# Understanding the Dirac delta

1. Apr 12, 2015

### FatPhysicsBoy

1. The problem statement, all variables and given/known data
Having trouble understanding dirac deltas, I understand what they look like and how you can express one (i.e. from the limiting case of a gaussian) but for the life of me I can't figure out why the results of some integrals featuring dirac deltas equate to what they do.

2. Relevant equations

N/A

3. The attempt at a solution
An example of one particularly baffling to me is the following (given $u=x+a$, and $du = dx$):

1) $\int^{\infty}_{-\infty}\delta(x-u)f(u-a)du = f(x-a)$,

especially when trying to compare to the 'standard' definition of how the dirac delta behaves under an integral:

2) $\int^{\infty}_{-\infty}\delta(x-x_{0})f(x)dx = f(x_{0})$,

in my mind it looks like 1) should be equal to $f(u) = f(x+a)$ because I look at it as $\int^{\infty}_{-\infty}\delta(x-u)f(x)dx$ and compare to 2).

2. Apr 12, 2015

### Orodruin

Staff Emeritus
Try letting $g(u) = f(u-a)$ and then apply (2) with $x = u$.

3. Apr 12, 2015

### FatPhysicsBoy

I don't follow? Why do I do this and also I don't understand, so I have $\int^{\infty}_{\infty}\delta(x-u)g(u)du$ but $x=u$ means I have $\delta(0)$?

4. Apr 12, 2015

### BruceW

I think Orodruin just meant equation (2), but replace x with u, i.e.
$$\int_\infty^\infty \delta (u-u_0) g(u) du = g(u_0)$$
now just use $g(u)=f(u-a)$ and continue from there.

5. Apr 13, 2015

### FatPhysicsBoy

$\int_\infty^\infty \delta(u-u_0)f(u-a)du = f(u_0 - a)$ I still don't understand? So $\int_\infty^\infty \delta(x-u)f(x-a)dx = f(-u-a) = f(-x-2a)$

I think I'm seriously missing something :S

6. Apr 13, 2015

### Orodruin

Staff Emeritus
Yes, you are doing the wrong replacements in the functions. When you have a $\delta(x-u)$ and integrate with respect to $x$, you need to replace all occurrences of $x$ in the other functions by $u$, not by $-u$ as you did here.

Also, when you integrate with respect to $x$, the result cannot possibly depend on $x$. Keep track of what is your integration variables and what are free variables!

7. Apr 13, 2015

### BruceW

This is the answer, but with $u_0$ replaced with $x$.

Uh wait... I'm re-reading your first post. You say given $u=x+a$, you want to find
$$\int^{\infty}_{-\infty}\delta(x-u)f(u-a)du$$
But $\delta(x-u)= \delta(-a)$ So this will always be zero (assuming $a$ is some nonzero constant). So equation 1) will just be equal to zero if you want to impose the constraint $u=x+a$. So I'm guessing you don't want to impose this constraint really, or the answer is not very interesting...

8. Apr 13, 2015

### FatPhysicsBoy

Is this because, $\delta(x-u)$ is a delta function centered on u? Is it sort of like an orthonormal function where if we turn the integral into a sum along $x$ then the only place where $\delta(x-u)dx$ is non-zero is at $x=u$? So if we're essentially summing over $\delta(x-u)g(u)$ then you have $0\times g(u)$ for all $u\neq x$? Then you get $1\times g(x)$? So the integral equals $g(x)$?

I understand it in this context - graphically, is this the only 'motivation' behind replacing blah with blahblah? I thought I was missing something analytic.

No I'm sorry I shouldn't have included $u=x+a$ that was used in the solutions to get to integral 1) and means nothing in the context of this question I suppose. Please see my above query to Orodruin regarding replacing things :)

9. Apr 13, 2015

### Orodruin

Staff Emeritus
Yes, this is essentially the way to understand it. In a more formal setting (distribution theory), the $\delta$ is simply defined such that
$$\int_{-\infty}^{\infty} \delta(x) \varphi(x) dx \equiv \varphi(0),$$
where $\varphi$ is a "nice" function (infinitely differentiable with compact support).
You can obtain the results for a delta offset from zero by a change of variables.

Edit: Just to mention, you do not get $1 \times g$, you informally get $\infty \times g$, but the infinity is only in one point and integrates to one. It therefore plays the same role as the Kronecker delta in countable sums.

10. Apr 13, 2015

### FatPhysicsBoy

Apologies that's what I meant, that the 'area' is then just 1 scaled by the 'area' of $g(u)$. Thanks for all your help guys!

11. Apr 13, 2015

### BruceW

good work! :) It's good to really think this stuff through.

12. Apr 13, 2015

### FatPhysicsBoy

Seriously, thanks for the help! I've got another question I've just posted over in advanced physics on the taylor expansion if you fancy it! :)