@Ilya Ehrenburg said:
@aka said:
Oh. So, what is integral[x1..x2] dx/x?
Provided
x1 and x2 are positive real numbers, it's log(x2) - log(x1), which of
course equals log(x2/x1). With a few caveats this also holds for
nonzero complex numbers. But what's that to do with the price of beer?
There are no units or dimensions involved.
@aka said:
...stuff about integrating dimensionful numbers
Ok, consider the physical area of the graph of y = 1/x where y and x are dimensions, say meters. This means the formula is really y (in meters) = 1 m^2 / x (in meters) - this is the only way the units work out. Given:
Area = integral[x1..x2] ( 1m^2 / x dx)
This works out to
1 m^2 * integral[x1..x2] ( 1/x dx) = 1 m^2 * (log(x2) - log(x1)).
Note this illustrates that the log terms must be dimensionless because the units come solely from the scaling constant.
Yes, physicists often use the logarithm expression about which we're conversing, but they don't "get away" with anything: they correctly keep track of where the units are in their equations.
Consider if instead of a symbol for units we used a number. Let's then define a type of number 'distance' D = d * u where 'd' is the magnitude of the distance (a 'pure' number) and 'u' is some arbitrary nonzero number (doesn't even have to be real) which represents meters.
So for our area integral where x1 and x2 are distances, we really have:
Area = 1m^2 * ( log(x2)-log(x1) ) = 1m^2 * ( log(d2*u) - log(d1*u) ) = 1m^2 * (log(d2)+log(u) - log(d1)-log(u) ) = 1m^2*(log(d2/d1) - log(u/u)) = 1m^2*(log(d2)-log(d1))
So, since a unit is really just a unitless scaling factor with a specific meaning, the transcendental functions really do just take unitless numbers.
I'll leave it as an exercise for the readers to show how thinking of units as a scaling factor works for addition and multiplication as well.