Consider the following expression $$ I(x) =\int_{x}^{0}f(y)\delta'(y-x)dy, \tag{1} $$ where $f(x)$ is some (say smooth) function. From partial integration the term $f(x)\delta(0)$ appears, which is either infinite or just ill-defined. However, is partial integration even the correct strategy for computing $I(x)$? If $I(x)$ is divergent, could one in some way regularize it?
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- $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Mathematics Meta, or in Mathematics Chat. Comments continuing discussion may be removed. $\endgroup$Shaun– Shaun ♦2025-09-11 09:29:15 +00:00Commented Sep 11 at 9:29
- 1$\begingroup$ The Dirac Delta is not a function, and the object over which it operates is not an integral. If the meaning of $I(x) =\int_x^0f(y)\delta'(y-x)\,dy$ is interpreted to mean $\langle \delta'_x(H_0-H_x),f\rangle$, this fails to exist. The product of distributions is, in general, not defined. If one were to set the "lower integration limit to $x-\varepsilon$, for $\epsilon>0$, then one could evaluate the functional as $-f'(x)$ since the singulari support of the derivative of the Dirac Delta $\delta'(y-x)$ and the Heaviside distribution $H(y-x+\varepsilon)$ do not overlap. $\endgroup$Mark Viola– Mark Viola2025-09-12 14:03:00 +00:00Commented Sep 12 at 14:03
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