# Delta Function

1. Jul 18, 2006

This isn't really homework, I'm just curious. So I'm dealing with the delta dirac function, and I was just wondering what would happen if we had two functions.

So the sampling property,

$$\int_{-\infty}^{\infty} f(t)\delta(t-a)\,\,dt = f(a)$$

$$\int_{-\infty}^{\infty} f(t)\delta(t-a)\delta(t-b)\,\,dt =?$$

What would happen?
My guess would be the following:

If $a = b$:
$$\int_{-\infty}^{\infty} f(t)\delta(t-a)\delta(t-b)\,\,dt =\int_{-\infty}^{\infty} f(t)\delta(a)\,\,dt =\int_{-\infty}^{\infty} f(t)\delta(b)\,\,dt$$

If $a \neq b$:
$$\int_{-\infty}^{\infty} f(t)\delta(t-a)\delta(t-b)\,\,dt = 0$$

Or is my answer or question just nonsense?

-Thanks

2. Jul 18, 2006

### benorin

$$\int_{-\infty}^{\infty} f(t)\delta(t-a)\delta(t-b)\,\,dt = \int_{-\infty}^{\infty} F(t)\delta(t-a)\,\,dt = F(a) = 0$$

where I have put $$F(t) = \delta(t-b)f(t)$$ so that $$F(a) = \delta(a-b)f(a)=0$$ if $$a\neq b$$

3. Jul 18, 2006

Thanks benorin!

That was (for lack of a better word) nifty how you just used the property that f(t)\delta(t-a)=f(a) by defining a function that encapsulated what was necessary to use the property Really cool.

Thanks again.

4. Jul 19, 2006

### HallsofIvy

Staff Emeritus
If you want to combine delta functions, of somewhat more importance is the fact that
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x,y)\delta(x-a)\delta(y-b)dxdy= f(a,b)$$

5. Jul 19, 2006

$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x,y)\delta(x-a)\delta(y-b)dxdy= \int_{\infty}^{\infty}dy \, \delta(y-b)\, [ f(a,y) ] = f(a,b)$$
Where, $\delta(y-b)$ is held constant while integrating through $dx$.