Antiderivative of $1/x$, with or without absolute value?

Question

Many textbooks include $\int \frac{1}{x} dx = \ln |x| + c$ in their list of antiderivative formulas, with the absolute value. Correspondingly, they do the same with the antiderivative of $\tan x$ or similar.

Is it productive to introduce the complication of the absolute value in a practically oriented course, when under time pressure? The alternative is to try to restrict ourselves to positive $x$.

My concerns are that one still needs to be careful not to include the singularity at 0 when writing definite integrals and the absolute value hides this. Some texts will omit the absolute value which can be confusing to students. So do most computer algebra systems, which operate with complex numbers by default.

Answer 1

Even $\int \frac{1}{x} \textrm{ d}x = \ln(|x|) + C$ is incorrect.

It should be

$$ \int \frac{1}{x} \textrm{ d}x = \begin{cases} \ln(x) + C_1 \textrm{ if $x > 0$}\\ \ln(-x) + C_2 \textrm{ if $x < 0$}\\ \end{cases} $$

Personally, I think that it would be best to include some problems where this issue arises naturally, and resolve it through thinking about it rather than memorizing another arcane aspect of a formula without understanding it.

Answer 2

Absolute value or no absolute value, there will be students that attempt to integrate over infinite discontinuities anyway. Leaving off the absolute value in $\int \frac{1}{x}\,\mathrm{d}x = \ln|x| + C$ may stop them from blindly attempting to evaluate $\int_{-1}^2 \frac{1}{x}\,\mathrm{d}x$, but what about $\int_{-1}^2 \frac{1}{x^2}\,\mathrm{d}x$?

I agree with Steven's approach of having students encounter a problem and, through resolving it, arrive at a deeper understanding. So I would suggest giving the formula $\int \frac{1}{x}\,\mathrm{d}x = \ln|x| + C$ and asking students to attempt to evaluate something like $\int_{-1}^2 \frac{1}{x}\,\mathrm{d}x$. Let them get it wrong, and have them realize that their answer is wrong by asking them to perform some sort of check.

It may be beneficial to start with $\int_{-1}^2 \frac{1}{x^2}\,\mathrm{d}x$, as this integral is unambiguously divergent. Whereas for $\int_{-1}^2 \frac{1}{x}\,\mathrm{d}x$, students might want to make some sort of symmetry argument, cf. Cauchy principal value.

Answer 3

One reason to discuss the anti-derivative of $\frac{1}{x}$ also for negative $x$ is that some (or many, depending on your audience) students will needs it later on when they learn about ordinary differential equations.

For instance, consider the very simple initial value problem $$ \begin{cases} \dot x(t) = x(t), \\ x(0) = x_0. \end{cases} $$ As we all know its solution is given by $x(t) = x_0 e^t$ for all $t \in \mathbb{R}$. But assume that you don't know the solution, yet (for instance, if you're a student taking an ODE course) and try to find it by using separation of variables. This gives $$ \int 1 \, \mathrm{d}t = \int \frac{1}{x} \, \mathrm{d}x. $$ Assume now that you only learnt that the anti-derivative of $\frac{1}{x}$ is $\ln x + c$ for positive $x$ - it happens to easily that you forget about the sign restriction and just blindly apply the formula. You thus get $$ t = \ln x(t) + c \qquad (*) $$ for a constant $c \in \mathbb{R}$ and hence, $x(t) = e^{-c} e^t$. The initial condition finally yields $e^{-c} = x_0$. But this cannot be true if $x_0$ is negative. (Unless one allows for complex $c$ - which is a pain here, since you then need to draw the complex logarithm of a negative number, which means that you cannot use the principal branch of the complex logarithm - in any case, this moves quite deep into the territory of complex analysis for such an innocent differential equation over the real field).

By the way, to add further to the confusion, the same problem does at first glance not arise if you do the separation of variables by using definite integrals instead of indefinite integrals. This gives you $$ \int_0^t 1 \, \mathrm{d}s = \int_{x_0}^{x(t)} \frac{1}{x} \, \mathrm{d}x, $$ thus (if makes again the mistake to believe that $\ln x$ were an anti-derivative of $\frac{1}{x}$ for postive and negative $x$) $$ t = \ln x(t) - \ln x_0 = \ln \frac{x(t)}{x_0}, \qquad (**) $$ and hence $x(t) = x_0 e^t$ - which is the correct solution formula. Yet, we made the same error as above, by taking $\ln x$ as a (presumed) anti-derivative of $\frac{1}{x}$ even for negative $x$. But the error cancels out for the following reason: Solutions cannot cross equilibrium points, so $x(t)$ always has the same sign as $x(0)$. So if $x(0)$ is negative, so is $x(t)$ and hence the correct computation in $(**)$ for negative $x(0)$ is actually $$ t = \ln (-x(t)) - \ln (-x_0) = \ln \frac{-x(t)}{-x_0} = \ln \frac{x(t)}{x_0}, $$ which just happens to give the same result as our incorrect (for $x(0) < 0$) computation above.

To sum up: If one learns the anti-derivative of $\frac{1}{x}$ only for positive $x$ and happens to forget about this restriction later on (which is a very easily made mistake), one gets a mess when working with initial value problems. The same method (separation of variables) can then lead either to correct or incorrect result, depending on how precisely one applies the method.

By the way, in case that you find the example discussed above too artificial (because the initial value problem is so simple) - a similar issue appears, for instance, for the logistic equation.

Answer 4

Note that the derivative of $\log x$ has a Dirac delta term: https://mathoverflow.net/questions/127601/does-the-derivative-of-log-have-a-dirac-delta-term

This involves the principal value.