ to $kN$. This is easily interpreted as the volume contained in between two $kN$ dimensional spheres of radii $\sqrt{2mE}$ and $\sqrt{2m(E-\triangle E)}$. But in this case what does it mean when $i$ runs from $k+1$ to $kN$?
EDIT: There is a condition which I forgot to mention: $$p_1^2+p_2^2+\cdots+p_k^2=constant=a^2 $$
 (To physicists, this means I am fixing the magnitude of momentum of one particle out of $N$ particles in a $k$-dimensional space)

A:

The integral can be rewritten as: 

$$\prod_{i=k+1}^{kN} \left(\int_{-\infty}^{\infty}dp_i\right)\times\left[ \Theta\left(2mE - a^2 - \sum_{i=k+1}^{kN}p_i^2\right) - \Theta\left(2m(E-\triangle E)-a^2 - \sum_{i=k+1}^{kN}p_i^2\right)\right]$$

Now this is simpy the volume contained between two $kN-k$ dimensional spheres. Since the volume of an $n$ dimensional sphere is $\frac{\pi^{n/2}}{\Gamma(n/2+1)}R^n$ where $R$ is the radius, the resulting integral simplifies to:
$$\frac{ \pi^{ \frac{kN-k}{2} } }{ \Gamma(\frac{kN-k}{2}+1) }\times \left[ (2mE-a^2)^{\frac{kN-k}{2}}-(2m(E-\triangle E)-a^2)^{\frac{kN-k}{2}} \right]$$

