Property of the dirac delta function

[jk89]

Hello team!

I saw the other day in a textbook that the Dirac delta function of the form d(x-a) can be written as d(a-x) but the method was not explained. I was wondering if anyone know where this comes from. I've been googling but can seem to find it out. Any help would be appreciated.

Cheers!
Jonathan

 

[micromass]

How did you define the Dirac delta function?? Did you do it by distributions?

 

[jk89]

The formal definition preceding the statement was:
f(a) = ∫f(x)d(x-a)dx the integral goes from inf to -inf
I was thinking maybe you make a change of variable.

 

[sweet springs]

Hi. So you would like to prove delta function is even function, I.e. delta x =delta -x.

 

[jk89]

Yes, i guess this is the crux of the problem.

 

[sweet springs]

Hi.

The formal definition preceding the statement was:
f(a) = ∫f(x)d(x-a)dx the integral goes from inf to -inf

∫f(x)Δ(a-x)dx, x[-∞,+ ∞]
=∫f(-t)Δ(a+t)dt, t[-∞,+ ∞]
=f(a)

Subtracting each other,
for any f(x) ∫f(x){Δ(x-a)-Δ(a-x)}dx=0
so Δ(x-a)=Δ(a-x).

Regards.

 

[knowlewj01]

I'm also interested in this proof.

if i start out with

f(a)=\int_{-\infty}^{\infty}f(x)\delta(a-x) dx [1]

and make the change of variable x\rightarrow -t

\Rightarrow dx\rightarrow -dt

then

f(a)=-\int_{-\infty}^{\infty}f(-t)\delta(a+t)dt [2]


i'm a bit confused how you then get one of them in the form δ(x-a)

 

[Avatrin]

You forgot another part of the definition; The dirac delta function is a function that is 0 everywhere except at zero and:
Δ(0) = infinity

With this in mind:
x-a = a-x when x = a

 

[Mute]

You forgot another part of the definition; The dirac delta function is a function that is 0 everywhere except at zero and:
Δ(0) = infinity

With this in mind:
x-a = a-x when x = a

More precisely, the dirac delta function has the property that

\int_b^c dx f(x)\delta(x-a) = \left\{\begin{array}{c} f(a),~a \in [b,c] \\ 0,~\mbox{otherwise} \end{array}\right.

The delta function doesn't really have a well-defined meaning outside of an integral, so as far as we're concerned, if integrating f(x) against \delta(x-a) or \delta(a-x) gives you the same result, then \delta(x-a) = \delta(a-x).

 

[sweet springs]

Ｈｉ．

f(a)=-\int_{-\infty}^{\infty}f(-t)\delta(a+t)dt [2]

f(a)=-\int_{\infty}^{-\infty}f(-t)\delta(a+t)dt [2]
isn't it? Regards.