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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
x 0 x 1 f ( x ) d d x ( δ ( x y ) ) d x
by:
= [ f ( x ) δ ( x y ) ] x 0 x 1 x 0 x 1 δ ( x y ) d f d x d x
I have a really simple question, how can we assume that [ f ( x ) δ ( 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 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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Answers (2)

Sawyer Day
Answered 2022-06-25 Author has 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 R and f is a smooth compactly supported function then δ a ( f ) = f ( a ).

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

People have found it notationally convenient to denote δ a ( f ) using the notation
δ a ( f ) = f ( x ) δ a ( x ) d x = f ( x ) δ ( x a ) d x
and similarly
δ a ( f ) = f ( x ) δ a ( x ) d x = f ( x ) δ ( x a ) d x .

Thus by definition
f ( x ) δ ( x a ) d x = δ a ( f ) = δ a ( f ) = f ( x ) δ ( x a ) d x .

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Sattelhofsk
Answered 2022-06-26 Author has 5 answers
Oh, maybe The integral is over the domain in question. Since δ is a distribution it acts on functions compactly supported in that domain.

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