# Dirac delta function proof

1. Apr 6, 2009

### zandria

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

3. 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.

Last edited: Apr 6, 2009
2. Apr 6, 2009

### StatusX

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

3. Apr 6, 2009

### 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'.

4. Apr 6, 2009

### 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?

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

5. Apr 6, 2009

### quZz

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

6. Apr 6, 2009

### Jimmy Snyder

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. Apr 6, 2009

### 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.

8. Apr 6, 2009

### Jimmy Snyder

Yes, that is correct. delta of x thuds to zero immediately x leaves 0, long before x gets to infinity.