How can integration by parts be used to prove the Dirac delta function?

[zandria]

1. The problem statement
Show that:
\int_{-\infty}^{\infty} f(x) \delta^{(n)}(x-a) dx = (-1)^n f^{(n)}(a)

The Attempt at a Solution

I am trying to understand how to prove:
\int_{-\infty}^{\infty} f(x) \delta '(x) dx =- f'(x)
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.

 

[StatusX]

u is f(x), dv is \delta'(x) dx.

 

[Jimmy Snyder]

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'.

 

[zandria]

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?

\boxed{[f(x)\delta(x)]_{-\infty}^{\infty}} - \int_{-\infty}^{\infty} f'(x)\delta(x) dx

 

[quZz]

I am trying to understand how to prove the simple statement:
\int_{-\infty}^{\infty} f(x) \delta '(x) dx =- f'(x)

I thought it was a definition of delta-function derivative... could someone tell me if I'm wrong?

 

[Jimmy Snyder]

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?

 

[zandria]

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.

 

[Jimmy Snyder]

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.