I am having troubles in understanding how to correctly perform the continuum limit of a double sum containing a Kronecker delta.

Imagine to integrate a function depending on $t$ and $t'$, both ranging from $0$ (initial time) to $T$ (final time):

$$I:=\int_0^Tdt_1\int_0^Tdt_2 f(t_1,t_2).\tag{1}$$

The corresponding Riemann sums, dividing the time intervals in slices of width $\epsilon=T/N$ is:

$$I_{disc}:=\sum_{j,j'=1}^N \epsilon^2 f(j\epsilon,j'\epsilon).\tag{2}$$

Obviosly lim$_{N\rightarrow\infty}I_{disc}=I$. Now consider the case when only the diagonal elements of the double integral are different from zero i.e.

$$J_{disc}:=\sum_{j,j'=1}^N \delta_{j,j'}\epsilon^2 f(j\epsilon,j'\epsilon).\tag{3}$$

I would expect that, in the continuum limit $N\rightarrow \infty$ ($\epsilon\rightarrow 0$) it becomes

$$J:=\int_0^Tdt\int_0^Tdt'\delta(t-t') f(t,t'),\tag{4}$$

where $\delta(t-t')$ is a Dirac delta.

Here is the problem: the Kronecker delta is adimensional, while the Dirac Delta has the dimensions of seconds$^{-1}$. This implies that $J_{disc}$ and $J$ have different dimensions, which does not make any sense. Therefore, there must be some mistake I am doing in going from the discrete to the continuum version of $J$. Could you help me spotting it and, more important, suggest the way to do this limit correctly?

  • $\begingroup$ Maybe $\delta\left(\frac{t-t'}{T}\right)$? $\endgroup$
    – Cryo
    May 14, 2019 at 0:06

1 Answer 1


The translation between the Kronecker delta function and the Dirac delta distribution is

$$ \frac{1}{\epsilon}\delta_{j,j^{\prime}}\qquad\longrightarrow\qquad\delta(t-t^{\prime}),\tag{A}$$

where $\epsilon$ is the "volume" of a unit-cell in the discretization. See e.g. this related Phys.SE post.

In particular, the rhs. of OP's eq. (3) should be divided with $\epsilon$ to have a finite continuum limit.

  • $\begingroup$ Thank you very much, I think this clarifies! To summarize, if I just take (3) as it is and perform the limit I get zero, which is also consistent of doing an integral of a finite quantity (the function f) on a set with null measure (the line s=s'). $\endgroup$
    – Sandro
    May 15, 2019 at 14:01
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic
    May 15, 2019 at 16:49

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