# Integral of $1/x$ about the asymptote with Cauchy versus complex

In short, the integral of $1/x$ from $-2$ to $1$ or some other such range is being confusing for me. By means of the Cauchy principal value it receives a value of $-\ln 2$. Whereas with complex integration it would seem I'd get $-\ln 2 + \pi i$. And it can be shown decidedly that $\ln(|x|)$ does not have the derivative of $1/x$, but rather $\operatorname{sgn}(x)/x$. Basically what's the correct integral here?

• Divergent integrals never have one "correct" answer. You have to choose the answer that makes the most sense for your application. Oct 23, 2013 at 3:34
• @icurays1 Why shouldn't they? Just because this has a singularity what makes it so special? Why not consider functions which have an 'asymptote' as you approach infinity. Something like $\frac{1}{1 - \frac{x}{x + 1}}$ is only undefined for an infinitely large x (and x <= -1 but ignore that for this example) Oct 23, 2013 at 20:31

For a function with a singularity, integrals in the complex plane are in general path-dependent. Depending on what contour you take, you could get $-\ln 2 + k \pi i$ for any odd number $k$.
As for $\ln |x|$, that does not have a derivative at all in the complex sense. The (complex) antiderivative of $1/x$ is $\ln(x) + C$, but this is multi-valued: there is no antiderivative that is analytic on all of ${\mathbb C} \backslash \{0\}$.