Replace a sum with an integral $\sum\rightarrow \int$ How can one turn a sum to an integral. Example
$$\sum_k f(k) \approx N\cdot\int_k dk\, f(k). $$
How do you find the factor $N$?
The quantities should be approximately equal.
Example form Peskin and Schroeder page $374$:

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

Why does this $VT$ show up in equation $(11.71)$?
 A: This is a physics question (or at least the context is) so be warned: non-rigorous explanations to follow.
For 1D. Take a (real space) box with volume $V = L$ and discretize it on a lattice using $n$ points. This gives rise to following Fourier space lattice: $k = \frac{i}{n}k_{\rm max}$ for $i=-n,\ldots,n$ where $k_{\rm max} = \frac{2\pi n}{L}$. This is the setup of the problem.
To find $N$ s.t.
$$\sum_k \approx N \int dk$$
we first note that
$$\sum_k = 2n$$
is the number of $k$ modes that we can fit onto our lattice. Further we have (note that the integral here is to be interpreted as the integral over the modes we have availiable so we only integrate over $[-k_{\rm max},k_{\rm max}]$)
$$\int dk = 2k_{\rm max}$$
so
$$N = \frac{n}{k_{\rm max}} = \frac{V}{2\pi}$$
This was for 1D, but the procedure above can be done for the general case giving 
$$N = \frac{V_{4D}}{(2\pi)^4} = \frac{V T}{(2\pi)^4}$$
A: Define a measure $\mu$ on $\mathbb R$ by $\mu(A)=|A\cap\mathbb Z|$. 
Then $\sum_{k\in\mathbb Z}f(k)=\int f\operatorname{d}\mu$. 
This way a sum is turned into an integral.
A: If $f$ is nonincreasing, $$\int_0^{n}f(x)\,\mathrm dx+f(n)-f(0)\leqslant\sum_{k=1}^nf(k)\leqslant\int_0^nf(x)\,\mathrm dx.$$If $f$ is nondecreasing, $$\int_0^{n}f(x)\,\mathrm dx\leqslant\sum_{k=1}^nf(k)\leqslant\int_0^nf(x)\,\mathrm dx+f(n)-f(0).$$ No prefactor $N$ in sight...
