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

• binbagsss
In summary: But it's not clear what you're trying to do. You've set up an integral to evaluate, but it's not clear what the integral is supposed to equal. And you're doing that by parts, but it's not clear what you're trying to evaluate by parts.
binbagsss

## Homework Statement

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...).

above.

## 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.

binbagsss said:

## Homework Statement

I am trying to show that ## \int \delta (x-a) \delta (x-c) dx = \delta (-a-c) ##

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!

binbagsss said:

## Homework Statement

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...).

above.

## 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.

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)##.

PeroK
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)##

#### Attachments

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binbagsss said:
See attached - quantum theory. first line.
Well, that step is wrong. Without seeing the rest of the argumentation it is difficult to tell you what is going on.

binbagsss said:
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)##

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!

binbagsss said:

## Homework Statement

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...).

above.

## 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.

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.

binbagsss said:
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.

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 .

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.

## 1. What is the delta property in integration by parts?

The delta property in integration by parts is a mathematical property that states that the integral of a product of two functions can be rewritten as the product of one of the functions and the integral of the other function. In other words, it allows us to simplify the integration process by breaking down the original integral into smaller, more manageable integrals.

## 2. How is the delta property used in integration by parts?

The delta property is used in integration by parts to simplify the integration process. It allows us to break down a complex integral into smaller, more manageable integrals that can be easily solved using basic integration techniques. By using this property, we can save time and effort while solving integrals.

## 3. What is the Heaviside simple property in integration by parts?

The Heaviside simple property in integration by parts is a mathematical property that states that the integral of a Heaviside function can be rewritten as the product of the Heaviside function and the integral of the derivative of the Heaviside function. This property is useful in solving integrals involving Heaviside functions.

## 4. How is the Heaviside simple property used in integration by parts?

The Heaviside simple property is used in integration by parts to simplify the integration process involving Heaviside functions. By using this property, we can rewrite the integral in a simpler form, making it easier to solve using basic integration techniques. It is especially useful when dealing with integrals that involve step functions.

## 5. How can the delta property and Heaviside simple property be proved?

The delta property and Heaviside simple property can be proved using basic integration techniques and the properties of the delta function and Heaviside function. These proofs involve breaking down the original integral into smaller integrals and using substitution and integration by parts to simplify them. With careful manipulation and algebraic techniques, the properties can be proven to hold true.

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