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 $(11.71)$?

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    $\begingroup$ What does the arrow mean? $\endgroup$ Oct 28, 2014 at 8:46
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    $\begingroup$ Do you want the quantities to be equal? Are you familiar with measure theory? $\endgroup$
    – DanZimm
    Oct 28, 2014 at 8:50
  • $\begingroup$ I updated the question slightly. I'm not very familiar with measure theory (not taken any formal course). $\endgroup$ Oct 28, 2014 at 8:57
  • $\begingroup$ Look at Riemann sums. $\frac{1}{n+1} \sum_{k=0}^n f(\frac{k}{n+1})$ converges to $\int_0^1 f(k) dk$ (in good cases, but since this is a physics question it doesn't really matter). $\endgroup$ Oct 28, 2014 at 9:39
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    $\begingroup$ To avoid more confusion you should add more details to the question (or maybe move it to Physics SE). Important things here: 1) $k$ here is the Fourier-space wave-number. 2) The method used by the authors is: discretize the space (and thereby Fourier space), solve equations etc. (this gives you sums), consider the large $n$ limit to get from sums to the integral you are interested in. 3) The integrals are implicitly assumed to be only over wavelengths availiable to us from the discretization. All these things are needed to properly answer the question in the context given here. $\endgroup$
    – Winther
    Oct 28, 2014 at 10:12

3 Answers 3


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 $[-k_{\rm max},k_{\rm max}]$)

$$\int dk = 2k_{\rm max}$$


$$N = \frac{n}{k_{\rm max}} = \frac{V}{2\pi}$$

This was for 1D, but the procedure above can be done for the general case giving

$$N = \frac{V_{4D}}{(2\pi)^4} = \frac{V T}{(2\pi)^4}$$

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    $\begingroup$ Is it correct if one uses the mnemonic: normalize the sum such that $$1:=\sum_{k_x}=N_x\underbrace{\int dk_x}_{\text{volume of ball of allowed }k-values=\frac{2\pi}{L_x}}\Rightarrow N_x = \frac{L_x}{2\pi}.$$ Generalizing to four dimension gives the result. $\endgroup$ Oct 28, 2014 at 10:54
  • $\begingroup$ @Winther, could you please also explain what $N_x$ should be for the case of the box being of size $L$, while momenta belonging to the interval ${(0,+\infty)}$? (This happens when quantising a relativistic theory on the light cone.) $\endgroup$
    – mavzolej
    Jun 8, 2019 at 5:11

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.

  • $\begingroup$ How do you define that measure and is the equal sign really an equal sign? $\endgroup$ Oct 28, 2014 at 8:56
  • $\begingroup$ I describe in my answer how you define the measure: $\mu(A)=|A\cap\mathbb Z|$ i.e. the number of integers $n$ that satisfy $n\in A$. Equal signs are equal signs just the way as apples are apples. $\endgroup$
    – drhab
    Oct 28, 2014 at 9:01
  • $\begingroup$ Thanks @drhab, good to know that. I have updated the question to be more specific. Could you have a look please? $\endgroup$ Oct 28, 2014 at 9:02
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    $\begingroup$ Rely on the answer of @Did. I am leaving now. Cheers ;) $\endgroup$
    – drhab
    Oct 28, 2014 at 9:05

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...


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