# Switching from sum to integral

I'm specifically asking about an equation in An Introduction to Quantum Field Theory, by Peskin and Schroeder. Example from page 374:

$$\mathrm{Tr} \log (\partial^2+m^2) = \sum_k \log(-k^2+m^2)$$ $$= (VT)\cdot\int\frac{\mathrm{d}^4k}{(2\pi)^4}\log(-k^2+m^2),\tag{11.71}$$ The factor $VT$ is the four-dimensional volume of the functional integral.

Why does this $VT$ show up in equation $(11.71)$?

One may only talk about a discrete sum over $k^\mu$ vectors if all the spacetime directions are compact. In that case, $k^\mu$ is quantized.
If the spacetime is a periodic box with periodicities $L_x,L_y,L_z,L_t$, then $V=L_x L_y L_z$ and $T=L_t$. The component $k^\mu$ in such a spacetime is a multiple of $2\pi \hbar / L_\mu$ (I added $\hbar$ to allow any units but please set $\hbar=1$) because $\exp(ik\cdot x / \hbar)$ has to be single valued and it's single-valued if the exponent is a multiple of $2\pi i$.
So the total 4-volume in the $k$-space that has one allowed value of $k^\mu$ – one term in the sum – is $(2\pi)^4 /(L_x L_y L_z L_t) = (2\pi)^4 / (VT)$. It means that if one integrates over $\int d^4 k$, one has to divide the integral by this 4-volume, i.e. multiply it by $(VT)/(2\pi)^4$, to get the sum – to guarantee that each 4-dimensional box contributes $1$ as it does when we use the sum. In the limit $L_\mu \to \infty$, the integral divided by the 4-volume of the cell and the sum become the same – via the usual definition of the Riemann integral.