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

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 &#x2208; R and f is a smooth compactly supported function then &#x03B4; a ( f ) = f ( a ).The derivative of a distribution T is defined to make the integration by parts formula work. That is, T &#x2032; ( f ) = &#x2212; T ( f &#x2032; ) for any smooth compactly supported function. In the particular case of the delta function you get

Question: "MathJax(?): Can't find handler for document MathJax(?): Can't..."

gvaldytist

Answered question

2022-06-24

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

\int_{x_{0}}^{x_{1}} f (x) \frac{d}{d x} (δ (x - y)) d x

by:

= [f (x) δ (x - y)]_{x_{0}}^{x_{1}} - \int_{x_{0}}^{x_{1}} δ (x - y) \frac{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}

x_{1}

was equal to y?

Is the answer simply 'we must assume separately that

x_{0}, x_{1} \neq y

, or is there something obvious that I'm missing, or is there some measure theory reason why we can say it is zero?

Answer & Explanation

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 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) = \int f (x) δ_{a} (x) d x = \int f (x) δ (x - a) d x

and similarly

δ_{a}^{'} (f) = \int f (x) δ_{a}^{'} (x) d x = \int f (x) δ^{'} (x - a) d x .

Thus by definition

\int f (x) δ^{'} (x - a) d x = δ_{a}^{'} (f) = - δ_{a} (f^{'}) = - \int f^{'} (x) δ (x - a) d x .

Sattelhofsk

Beginner2022-06-26Added 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.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Pre-Algebra

Didn't find what you were looking for?