Integrating 1/x with units (logarithm)

Hi.

What exactly is happening mathematically when you integrate ##\frac{1}{x}##
$$\int_a ^b \frac{1}{x} dx=\ln{b}-\ln{a}=\ln{\frac{b}{a}}$$
if there's units? Sure, they cancel if you write the result as ##\ln{\frac{b}{a}}##, but the intermediate step is not well-defined, so why should log rules even be justified?

greypilgrim said:

Hi.

What exactly is happening mathematically when you integrate ##\frac{1}{x}##
$$\int_a ^b \frac{1}{x} dx=\ln{b}-\ln{a}=\ln{\frac{b}{a}}$$
if there's units?

You multiply the dimension of ##\left[\dfrac{1}{x}\right]## with the dimension of ##\left[dx\right]## ...

greypilgrim said:

Sure, they cancel if you write the result as ##\ln{\frac{b}{a}}##, but the intermediate step is not well-defined, so why should log rules even be justified?

... and the intermediate step is even more directly an area built from ##1/x## and ##x## in the Riemann sums.

We could substitute ##x=y \times 1##, where ##y## is unitless. This would make the entire expression unitless.

greypilgrim said:

Hi.

What exactly is happening mathematically when you integrate ##\frac{1}{x}##
$$\int_a ^b \frac{1}{x} dx=\ln{b}-\ln{a}=\ln{\frac{b}{a}}$$
if there's units? Sure, they cancel if you write the result as ##\ln{\frac{b}{a}}##, but the intermediate step is not well-defined, so why should log rules even be justified?

You think we should just throw out a Mathematical result because the units are somewhat troublesome? Yeeouch!

Let y = cx where c is a constant factor to make y unitless. Then do the integration.

-Dan

fresh_42 said:

You multiply the dimension of ##\left[\dfrac{1}{x}\right]## with the dimension of ##\left[dx\right]## ...

... and the intermediate step is even more directly an area built from ##1/x## and ##x## in the Riemann sums.

Sure, the result is dimensionless, but if ##\left[dx\right]=\left[a\right]=\left[b\right]\neq 1##, then ##\ln{b}-\ln{a}## is ill-defined, isn't it?

topsquark said:

You think we should just throw out a Mathematical result because the units are somewhat troublesome?

Of course not, it's more that algebra with units usually works out perfectly except in this one case (at least this is the only exception I can think of).

I think the fix is that the proper antiderivative of ##\frac{1}{x}## isn't actually ##\ln{\left|x\right|}+C## but ##\ln{\frac{\left|x\right|}{\left[x\right]}}+C##.

greypilgrim said:

Sure, the result is dimensionless, but if ##\left[dx\right]=\left[a\right]=\left[b\right]\neq 1##, then ##\ln{b}-\ln{a}## is ill-defined, isn't it?

You say that ##a=v(a)\cdot u(a)## and ##b=v(b)\cdot u(b)## where ##v(\cdot)## is the value and ##u(\cdot)## is the unit. Now you assume ##u(a)=u(b)## and compare the dimensionless quantity
$$
\log\left(\dfrac{a}{b}\right)=\log\left(\dfrac{v(a)u(a)}{v(b)u(b)}\right)=\log\left(\dfrac{v(a)}{v(b)}\right)=\log v(a)-\log v(b)
$$
with
\begin{align*}
\log\left(\dfrac{a}{b}\right)&=\log(a)-\log(b)=\log(v(a)\cdot u(a))-\log(v(b)\cdot u(b))\\&=
\left(\log v(a) + \log u(a)\right)-\left(\log v(b)+\log u(b) \right)=\log v(a)-\log v(b)
\end{align*}
where the units cancel, too.

Your confusion comes only because you leave out the details.

greypilgrim said:

Hi.

What exactly is happening mathematically when you integrate ##\frac{1}{x}##
$$\int_a ^b \frac{1}{x} dx=\ln{b}-\ln{a}=\ln{\frac{b}{a}}$$
if there's units? Sure, they cancel if you write the result as ##\ln{\frac{b}{a}}##, but the intermediate step is not well-defined, so why should log rules even be justified?

