How do I understand this symmetry of the Dirac delta function? [closed]

So I was just studying classical electromagnetic theory and I just learnt about the dirac delta (density) function. Out of curiosity, I just looked up why we choose to write the dirac delta function as $\delta(x-x_0)$ rather than just $\delta(x,x_0)$ or even $\delta_{x_0}(x)$ and I encountered a rather interesting answer.

It seems that the dirac delta function has a certain symmetry associated with it, namely that -

$$\int f(x)\delta_{x_0}(x) dx = \int f(x_0)\delta_{x_0}(x)dx_{0}$$

Can someone tell me why this is true. I know that perhaps its proof would require knowledge of some complex distribution theory or measure theory and what not which I do not possess but some intuition would be helpful.

closed as unclear what you're asking by Qmechanic♦Apr 28 '17 at 21:26

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• Yes it is. Essentially integrating $\delta_{x_0}(x)$ from $[a,b]$ is 1 if $x_0 \in [a,b]$ and zero otherwise. – Appolo Bozec Apr 28 '17 at 21:17
• This is an abuse of notation. On the LHS the dummy variable is $x$, so it has nothing to do with the $x$ on the RHS. Similarly for $x_0$. – Damian Sowinski Apr 28 '17 at 21:17
• Sorry I actually wrote the wrong identity (as you pointed out). Anyways I think now it's correct. – Appolo Bozec Apr 28 '17 at 21:24
• Still an abuse of notation, you're using the same dummy variables. – Damian Sowinski Apr 28 '17 at 21:29

What is relevant to us at the moment is that by definition $$\int_{-\infty}^{+\infty} f(x)\delta(x)dx = f(0)$$ It's easy to see that $$\int_{-\infty}^{+\infty} f(x)\delta(x-x_0)dx = \int_{-\infty}^{+\infty} f(x+x_0)\delta(x)dx = f(x_0)$$
So the function $\delta(x, x_0)$ you define is effectively equal to $\delta(x-x_0)$ in a literal sense.
Note that by substituting $y = -x$ in the first equation you can also show that $\delta(-x) = \delta(x)$, because of course $f(-0) = f(0)$. This in turn shows $\delta(x-x_0) = \delta(x_0 -x)$, i.e., in your notation, $\delta(x, x_0) = \delta(x_0, x)$.