gvaldytist

2022-06-24

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MathJax(?): Can't find handler for document
Often in physics we integrate by parts

$$

by:

$$

I have a really simple question, how can we assume that $[f(x)\delta (x-y){]}_{{x}_{0}}^{{x}_{1}}=0$?

Intuitively the delta function is zero except for at $x=y$, but what if either ${x}_{0}$ or ${x}_{1}$ was equal to y?

Is the answer simply 'we must assume separately that ${x}_{0},{x}_{1}\ne y$, or is there something obvious that I'm missing, or is there some measure theory reason why we can say it is zero?

$$

by:

$$

I have a really simple question, how can we assume that $[f(x)\delta (x-y){]}_{{x}_{0}}^{{x}_{1}}=0$?

Intuitively the delta function is zero except for at $x=y$, but what if either ${x}_{0}$ or ${x}_{1}$ was equal to y?

Is the answer simply 'we must assume separately that ${x}_{0},{x}_{1}\ne y$, or is there something obvious that I'm missing, or is there some measure theory reason why we can say it is zero?

Sawyer Day

Beginner2022-06-25Added 30 answers

MathJax(?): Can't find handler for document
MathJax(?): Can't find handler for document
I think the issue is mostly one of notation. The delta function is actually a distribution. If $a\in \mathbf{R}$ and $f$ is a smooth compactly supported function then ${\delta}_{a}(f)=f(a)$.

The derivative of a distribution $T$ is defined to make the integration by parts formula work. That is, ${T}^{\prime}(f)=-T({f}^{\prime})$ for any smooth compactly supported function. In the particular case of the delta function you get

$$

People have found it notationally convenient to denote ${\delta}_{a}(f)$ using the notation

$$

and similarly

$$

Thus by definition

$$

The derivative of a distribution $T$ is defined to make the integration by parts formula work. That is, ${T}^{\prime}(f)=-T({f}^{\prime})$ for any smooth compactly supported function. In the particular case of the delta function you get

$$

People have found it notationally convenient to denote ${\delta}_{a}(f)$ using the notation

$$

and similarly

$$

Thus by definition

$$

Sattelhofsk

Beginner2022-06-26Added 5 answers

Oh, maybe The integral is over the domain in question. Since ${\delta}^{\prime}$ is a distribution it acts on functions compactly supported in that domain.

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