Why is Log the Antiderivative of 1/x?

October 30, 2024

The power rule has a weird deficiency that everyone just learns to deal with. Recall, the power rule tells us that the antiderivative of $x^p$ is

which gives us our familiar relations: the antiderivative of $x$ is $\frac12x^2$, the antiderivative of $0$ is a constant, and the antiderivative of $x^{-1}$ is... $\log x$? What?! This logarithm just seems to come out of nowhere. It's easy to algebraically prove why this happens, but it still feels weird. Most students end up treating this as a rule to be memorized, but it's not hard to see why this happens.

Log as a Limit

(To be clear, the proof of the derivative of $\log$ is quite simple.^[1] My goal here is to bring a more intuitive understanding to the table.)

As with most edge cases/singularities in math, they can be better understood as a limit of local behavior. In this case, we can view the antiderivative of $x^{-1}$ as the limit of the antiderivatives of $x^{-1.1}$, $x^{-1.01}$, $x^{-1.001}$, and so on. That is, we should expect

Note that we need to pay special attention to the bounds of integration here. If we start the integral of $t^{-1}$ from $t = 0$, it diverges, so we need to start from $t = 1$ instead. Thus, for any positive input $x$, we want

Applying the power rule, expression on the right becomes

which, on a graph, actually looks pretty close to a logarithm!

So it would appear that our intuition is correct, and the local limiting behavior is as we expect. Can we do any better? Well, the next step would be to show that the above expression actually approaches a logarithm, i.e.

which can be rewritten as

Occasionally, this limit is used as a definition of $\log$ itself! But there's actually a very simple explanation. Remember that $\log x$ is the inverse of $e^x$, which is defined as

(Remember back to your introduction to $e$ as compound interest!) If we make the substitution $p = \frac1n$, the above expression becomes

Now, just as the defining relationship between $e^x$ and $\log$ is $\log e^x = x$, we can do the same for their limiting expressions:

and as you can check for yourself, the same applies the other way around, just like $e^{\log x}$. So it's not a surprise that $\lim_{p \to 0} \frac{x^p - 1}p = \log(x)$.

Another way to view the logarithm here is in terms of growth rate. $\log x$ is a subpolynomial function, which means that it grows slower than any polynomial.^[2] This is what we should expect from the power rule: we want the antiderivative of $x^{-1}$ to look something like $x^0$, which would also be subpolynomial (since it's a constant). And although we don't quite get that answer, its subpolynomial nature is echoed in the logarithm.

Similarly, note that $e^x$ is superpolynomial, i.e. it grows faster than any polynomial. And just like how the inverse of $x^p$ is $x^\frac1p$, we have that the inverse $\log x$ is $e^x$, or the inverse of a subpolynomial is a superpolynomial. Or, less precisely, the inverse of $x^0$ is $x^{\frac10} = x^\infty$, larger than any polynomial!

1. Take $y = \log x$, so $x = e^y$. Differentiating both sides with respect to $x$, we get $1 = e^yy'$, and therefore $y' = \frac1{e^y} = \frac1x$. ↩
2. If you're familiar with the notion of uniform convergence, perhaps from a real analysis course, then it's worth noting that the polynomials $\frac{x^p - 1}p$ don't converge to $\log x$ uniformly, precisely because of the logarithm's subpolynomial nature: for any $p$, the polynomial gets arbitrarily far away from the logarithm for large $x$. However, if we consider a positive bounded interval $(\varepsilon, \frac1\varepsilon)$ with $\varepsilon > 0$, then we do indeed get uniform convergence. ↩