# Delta property, integration by parts, heaviside simple property proof

1. Dec 23, 2016

### binbagsss

1. The problem statement, all variables and given/known data

I am trying to show that $\int \delta (x-a) \delta (x-c) dx = \delta (-a-c)$ via integeration by parts, but instead I am getting $\delta (c-a)$ (or $\delta (a-c)$ depending how I go...).

2. Relevant equations

above.

3. The attempt at a solution

I am trying to show that $\int \delta (x-a) \delta (x-c) dx = \delta (-a-c)$ via integeration by parts, but instead I am getting $\delta (c-a)$ (or $\delta (a-c)$ depending how I go...).

By parts $dv/dx= \delta (x-c)$
$v= H(x-c)$, $H(x)$ the Heaviside function
$u=\delta(x-a)$
$du/dx=d/dx ( \delta (x-a)$

Then

$\delta (x-a) \delta (x-c) dx = ^\infty _{-\infty} [ \delta (x-a) H (x-c) ] - \int H(x-c) d/dx (\delta (x-a)) dx$
First term vanishes,
=$- ^{\infty}_{c} [\delta (x-a) ] = \delta (c-a)$, if $c \geq a$
or $=0$ if $c<a$

The limit change from $H(x-c) \neq 0$ for $x-c > 0$.

I can't spot what's gone wrong.

thanks alot.

2. Dec 23, 2016

### PeroK

Are you sure about this? The answer you are getting of $\delta (a-c) = \delta(c-a)$ looks right.

3. Dec 23, 2016

### PeroK

As as aside, here's something I use to take some of the mystery out of the Dirac delta function. You replace the delta function with a finite spiked function. Any function will do, but two obvious candidates are:

$S_1(x) = \frac{1}{2\alpha} \ (-\alpha < x < \alpha) \ ; \ S_1(x) = 0,$ otherwise. For some small, postive $\alpha$

$S_2(x) = \frac{1}{\sqrt{2\pi}\sigma}exp(-\frac{x^2}{2\sigma^2})$. For some small, positive $\sigma$

Loosely, the Dirac delta function is the limit of either of these functions as $\alpha$ or $\sigma$ tends to $0$.

Now, in this example, you could replace the second Delta function with one of these finite spiked functions:

$\int \delta(x-a)S(x-c)dx = S(a-c) \approx \delta(a-c)$

You know this from the standard, defining property of the Delta function. $S(a-c)$ is just your spiked function $S(x)$ evaluated at $a-c$. In the finite case, this is effectively $0$ when $a$ is not close to $c$ and large when $a$ is close to $c$. In any case, it's a approximately $\delta(a-c)$.

This loosely shows that the standard property of the delta function can be applied, even when the second function is also a delta function - which wasn't immediately obvious. And that confirms that you are trying to prove an incorrect identity!

4. Dec 23, 2016

### Ray Vickson

You should have $\delta(c-a)$, not $\delta(-a-c)$.

Besides the method of Perok in post #3, you can instead look at the effect as a generalized function, applied to a test function. That is, let $F(c) = \int \delta(x-a) \delta(x-c) \, dx$, with $a$ considered as an input constant. We have
$$\int F(c) f(c) \, dc = \int_c \int_x f(c) \delta(x-a) \delta(x-c) \, dx \, dc\\ = \int_x \delta(x-a) \left(\int f(c) \delta(x-c) \, dc\right) \, dx = \int f(x) \delta(x-a) \, dx = f(a),$$
so considered as an operator on functions we have $F(c) = \delta(c-a)$.

5. Dec 24, 2016

### binbagsss

See attached - quantum theory. first line.
Setting $a=k_1+k_2$ and $c=-k_3-k_4$
$\delta ( -a -c )$ gives me the correct expression of $\delta(k_3+k_4-k_1-k_2) = \delta ( -k_3 - k_4 + k_1 + k_2)$
whereas
$\delta(a-c)$ gives me $\delta(k_1+k_2+k_3+k_4)$

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6. Dec 25, 2016

### Orodruin

Staff Emeritus
Well, that step is wrong. Without seeing the rest of the argumentation it is difficult to tell you what is going on.

7. Dec 26, 2016

### Ray Vickson

Fact:
$$\int \delta(x-a) \delta(x-c) \, dx = \delta(a-c) = \delta(c-a).$$
You have been shown at least two correct, independent proofs of that fact. If you are reading an article that says something else then that article is wrong!

8. Dec 27, 2016

### binbagsss

However can someone please help me see what is wrong with the step above where I have
=$- ^{\infty}_{c} [\delta (x-a) ] = \delta (c-a)$, if $c \geq a$
or $=0$ if $c<a$

I have $c \geq a$ since $a$ must be $\in$ the domain I am integrating over, however this doesn't seem to make sense because $\delta (c-a) = \delta (a-c)$, so if I do the above derivaiton the other way around I'd instead get the condition $=0$ if $a<c$ ... ?

Many thanks.

9. Dec 27, 2016

### Ray Vickson

Derivatives of $\delta$ do not behave the way you think they should. Formally, we may regard the derivative as
$$\delta'(x) = -\frac{1}{x} \delta(x).$$
This does not make much sense as a statement about a function; it really means that $x \delta'(x) = -\delta(x)$.

$\delta(x)$ is not really an "ordinary" function, so using ordinary methods on it may lead to disaster, as they have done in your case.

For more on this, see, eg., http://mathworld.wolfram.com/DeltaFunction.html
or https://en.wikipedia.org/wiki/Dirac_delta_function .

10. Jan 1, 2017

### binbagsss

so rather above where i wrote:

=$- ^{\infty}_{c} [\delta (x-a) ] = \delta (c-a)$, if $c \geq a$
or $=0$ if $c<a$

The limit change from $H(x-c) \neq 0$ for $x-c > 0$.

It should have been $= - ^{\infty}_{c} [\delta (x-a) ] = \delta (c-a)$ for all $c$,$a$, i.e. without any condition?
what's the proper way I should have wrote this?

thanks alot.