The power rule for integration and the logarithm

An unsatisfying rule

When I first learnt the power rule for integration in school, namely

∫xn ⁣dx={xn+1n+1+const,n≠−1log⁡x+const,n=−1,\int x^n \td x = \begin{cases} \dfrac{x^{n+1}}{n + 1} + \const, & n \ne -1 \\[\tallspace] \log x + \const, & n = -1, \end{cases}

it always bugged me how the special case of n=−1n = -1 required separate treatment. In my mind, there was no reason for

∫x−1 ⁣dx=x00+const\int x^{-1} \td x = \frac{x^0}{0} + \const

to not work.

Divine revelation

Many years later, it suddenly occurred to me that it does indeed work; only the constant of integration must be infinite. Specifically, −10/0-1^0/0 plus something finite:

∫x−1 ⁣dx=(x00−100)+const.\int x^{-1} \td x = \roundbr{\frac{x^0}{0} - \frac{1^0}{0}} + \const.

The bracketed portion is assuredly log⁡x\log x, as it vanishes when x=1x = 1. As a side-effect, we immediately see why a logarithm grows slower than any polynomial: because a logarithm is a polynomial with infinitesimal degree.

If all of this is sacrilege unto you, consider reading some Euler (see translations by Ian Bruce at <www.17centurymaths.com>).

For haters of infinity

If you're still not convinced, compute

lim⁡ε→0(xεε−1εε),\lim_{\eps \to 0} \roundbr{\frac{x^\eps}{\eps} - \frac{1^\eps}{\eps}},

and be satisfied when you get log⁡x\log x at the end.

Cite this page

Conway (2022). The power rule for integration and the logarithm. <https://yawnoc.github.io/math/power-rule-log> Accessed 2025-06-15.