During the course of a textbook problem, I obtain the following (simplified to keep only important elements) :

$$\int^{b}_{-b}dy\int^{b}_{-b}dy' \space exp\{A(y^{2}-y'^{2})\} \space \delta(y-y')$$

where the answer is $2b$

My instinct would be to treat the problem as such :

$$\int^{b}_{-b}dy\int^{b}_{-b}dy' \space f(y, y') \space \delta(y-y')$$

where $\space f(y, y') = exp\{A(y^{2}-y'^{2})\}$

computing the first integral would yield

$$\int^{b}_{-b}dy \space f(y=y'|-b\leq y \leq b) = \int^{b}_{-b}dy = 2b$$

I understand the dirac distribution if very often misused and that bad shortcuts are often taken, I apologise for writing what is probably horrible mathematics.

My question however is whether the first double integral is well-defined, and whether the Dirac delta function can ever make any sense under an integral whose limits aren't infinity, for example in the case :

$$\int^{a}_{b}dy\int^{c}_{d}dy' \space f(y, y') \space \delta(y-y')$$

  • 2
    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$ – Qmechanic Sep 15 '20 at 13:17
  • $\begingroup$ Possibly, I wasn't too sure myself but the exercise in question came up in a QM textbook and I felt this type of question is often dealt with by physicists without going into too intricate mathematical detail. Very happy to move if this is the wrong place though $\endgroup$ – Mr Lolo Sep 15 '20 at 13:26

The double integral makes sense but its value depends on the intersection (if any) of the intervals $[b,a]$ and $[d,c]$. If they intersect on the interval $[f,e]$ then

$\int^{a}_{b}dy\int^{c}_{d}dy' \space f(y, y') \space \delta(y-y') = \int^{e}_{f} f(y,y) \space dy$

If they do not intersect then there is no point within the domain of the double integral at which $y=y'$ so

$\int^{a}_{b}dy\int^{c}_{d}dy' \space f(y, y') \space \delta(y-y') = 0$

In your example the two intervals are the same so

$\int^{b}_{-b}dy\int^{b}_{-b}dy' \space exp\{A(y^{2}-y'^{2})\} \space \delta(y-y') = \int^{b}_{-b} \space exp\{A(y^2-y^2)\} \space dy = \int^{b}_{-b} dy = 2b$


The Dirac distribution $\delta_x$ (the notation $\delta(x-y)$ is obtained by representing it as an integral kernel) is defined as the linear functional that maps a smooth function $f$ to the value $f(x)$. This definition does not really depend on the bounds of the domain of definition. Especially since in mathematical terms the Dirac distribution has as support the origin (or a translate thereof, i.e. any point $x$). So the behaviour of functions away of this point of support does not matter.

PS: I did not go to deep into the mathematical details, since only the application seemed to be relevant.

  • $\begingroup$ I see. So is it correct to say in my double integral example that what I really have is an integral over y whereby the Dirac distribution, here represented as the integral over y', maps the integrand (the function f(y, y')) to f(y, y). The process works since each time the support (in each case y) lies in [-b, b], the bounds of the integral representation of the Dirac distribution. $\endgroup$ – Mr Lolo Sep 15 '20 at 14:52
  • $\begingroup$ Yes, that is exactly it. $\endgroup$ – NDewolf Sep 15 '20 at 15:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.