Dirac Delta Function: Integral at x=a

[pivoxa15]

Homework Statement

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?

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).

 

[Gib Z]

Think about it this way.

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

 

[HallsofIvy]

Homework Statement

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?

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

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).

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.

 

[pivoxa15]

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

 

[Gib Z]

Yup, choose your a.

 

[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,..!

 

[HallsofIvy]

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).