dimensional analysis - Units INSIDE of a Dirac Delta function

I know that the units of a Dirac Delta function are inverse of it's argument, for example the units of $\delta(x)$ if $x$ is measured in meters is $\frac{1}{meters}$.

But, my question is what are the units inside the Dirac Delta?

For example if you have $\delta (x - 1)$ is the exact expression $x$ inside the dirac delta unitless? If not, that means that the $1$ would actually have to have units of meters. That seems odd to me, which is why I think that inside the units are unitless and as a whole you give the dirac delta inverse units of the argument but I'm not sure. If someone could clear this up for me I'd really appreciate it, thanks!

Answer

Let's clear up one thing first: you can never add (or subtract) two expressions with different units. For example, in $\delta(x - 1)$, the 1 is unitless, which means $x$ has to be unitless as well.

Then there's the separate issue of the units of the expression inside the Dirac delta. The expression inside the delta is its argument, and as you know, the argument does not have to be unitless. So there's your answer. You can write $\delta(x - 1\text{ m})$, for example, and since $x - 1\text{ m}$ has units of length, the delta function itself will have units of inverse length.

The reason the arguments of many functions have to be unitless is that those functions can be expressed as a power series,

$$f(x) = a_0 + a_1 x + a_2 x^2 + \cdots$$

where the $a_i$ are just numbers.¹ For example,

$$\begin{align} \exp(x) &= 1 + x + \frac{1}{2}x^2 + \cdots \\ \sin(x) &= x - \frac{x^3}{6} + \cdots \end{align}$$

If $x$ had units of, say, length, then you would be adding a number to a length to an area (length squared) to a volume etc., and as I said, that can't happen.

But the delta function is not one of these functions that can be expressed as a power series. In fact, it's not really a function at all. It's a distribution, defined implicitly by the integral

$$\int_a^b f(x) \delta(x - x_0)\mathrm{d}x = \begin{cases}f(x_0), & a \le x_0 \le b \\ 0,& \text{otherwise}\end{cases}$$

In order for this integral to come out with the same units as $f(x_0)$, the rest of the expression being integrated - that is, the combination $\delta(x - x_0)\mathrm{d}x$ - has to be unitless. And since $\mathrm{d}x$ has the same units as $x$, $\delta(x - x_0)$ has to have the unit that cancels that unit out.

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¹One might argue that you can define a function $f$ as a power series where the coefficients $a_n$ do have units of the appropriate type, and in that case the argument $x$ would have units. But you can always generalize such a function to a different function of a unitless variable: just write $a_n = f_0 b_n/x_0^n$ where $b_n$ is unitless and $x_0$ has the same dimension as $x$, and then express the function as $$f(x) = \sum a_n x^n = f_0\sum b_n (x/x_0)^n = f_0 g(x/x_0)$$ where $g(y) = \sum b_n y^n$. The function $g$ expresses the same functional relationship as $f$, but in terms of a unitless variable, and is thus more generally useful.