 # Often in physics we integrate by parts <math xmlns="http://www.w3.org/1998/Math/MathML" "> <ms gvaldytist 2022-06-24 Answered
MathJax(?): Can't find handler for document 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 $\left[f\left(x\right)\delta \left(x-y\right){\right]}_{{x}_{0}}^{{x}_{1}}=0$?

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?
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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 ${\delta }_{a}\left(f\right)=f\left(a\right)$.

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

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

and similarly

Thus by definition

We have step-by-step solutions for your answer! Sattelhofsk
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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