Can the differential be unitless while the variable have an unit in integration?

Question

Apologies for terminology inconsistencies, as I'm reading a Chinese statistics and probabilities textbook while looking up intrinsics on an English encyclopedia.

This arose when I was reading the definition of expected value (a.k.a. mean according to my textbook) for a continuous random variable $x$ which is:

$$E(X)=\int^{\infty}_{-\infty}xf(x)\text{d}x$$

Where $x$ is the random variable, and $f(x)$ is the probability density function.

At first, I wasn't able to make sense of it : it's the product of 3 variables (I know $\text{d}x$ is a special symbol, but let's assume we're doing a brute-force calculation by setting it to a really small value) - which means the unit must be very strange. By unit, I mean things like meters, seconds, kilograms, etc. (I know all 3 are supposed to be from the set of real numbers, but let's suppose they've got types like programming language variables).

But then I thought: the $\text{d}x$ doesn't need to have a unit - it just need to approach a very small value whilst making sure everything's consistent.

So Q: Is my interpretation of the formula for expected value reasonable and sound? Can variable and its differential have different unit (as in one is of a unit while another can have another or being completely unitless)?

Answer 1

No, even if $\mathrm{d}x$ represents an infinitesimal quantity, it must have the same units as $x$. Moreover, as a density, $f(x)$ will display the inverse units of $x$. In consequence, if $[X]$ denotes the units of the variable $X$, then one has : $$ [\Bbb{E}(X)] = [x]\cdot[f(x)]\cdot[\mathrm{d}x] = [X]\cdot[X]^{-1}\cdot[X] = [X] $$ Unsurprisingly, the average value of $X$ possesses the same units as $X$.

Answer 2

Like everything in mathematics, it depends on the definitions and rules we set. Here, we need to remember that derivatives and integration are limits and it seems reasonable that

If a quantity has U for unit, then its limits has also U for limit. In particular,

1. If $x_n$ is a convergent sequence with unit U, then $\lim_{n\to\infty}x_n$ has also U for unit
2. If $y=g(x)$ is a function with unit U (on the output), then $\lim_{x\to a} g(x)$ has also limit U (when the limit exists).

Together with the other, universally-accepted rules for units

1. If $a$ and $b$ have the same unit U, then $a+b$ and $a-b$ have also U for unit.
2. If $a$ has U for unit and $b$ has V for unit, then $ab$ have UV for unit
3. If $a$ has U for unit and $b$ has V for unit, then $\frac{a}{b}$ have U/V for unit.

we get that if $f(x)$ is a function where the input $x$ has unit V and the output $f(x)$ has unit U, then $\frac{f(x)-f(a)}{x-a}$, hence the derivative $f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$, has unit U/V

On the other hand, the Riemann sum $S_n=\sum_{k=1}^nf(x_k^\ast)\Delta$, hence the definite integral $\int_a^b f(x)\ dx=\lim_{n\to\infty}S_n$, has unit UV.

Answer 3

In the case you've explicitly asked about,

$$ \int x f(x) \, dx $$

The answer is probably no: $dx$ has the same units as $x$, and if $x$ has units, then $dx$ has units. The reason I say probably is that if $x$ is unitless, then $dx$ is similarly unitless.

In the more general case which your title text may reference:

$$ \int x(z) \, f(y(z)) \,dz $$

The clear answer is a resounding yes. In physics, they often integrate over states which would give $dz$ units of count (which is unit-less) even though the integral as a whole may have meaningful units like temperature or pressure.

Answer 4

Interesting View-Point , though there are certain logical issues like what you encountered.

"Can the differential be unitless while the variable have an unit in integration?" : No

We have $\Delta x$ tracking the small change in $x$. That lets us calculate $x + \Delta x$
When volume $v$ in $m^3$ varies , the small change $\Delta v$ must be in $m^3$ , which will let us calculate $v + \Delta v$ in $m^3$
When time $t$ in $s$ varies , the small change must be in $s$ , which will let us calculate $t + \Delta t$ in $s$

In Case the units do not match , such calculations will be meaningless : We can not add a number with units to a number without units.

We can avoid that issue with 2 ways.

(1) Both have same units :

$x$ & $\Delta x$ have the same units like $m^3$ , $s$ , $\cdots$
That will let us add $x$ & $\Delta x$
While certain things will work out , it will complicate calculations like Series $1+x+x^2+x^3\cdots=1/(1-x)$
That complication can be resolved , though more work is required.
Basically , we have to ensure that there are certain co-efficients like $a_1x+x_2x^2+a_2x^3\cdots$ where these co-efficients (which should have the necessary units) will convert all the terms to some common unit , letting us add the terms.

Your Integral can use the same thinking :
In your Integral , $x$ & $\Delta x$ will have same units , while the "Density" function must be in reciprocal units.
Hence , the over-all units will be consistent : $E(x)$ & $x$ will have same units.

(2) Both are unitless :

Alternate way is to take all variables to not have units. When we want to calculate area or volume or what-ever , we remove that units from the variables , then we make the calculations , then we add the units back.
There are no complication with calculations like Series $1+x+x^2+x^3\cdots=1/(1-x)$

Your Integral can use the same thinking :
All terms $x$ , $\Delta x$ , $f(x)$ have no units.
Make the calculations , then introduce the relevant units when required.
This way lets us talk about arbitrary terms like $E(x+x^2)$ , $E(x+1/x)$ , $E(\sqrt{x}+e^x)$ , for example , without losing consistency.

Both ways are common & used where-ever appropriate.