# Does $\int \frac{1}{u} du = \ln|u| + C$ also work when $u$ is complex?

Does $\int \frac{1}{u} du = \ln|u| + C$ also work when $u$ is complex? I was taught this in calculus but I'm not sure if it generalizes to complex variables.

Thank you!

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Well, the absolute value in the right should not be there if you're considering complex values... – J. M. Apr 20 '12 at 5:49

The logarithm in the complex plane $\Bbb C$ is a funny thing:

$\hskip 2.3in$

If you extend a ray outwards from the origin (here the ray is placed on the negative real axis), and begin traversing a contour around $0$ counterclockwise, by the time you get back to where you started, the value of the logarithm will have a surplus of $2\pi i$. That is, its value when you get back is the same as the original value it was but with $2\pi i$ added to it!

Why is this? You might recall the formula $e^{\pi i}+1=0$, which entails $e^{2\pi i}=1$, and thus for any integer $n$, $\exp(b)=\exp(b+2\pi in)$. We see from this that the inverse of the exponential function is multi-valued. How do we get around this so that $\log$ is single-valued and therefore a bona fide function? We define a branch cut: as before, we cut out a ray of the complex plane and stop the logarithm from getting that surplus when we cross the line. (The axis shown in the picture is the standard choice of branch cut for logarithms.)

Clearly $\log$ is not continuous on this ray, or the origin itself, but it is still holomorphic (complex-differentiable) everywhere else. I should also point out that the existence of a complex derivative is much stronger than that of a real derivative, because in the former the limit must exist for any path in $\Bbb C$ that $\epsilon$ can take to tend towards $0$, whereas in the latter it need only exist for $\epsilon\to0$ from the left and right. (The $\epsilon$ and limit I refer to are in the definition of the derivative.)

It turns out that

$$\frac{d}{du} \log u = \frac{1}{u}$$

is true on $\Bbb C$ minus the origin and branch cut (the inverse function theorem applies), which means that the negative side of the real axis is out of luck. However, one can observe that antiderivatives are only unique up to a constant, and that with our branch cut

$$\log(-x)=\pi i +\log x$$

for $x>0$ real. Thus $\log u$ and $\log |u|$ differ by $0$ on the positive reals and differ by $\pi i$ (which is a constant) on the negative reals, as we might expect. (More generally, primitives may differ by functions that are constant on the connected components of the integrand's domain, but not necessarily the whole domain. Here the domain of $1/x$ has two components: $(-\infty,0)$ and $(0,\infty)$.)

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$^\dagger$Technically left and right derivatives can exist and be unequal, so their existence does not suffice for the existence of "the" derivative. – anon Apr 20 '12 at 6:07
how to draw this picture?which software do you use? – noname1014 Apr 20 '12 at 6:16
@Tao Hong: The image was nabbed from Wikipedia, but Homann actually made it using Mathematica. See here. – anon Apr 20 '12 at 6:19