6
$\begingroup$

Say I have equations of the form $f(x) = 1/x$ (to illustrate the simplest case of the problem) or $g(x) = \frac{1}{a - b \cdot x/c}$ and I want to find the integrals.

Normally, these would both be pretty straightforward. $\int f(x)dx = \ln(x) + C$, and $\int g(x)dx = -\frac{c}{b}\ln(a \cdot c - b \cdot x) + C$.

Suppose now, however, that $a$, $b$, $c$, and $x$ are dimensionful quantities; more specifically, $a$ has the same units as $b$ and $x$ has the same units as $c$. We would expect that $\int \frac{dx}{x}$ should be dimensionless, and indeed the $\ln$ function does have a pure number result... but the argument to $\ln$ should also be dimensionless, which $x$ is not, which leads to a problem in interpretation.

Similarly, $\int g(x)dx$ should have units of $type(x)$ over $type(a)$, which indeed it does (since we defined $x$ & $c$ to have the same units, and $a$ and $b$ to have the same units), but it also involves a non-dimensionless argument to $\ln$.

So, how can this be resolved? How does one go about integrating inverse units in a way that makes sense?

  • 1
    $\int_a^b\,dx/x=\ln b- \ln a = \ln(b/a)$ – 2017-01-06

2 Answers 2

4

Part of the problem here is that indefinite integration does not really make much sense; it is best regarded as a notational shorthand. The meaningful quantity is a definite integral: $\int_a^b f(x) dx$. For $f(x)=1/x$, this does make dimensional sense, because as long as $a,b$ have the same sign, $\int_a^b 1/x dx = \ln(b/a)$. (If $a,b$ have opposite signs then the integral just plain fails to converge classically.)

Back in terms of $\ln(|b|)-\ln(|a|)$, this still makes sense for dimensional $a,b$, because for any two choices of units of $a,b$, the two differ by a multiplicative constant. And $\ln(cx)-\ln(cy)=\ln(c)+\ln(x)-\ln(c)-\ln(y)=\ln(x)-\ln(y)$ (assuming $c,x,y>0$ for ease of presentation). So you have invariance under change of units, which is what we should have for a dimensionless quantity.

  • 0
    I suppose that makes sense, but it doesn't quite get me to solving my problem, as I am now stuck wondering what limits to use for a definite integral. E.g., if I have a separable DE of the form $h(y)dy = \frac{dx}{a-b\dot x/c}$, would I get a valid solution just picking any arbitrary limits $j$ & $k$ and calculating $\int_j^k h(y)dy = \int_j^k \frac{dx}{a-b\dot x/c}$? – 2017-01-06
  • 0
    @LoganR.Kearsley When solving a separable differential equation, the correct way to write the solution (assuming it's an IVP $y(x_0)=y_0$) is $\int_{y_0}^y f(z) dz = \int_{x_0}^x g(u) du$. For example, in $y'=y$, you can write $\int_{y_0}^y 1/z dz = \int_{x_0}^x du$, so $\ln(y/y_0)=x-x_0$. If you don't actually have an IVP in mind and just want a general solution you can still use $x_0,y_0$ as free parameters. – 2017-01-06
  • 0
    Ah, I see. So it seems, then, that I need to know at least one point in the solution curve ahead of time in order to find the complete solution. That is somewhat bothersome. Or is it the case that the solution set is parameterized over the initial value, such that any choice of values for $y_0$ and $x_0$ would produce a valid solution? – 2017-01-06
  • 1
    @LoganR.Kearsley Unless domain issues arise, you can pick $x,y,x_0,y_0$ freely. (In this example, $y_0$ has to be nonzero and $y$ has to have the same sign as $y_0$.) So in particular, the separable solutions to a separable differential equation are parametrized by $x_0,y_0$ in the way I described. (Note that there are quite frequently nonseparable solutions to a separable differential equation.) – 2017-01-06
2

normally we say that the argument to any of the transcendental functions

$$f(x) = \sum\limits_{n=0}^{\infty} a_n \, x^n $$

must be dimensionless because of the Maclauren series used to evaluate it has many different powers of $x$, the coefficients $a_n$ are dimensionless, and the only way you can meaningfully add those terms is if they are all dimensionless, which means $x$ must be dimensionless.

power functions

$$ f(x) = x^p $$

can take a dimensionful argument and will return another dimensionful result that is the dimension of the argument raised to the $p$ power.

the $\log(\cdot)$ function might be a little different. it's not particularly good form but since

$$ \log(a \, b) = \log(a) + \log(b) $$

the $\log(x \cdot \textsf{})$ where $x$ is dimensionless and multiplies the unit token, can be thought of the sum of the $\log(x)$ and the $\log(\textsf{})$. This sorta works out with, say:

$$\begin{align} \log \left( \frac{1 \textsf{ mile}}{1 \textsf{ km}} \right) &= \log(1 \textsf{ mile}) - \log(1 \textsf{ km}) \\ &= \log(1) + \log(\textsf{ mile}) - (\log(1) + \log(\textsf{ km})) \\ &= \log(1.609344) \\ & \approx 0.47582664254 \\ \end{align}$$

which is dimensionless. so somehow the $\log(\textsf{mile})$ and $\log(\textsf{km})$ must differ by a dimensionless constant. but i would not express it that way.

i would generally insist that only like-dimensioned quantities may be added, subtracted, equated, or compared. that multiplication of dimensioned quantities multiplies their dimensions and that division and power (or root) functions reflect that convention, and that transcendentals must have dimensionless arguments.