Why do integrals featuring Dirac deltas equate to certain values?

[FatPhysicsBoy]

Homework Statement

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.

Homework Equations

N/A

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).

 

[Orodruin]

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

 

[FatPhysicsBoy]

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

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 $\delta(0)$?

 

[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.

 

[FatPhysicsBoy]

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.

$\int_\infty^\infty \delta(u-u_0)f(u-a)du = f(u_0 - a)$ 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

 

[Orodruin]

I think I'm seriously missing something :S

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!

 

[BruceW]

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

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...

 

[FatPhysicsBoy]

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!

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.

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...

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 :)

 

[Orodruin]

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.

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.

 

[FatPhysicsBoy]

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.

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!

 

[BruceW]

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

 

[FatPhysicsBoy]

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

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! :)