What's the unit of delta function? For dimensional analysis of delta function, what should be the unit of $\delta\left(x-x^{\prime}\right) \delta(t-\tau)$? Is it time*space or the delta function is non-dimensional?
Also similar question, what's unit of $e^x$, where $x$ is in meter?
I'm new to mathematical physics so I'm a little confused about the units. Any help is appreciated.
 A: The way we write these things in physics is quite loose. In essence, $e^x$ is not defined if $x$ is in meters!
In order to see this, consider the definition of $e^x = 1 + x + x^2/2! +\ldots$.
It is quite clear that you can't add $1$ to $x$ as each term has different dimensions!
The way to write it would be $e^{x/k}$ where $k = 1m$. This cancels the dimension and the expression is well-defined. Normally, you won't encounter $e^x$ type expression if you had started with a meaningful expression. The only way I can imagine arriving at $e^x$ and not at $e^{x/k}$ is in a textbook exercise that says that the trajectory of some contrived particle is $(x,e^x)$. Of course, normally you won't see this artificial constant $k$ but some combination of common constants like $hc/2\pi$ (or something of appropriate dimension).
See this answer to get a better understanding of what units really are.
A: For the Dirac delta, we must have
$$
\int_{\Bbb R} \delta(x)\, \text{d}x = 1\, .
$$
So if $x$ is expressed in $\text{m}$, the Dirac delta $\delta(x)$ is expressed in $\text{m}^{-1}$. Therefore, $\delta(x-x')\delta(t-\tau)$ has physical unit $[x]^{-1}[t]^{-1} = /\text{ms}$. For exponentials, one might rather introduce a characteristic length $\ell$ and compute dimensionless expressions such as $e^{x/\ell}$. To avoid potential errors, it is often useful to work in dimensionless coordinates.
