# Dirac Delta fn

1. Apr 14, 2007

### pivoxa15

1. The problem statement, all variables and given/known data
int[d(x-a)f(x)dx]=f(a) is the dirac delta fn

but is int[d(a-x)f(x)]=f(a) as well? If so why?

3. The attempt at a solution
Is it because at x=a, d(0)=infinite and integrate dirac delta over a region including x=0 when d(0) is in the value in the integral will produce 1 hence f(a).

2. Apr 14, 2007

### Gib Z

Since
$$\int f(x) d(a-x) = -\int f(x) d(x-a) = -f(a)$$ your asking is f(a)=-f(a).

3. Apr 14, 2007

### HallsofIvy

Staff Emeritus
No, $\int_{-\infty}^{\infty} \delta (a- x)f(x)dx= -f(a)$. Use the substitution u= a-x.

That's a very rough way of putting it. More correctly the dirac delta function (actually a "generalized function" or "distribution") is DEFINED by the property that $\int \delta(x) f(x)dx = f(0)$ as long as the integration include x= 0.

Last edited: Sep 20, 2009
4. Apr 14, 2007

### pivoxa15

So it can be but in some cases it is not.

5. Apr 14, 2007

### Gib Z

6. Sep 20, 2009

### Shafikae

is it possible to explain step by step including the integration of the following:

(1) f(x) d(x-a) = f(a) d(x-a)

(2) \int f(a) d(a-x) = f(x)

Thank you,..!!!

7. Sep 20, 2009

### HallsofIvy

Staff Emeritus
Are you changing symbols here? does "d(x-a)" still mean the delta function? Are you asking why
(1) $$\int f(x)\delta(x- a)dx= \int f(a)\delta(x-a)dx$$?

That is because the definition of $\delta(x-a)$ require that the
$$\int f(x)\delta(x-a)dx= f(a)$$
for any function f as long as the integral of integration includes x= a.
Of course,
$$\int f(a)\delta(x-a) dx= f(a)\int \delta(x-a)dx= f(a)(1)= f(a)$$

As for
(2)$$\int f(a)d(a- x) dx= f(x)$$
That's not true. In fact,the right hand side cannot be a function of x at all. The integral is just as in (1).