The n-Category Café

The perspective in question is complex analysis. Most of the functions encountered in elementary calculus are actually complex-analytic — the only real counterexamples are explicit “piecewise” functions and things like ${|x|}$, which are mainly introduced as counterexamples to illustrate the meaning of continuity and differentiability. Therefore, it’s not unreasonable to interpret the indefinite integral $\displaystyle\int f(x)\,dx$ as asking for the most general complex-analytic antiderivative of $f$. And the complex domain of $\frac{1}{z}$ and $\frac{1}{z^2}$ is $\mathbb{C}\setminus \{0\}$, which is connected!

Thus, for instance, since $\frac{d}{d z}\left[-\frac{1}{z}\right] = \frac1{z^2}$, it really is true that the most general (complex-analytic) antiderivative of $\frac{1}{z^2}$ is $-\frac{1}{z}+C$ for a single arbitrary constant $C$, so we can write $\displaystyle\int \frac{1}{z^2}\,dz = -\frac{1}{z} + C$. Note that any such antiderivative has the same domain $\mathbb{C}\setminus \{0\}$ as the original function.

In addition, the dodgy calculation

$$\int_{-1}^1 \frac{1}{z^2}\,dz = \left.-\frac{1}{z}\right]_{-1}^1 = -2$$

is actually correct if we interpret $\int_{-1}^1$ to mean the integral along some (in fact, any) curve in $\mathbb{C}$ from $-1$ to $1$ that doesn’t pass through the singularity $z=0$. Of course, this doesn’t offend against signs because any such path must pass through non-real numbers, whose squares can contribute negative real numbers to the integral.

The case of $\displaystyle\int \frac{1}{z}\,dz$ is a bit trickier, because the complex logarithm is multi-valued. However, if we’re willing to work with multi-valued functions (which precisely means functions whose domain is a Riemann surface covering some domain in $\mathbb{C}$), we have such a multi-valued function that I’ll denote $\log$ (in contrast to the usual real-number function $\ln$) defined on a connected domain, and there we have $\frac{d}{d z}\left[ \log(z) \right] = \frac{1}{z}$. Thus, the most general (complex-analytic) antiderivative of $\frac{1}{z}$ is $\log(z)+C$ where $C$ is a single arbitrary constant, so we can write $\displaystyle\int \frac{1}{z}\,dz = \log(z) + C$.

What happened to $\ln{|x|}$? Well, as it happens, if $x$ is a negative real number and $Log$ denotes the principal branch of the complex logarithm, then $Log(x) = \ln{|x|} + i\pi$, hence $\ln{|x|} = Log(x) - i\pi$. Therefore, the antiderivative $\ln{|x|}$ for negative real $x$ is of the form $\Log(x)+C$, where $\Log$ is a branch of the complex logarithm and $C$ is a constant (namely, $-i\pi$).

Of course it is also true that for positive real $x$, the antiderivative $\ln{|x|} = \ln(x)$ is of the form $\Log(x)+C$ for some constant $C$, but in this case the constant is $0$. And changing the branch of the logarithm changes the constant by $2i\pi$, so it can never make the constants $0$ and $-i\pi$ coincide. Thus, unlike $-\frac{1}{x} +C$, the real-number function $\ln{|x|} + C$ in the “usual answer” is not the restriction to $\mathbb{R}\setminus \{0\}$ of any complex-analytic antiderivative of $\frac{1}{x}$ on a connected domain. This is what I mean by saying that $\displaystyle\int \frac{1}{x}\,dx = \ln{|x|}+C$ is now wrong for a different reason. And we can see that the analogous dodgy calculation

$$\int_{-1}^1 \frac{1}{x}\,dx = \ln{|x|}\Big]_{-1}^1 = 0$$

is also still wrong. If $\gamma$ is a path from $-1$ to $1$ in $\mathbb{C}\setminus \{0\}$, the value of $\int_{\gamma} \frac{1}{x}\,dx$ depends on $\gamma$, but it never equals $0$: it’s always an odd integer multiple of $i\pi$, depending on how many times $\gamma$ winds around the origin.

I’m surprised that no one in the previous discussion, including me, brought this up. Of course we probably don’t want to teach our elementary calculus students complex analysis (although I’m experimenting with introducing some complex numbers in second-semester calculus). But this perspective makes me less unhappy about writing $\displaystyle\int \frac{1}{x^2}\,dx = -\frac{1}{x} + C$ and $\displaystyle\int \frac{1}{x}\,dx = \log(x) + C$ (no absolute value!).