The physical units of the Dirac delta function

Question

I was given the following instruction to follow when we find the units of the $\delta(x)$ function:

We can deduce these by inspecting the expression for its area: $∫_{−∞}^{∞}()=1$

Clearly the right-hand side of this is dimensionless

On the other hand, if $$ is a position, then $$ has the dimension of length

So the delta function must cancel out those length dimensions in order to leave a dimensionless right-hand side, i.e. $()$ must have dimensions of $(ℎ)^{−1}$

So we see a very important thing here: the physical dimensions of the delta function are the inverse of the physical dimensions of its argument.

A delta function of time, $()$, has dimensions $()^{−1}$; a delta function of momentum, $()$, has dimensions $()^{−1}$ and so forth

I thought that the right hand has the units (for example if $x$ is position) $m^2$. as we are finding the area under the graph. So in general the right hand would have the units of $(units)^{2}$

Answer 1

If $a(t)$ is acceleration and $t$ is time, will $\int a(t) \, dt$ have dimension $\text{length}^2$ since we are finding the area under the graph? No, it will not; it will have dimension $\text{length}/\text{time}$.

The interpretation of integral as area should not be taken too far.

Answer 2

Your are correct that the physical dimension of the delta function is the inverse of its arguments dimension.

The dimension of an integral is the product of the dimensions of the integrand and the integration variable.