Dirac delta function proof

  • Thread starter zandria
  • Start date
  • #1
15
0
1. The problem statement
Show that:
[tex]\int_{-\infty}^{\infty} f(x) \delta^{(n)}(x-a) dx = (-1)^n f^{(n)}(a)[/tex]

The Attempt at a Solution


I am trying to understand how to prove:
[tex]\int_{-\infty}^{\infty} f(x) \delta '(x) dx =- f'(x)[/tex]
I know that we need to use integration by parts, but I'm looking for a more detailed explanation of how to use integration by parts (what is u and what is dv?). I think if I understand this, then I will be able to apply this to the problem above.
 
Last edited:

Answers and Replies

  • #2
StatusX
Homework Helper
2,571
1
u is f(x), dv is [itex]\delta'(x) dx[/itex].
 
  • #3
1,031
16
I think you left out a minus sign. There's only one thing there than can be 'u' and only one thing that can be 'dv'.
 
  • #4
15
0
thank you, i did leave out a minus sign. I will correct that above. Also, I think I was trying to make things more complicated than they actually are. But why does the first term when doing integration by parts go to zero?

[tex]\boxed{[f(x)\delta(x)]_{-\infty}^{\infty}} - \int_{-\infty}^{\infty} f'(x)\delta(x) dx[/tex]
 
  • #5
125
1
I am trying to understand how to prove the simple statment:
[tex]\int_{-\infty}^{\infty} f(x) \delta '(x) dx =- f'(x)[/tex]
I thought it was a definition of delta-function derivative... could someone tell me if i'm wrong?
 
  • #6
1,031
16
But why does the first term when doing integration by parts go to zero?
All that is necessary for the first term to go to zero is that delta goes to zero faster than f goes to infinity. How fast does delta go to zero?
 
  • #7
15
0
delta tends to zero very quickly when x does not equal a (which is zero in this case) by definition of the generalized function . If that's right, then I think I get it. Thank you for your help.
 
  • #8
1,031
16
delta tends to zero very quickly when x does not equal a (which is zero in this case) by definition of the generalized function . If that's right, then I think I get it. Thank you for your help.
Yes, that is correct. delta of x thuds to zero immediately x leaves 0, long before x gets to infinity.
 

Related Threads on Dirac delta function proof

  • Last Post
Replies
10
Views
8K
Replies
5
Views
758
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
996
  • Last Post
Replies
7
Views
4K
  • Last Post
Replies
7
Views
620
Replies
3
Views
3K
  • Last Post
Replies
4
Views
2K
Replies
1
Views
2K
Top