2

All the Dirac delta integrals I have seen either contain the Dirac delta function strictly within the limits of the lower and upper limits of integration or do not contain the delta function. What happens when one of the limits is at the location of the Dirac delta.

Also, how does one solve $\int_{-\infty}^a f(x)\, \delta(x - a) \,\mathrm d x$? Does the sifting property work now?

LL 3.14
  • 12,457
Debbie
  • 840
  • See THIS ANSWER which addresses the non-existence of $\langle \delta,H\rangle =\int_{-\infty}^\infty H(x) \delta(x),dx=\int_{-\infty}^0 \delta(x),dx$ regardless of whether $0$ is included in the set or not included. – Mark Viola Oct 11 '22 at 17:53

1 Answers1

6

So the short answer is that the notation $$ \int_{-\infty}^a \delta(x-a)\,f(x)\,\mathrm d x $$ does not make sense. Do you have a definition for it?

Here is a longer answer. So first the Dirac delta $\delta_a(x) = \delta(x-a)$ is not a function, it is either defined as a distribution of order $0$ or as a bounded measure. As a measure, it is defined as acting on a set $A$ by $$ \delta_a(A) = \left\{ \begin{array}{} 1 &\text{ if } a∈ A \\ 0 &\text{ if } a\notin A. \end{array}\right. $$ Similarly as for the Lebesgue measure $\mathrm d x$, one can then define the integration with respect to this new measure $\delta_a(\mathrm d x)$ by $$ ∫_{\mathbb{R}} \varphi(x)\, \delta_a(\mathrm{d}x) = \varphi(a) $$ for any nice function $\varphi$. In the same way if $f∈L^1$ is a classical function, it can be identified with the measure $\mu_f(\mathrm{d}x) = f(x)\,\mathrm{d}x$ verifying $$ ∫_a^b \varphi(x)\,\mu_f(\mathrm{d}x) = ∫_a^b \varphi(x)\,f(x)\,\mathrm{d}x = ∫_{[a,b]} \varphi(x)\,f(x)\,\mathrm{d}x = ∫_{(a,b)} \varphi(x)\,f(x)\,\mathrm{d}x $$ and this is the reason why the Dirac delta is often identified to a function. Notice however that the last identity is a special feature of locally integrable functions (coming from the fact that the integral is the same if we remove a set of measure $0$) which allows to use the notation $\int_a^b$. This is not the case of the dirac measure. Therefore we have to specify if we are looking at the integral of the dirac over $[a,b]$ or $(a,b)$. And then we have $$ \begin{align*} \int_{[a,b]} \varphi(x)\,\delta_a(\mathrm{d}x) &= \varphi(a) \\ \int_{(a,b)} \varphi(x)\,\delta_a(\mathrm{d}x) &= 0. \end{align*} $$

So the full answer is $$ \begin{align*} \int_{-\infty}^a \varphi(x)\,\delta_a(\mathrm{d}x) &\text{ is not a clear notation} \\ \int_{(-\infty,a]} \varphi(x)\,\delta_a(\mathrm{d}x) &= \varphi(a) \\ \int_{(-\infty,a)} \varphi(x)\,\delta_a(\mathrm{d}x) &= 0. \end{align*} $$

LL 3.14
  • 12,457
  • Personally I would disagree with the the last two assertions. The problem is that they don't respect approximate identities; indeed no fixed definition will respect every approximate identity. In some physical contexts you have a reason to think of approximate identities as symmetric, and in these settings you should really consider $\int_{-\infty}^a \delta_a(x) f(x) dx$ to be $f(a)/2$. – Ian Apr 15 '22 at 00:20
  • Justified comment. But I think you are just telling that approximations of the identity might not converge to the Dirac delta in the good topology here ... indeed, the Dirac is a continuous functional on the set of continuous functions, so if you take approximations of the identity $\varphi_n$ that converge to $\delta_a$ in $(C^0)'$, then you might have $$ ⟨φn,\mathbf{1}{(−∞,a)}f⟩→f(a)/2 $$ because your test function is discontinuous. – LL 3.14 Apr 15 '22 at 00:49
  • Told differently, in mathematics a fixed definition is preferred when we choose to define an object. The measure definition of the Dirac still gives a unique definition in this case (giving $\delta_a([-\infty,a]) = a$) while the distributional definition is not applicable (the test function is not continuous in $a$). I might be wrong but approximations of the identity might converge to different objects in $(L^\infty)'$, right? – LL 3.14 Apr 15 '22 at 00:49
  • (+1) for the well-written solution (as always)! I want to also share THIS ANSWER which addresses a similar question often asked about the existence of $\langle \delta,H\rangle$. – Mark Viola Oct 11 '22 at 17:51