What units!!! The problem has no units in it.

greypilgrim said:

Of course not, it's more that algebra with units usually works out perfectly except in this one case (at least this is the only exception I can think of).

I think the fix is that the proper antiderivative of ##\frac{1}{x}## isn't actually ##\ln{\left|x\right|}+C## but ##\ln{\frac{\left|x\right|}{\left[x\right]}}+C##.

Your first expression is correct, the second makes no sense what-so-ever. It isn't the correct antiderivative!

Look again at my comment about defining a constant c such that cx is unitless.

-Dan

fresh_42 said:

\begin{align*}
\left(\log v(a) + \log u(a)\right)-\left(\log v(b)+\log u(b) \right)=\log v(a)-\log v(b)
\end{align*}

This step requires ##\log u(a)=\log u(b)## which requires ##\log u(\cdot)## to be defined, so how is the logarithm of a unit defined?

greypilgrim said:

This step requires ##\log u(a)=\log u(b)## which requires ##\log u(\cdot)## to be defined, so how is the logarithm of a unit defined?

You confuse different aspects again. The logarithm of a unit ##u## is simply ##\log u,## a formal concept. If you ask for physical relevance you need to come up with an example for ##a=v(a)\cdot u(a) =e^{b}=e^{v(b)\cdot u(b)}## first. You cannot switch the burden of proof. I am fine with dimensionless ##a## and ##b.##

greypilgrim said:

Hi.

What exactly is happening mathematically when you integrate ##\frac{1}{x}##
$$\int_a ^b \frac{1}{x} dx=\ln{b}-\ln{a}=\ln{\frac{b}{a}}$$
if there's units? Sure, they cancel if you write the result as ##\ln{\frac{b}{a}}##, but the intermediate step is not well-defined, so why should log rules even be justified?

This is discussed in some depth here:

https://math.ucr.edu/home/baez/physics/General/logs.html

fresh_42 said:

The logarithm of a unit is simply a formal concept. If you ask for physical relevance

I'm not asking for physical relevance (which is why I posted in the maths forum), just a clean mathematical definition. This is analysis, where ##\log## is often defined by its series. Well I guess you can formally write down the series of the log of a unit, but for subtraction questions about convergence arise.

PeroK said:

This is discussed in some depth here:

https://math.ucr.edu/home/baez/physics/General/logs.html

Thank you, this seems to exactly address the issue.

topsquark said:

Your first expression is correct, the second makes no sense what-so-ever. It isn't the correct antiderivative!

Look again at my comment about defining a constant c such that cx is unitless.

My second expression is exactly equation (7) in that link (except that I focused on the special case ##x_0=1\cdot [x]##).

But I just googled and apparently the use of square brackets ##[\cdot]## isn't consistent, some sources use it for dimension, others for the unit (which obviously depends on the unit system). I meant the latter, unit.

greypilgrim said:

I'm not asking for physical relevance (which is why I posted in the maths forum), just a clean mathematical definition. This is analysis, where ##\log## is often defined by its series. Well I guess you can formally write down the series of the log of a unit, but for subtraction questions about convergence arise.

But there are no units in mathematics!

martinbn said:

But there are no units in mathematics!

I'm really confused by that statement and have never heard of anything like that before. So all the calculations in physics are not mathematical? Maths isn't the language of science after all? What about all the algebraic operations done with units as if they were variables, like multiplying and cancelling?

The question about what kind of mathematical object units are is apparently not such an easy one though, Terry Tao wrote an article about this:
https://terrytao.wordpress.com/2012/12/29/a-mathematical-formalisation-of-dimensional-analysis/

greypilgrim said:

My second expression is exactly equation (7) in that link (except that I focused on the special case ##x_0=1\cdot [x]##).

Oh. I missed the [ ] in the denominator. My bad.

Still, the y = cx is the standard prescription for this.

-Dan