- #1

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## Homework Statement

Good day.

May I know, for Dirac Delta Function,

Is δ(x+y)=δ(x-y)?

## The Attempt at a Solution

Since δ(x)=δ(-x), I would say δ(x+y)=δ(x-y). Am I correct?

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- Thread starter Poligon
- Start date

- #1

- 23

- 0

Good day.

May I know, for Dirac Delta Function,

Is δ(x+y)=δ(x-y)?

Since δ(x)=δ(-x), I would say δ(x+y)=δ(x-y). Am I correct?

- #2

- 17,572

- 8,567

[tex]\int_{\mathbb{R}} \mathrm{d} x \delta(x) f(x)=f(0),[/tex]

where [itex]f[/itex] is a function of an appropriate space of functions (e.g., the Schwartz space of quickly vanishing functions or the space of functions with compact support, etc.).

You can prove properties of the distribution by formally handling this kind of integral as if the [itex]\delta[/itex] distribution were a usual function. E.g., you have

[tex]\int_{\mathbb{R}} \mathrm{d} x \delta(x-y) f(x) = \int_{\mathbb{R}} \mathrm{d} x' \delta(x') f(x'+y)=f(0+y)=f(y).[/tex]

Now you can evaluate yourself, what [itex]\delta(x+y)[/itex] must be and also prove the (correct) statement [itex]\delta(-x)=\delta(x)[/itex].

- #3

arildno

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Thus, you must ask yourself:

Is x+y=x-y an identity or an equation?

- #4

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Thanks a lot for your prompt reply!

I see. So ∫dx δ(x+y)f(x) will equal to f(-y),

if f(x) is an even function, then

∫dx δ(x+y)f(x) = f(y)

if f(x) is an odd functoin, then

∫dx δ(x+y)f(x) = -f(y)

Am I correct? I am sorry if I still not able to get the concept, do allow me to make some mistakes so I can learn from it. While for arildno's question, I am sorry that I am not able to answer as it's little deep for me (I can't understand what equation and identity are in this context). I will dig into that.

- #5

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Yes, that's correct. So you see that in general [itex]\delta(x+y) \neq \delta(x-y)[/itex].

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