# Dirac delta function homework help

1. Oct 5, 2006

### shoehorn

Suppose that we take the delta function $$\delta(x)$$ and a function f(x). We know that

$$\int_{-\infty}^{\infty} f(x)\delta(x-a)\,dx = f(a).$$

However, does the following have any meaning?

$$\int_{-\infty}^{\infty} f(x)\delta(x-a)\delta(x-b)dx,$$

for some constants $$-\infty<a,b<\infty$$.

2. Oct 5, 2006

### HallsofIvy

Staff Emeritus
Remember that $\delta$ is not a true function. It is a "distribution" or "generalized function". The first integral is defined for distributions but the second isn't. (If we were to force a meaning on it, the only reasonable value it could have would be 0.)

You could do that in more than one dimension:
$$\int_{x=-\infty}^{\infty}\int_{y=-\infty}^{\infty}f(x,y)\delta(x-a)\delta(y-b)dydx= f(a,b)[/itex] 3. Oct 5, 2006 ### shoehorn Here's what I was thinking about it. The integral [tex]\int_{-\infty}^\infty \, f(x)\delta(x-a) dx$$

integrates a function f(x) times a distribution $$\delta$$, so the whole thing is an integral of a generalized function, dependent on x. Suppose now that I define a generalized function $$g(x)$$ according to

$$g(x) \equiv f(x)\delta(x-a).$$

Then the second integral becomes

$$\int_{-\infty}^\infty f(x)\delta(x-a)\delta(x-b)\,dx = \int_{-\infty}^\infty g(x) \delta(x-b).$$

My instinct is to then conclude that

$$\int_{-\infty}^\infty g(x)\delta(x-b)\, dx = g(b) = f(b)\delta(b-a)$$

Thus, the second integral is itself something which has meaning only if it appears within an integral. Yet you don't think that this is correct?

Last edited: Oct 5, 2006
4. Oct 5, 2006

### daniel_i_l

the function g(x) is basiclly a constant (f(a)) times the dirac delta function. so in your second integral you want the integral of the product of two different DD functions times a constant, that would equal 0

5. Oct 5, 2006

### shoehorn

I'm afraid I don't see how you conclude that g(x) is a constant. Going with the definition I gave above,

$$g(x)\equiv f(x)\delta(x-a)$$

for some function f(x). The only relationship between g(x) and f(a) is

$$\int_{-\infty}^\infty g(x)\,dx = \int_{-\infty}^\infty f(x)\delta(x-a)\,dx = f(a),$$

i.e., the integral of g(x) is equal to f(a).

6. Oct 5, 2006

### daniel_i_l

I didn't say that it was a constant, i said that is was equal to a constant times the DD function. what i meant to say that was in this case you could look at g(x) as being g(x) = f(a)*DD(x-a) when you're doing the integral. so you're right, it doesn't really equal f(a)*DD(x-a) but when you integrate you get the same resault when you interchange the two.
and anyway f(b)DD(b) = 0.

Last edited: Oct 5, 2006
7. Oct 5, 2006

- Shoehorn according to "Schwartz theorem2 you can't multiply 2 distribution..this is really "nasty" since avoiding this (if we could) we would have solved divergences in the form $$\int_{0}^{\infty}dx x^{n}$$ using "Casual perturbation theory2 (see wikipedia) although i believe that for big n:

$$\delta (x-a) \delta (x-b) =n^{2}\pi e^{-n^{2}((x-a)^2 +(x-b)^2)}$$ for n tending to $$\infty$$

8. Oct 5, 2006

### matt grime

How do you conclude something that is not even a function is actually a lebesgue square measurable function?

9. Oct 5, 2006

### shoehorn

Oops, sorry. I was editing a post earlier and pasted that in by accident. Please ignore it - it's obviously not true. :-)

10. Oct 5, 2006

### Hurkyl

Staff Emeritus
No.

However, if a (or b) was a variable, then this expression could be considered a distribution. Convolving it with h(a) gives:

$$\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} f(x) h(a) \delta(x-a) \delta(x-b) \, da \, dx = f(b) h(b)$$