$\int dx\, dy\, dz\, d p_x\, dp_y\, dp_z$ Does it have any physical meaning? I was reading a Physics book. Then I saw an equation which was looking like this :
$$\int dx\, dy\, dz\, d p_x\, dp_y\, dp_z$$
I was thinking it from just a Calculus book. I can see lots of variable (momentum) changing (adding all the "pieces") respect to their position. When I saw the equation a question came to my mind which is : Does it have any physical meaning if I just think it from Mathematics? Is there possible way to evaluate it?
Usually the book wrote that
$$\int dx\, dy\, dz\, d p_x\, dp_y\, dp_z=V4\pi (\frac{hv}{c})^2 h\frac{dv}{c}$$
I am not sure if I wrote it correctly cause I had taken the image of that book illegally that's why I couldn't take whole equation pic but I didn't notice it that time.
I wonder which I was reading that wrote single integral. But it I was directly searching through online I found that another had wrote 6 integral
$$\int\int\int\int\int\int dx\, dy\, dz\, d p_x\, dp_y\, dp_z$$
which proves that they are taking integral for each function.
Note : I am honestly saying I don't understand anything of it now. And I may not understand answer properly also. But I am just leaving the question to read in future.. :)
 A: It is a common notation to indicate the integration on the 6-dimensional phase space $(PS)$  of a single particle with finite mass (otherwise the momentum $\mathbf{p}$ is not the canonical conjugate of the position $\mathbf{x}$).
In the end, $d^3x \, d^3p$ is just a way to indicate the phase-space volume element (or, better, the volume 6-form that provides the natural integration measure on the PS).
Edit: the integration bounds are omitted because they are physically "obvious": the whole volume $V \subset \mathbb{R}^3$ in which the position $\mathbf{x}$ of the particle is allowed to vary and the physically possible values of the conjugate momentum $\mathbf{p}$. Typically $\mathbf{p}\in \mathbb{R}^3$ and $\mathbf{x}\in V$. We could formally write that $PS = V\times \mathbb{R}^3 $, where $\times$ is the usual cartesian product.
Therefore: $\int d^3x = V$, where $V$ indicates also the measure of the position domain $V \subset \mathbb{R}^3$. Some examples of how you should interpret this kind of integrations (that are just standard Riemann multidimensional integrals):
$$
\int_{PS} f(\mathbf{x}, \mathbf{p}) d^3x d^3p = \text{Integrate $f$ over the whole PS.}
$$
$$
\int_{PS} f( \mathbf{p}) d^3x d^3p =  V \int_{\mathbb{R}^3} f( \mathbf{p})  \, d^3p
$$
$$
\int_{PS} f(p) d^3x d^3p =  4 \pi V \int_{\mathbb{R}^+} f( p) \,  p^2dp \quad (\text{where } p=|\mathbf{p}|)
$$
Note that the last equality should help you to understand the example reported in the question (even though I do not understand the factors $c$ and $h$ there and the meaning of $v$, but I guess that $p = \frac{h}{c}v$):
$$
\int_{PS} f(p) d^3x d^3p =   \int_{\mathbb{R}^+} f( p) \, V4\pi \left(\frac{hv}{c}\right)^2 \frac{h\, dv}{c} \quad (\text{where } p=|\mathbf{p}|=\frac{h\, v}{c})
$$
Final note: here are some Terence Tao's notes on the phase space (that is a useful concept in classical mechanics, kinetic theory, statistical physics, quantum mechanics, chaos theory, dynamical systems).
