# Splitting Integrals with Complex Limits

We all know from school, that $$\int_a^b f(x) dx=F(b)-F(a),$$ with $F(x)$ being the antiderivative of $f(x)$. It is also obvious, if we deal with real $a$,$b$ and $f(x)$, that we can split in the following way: $$\int_a^b f(x) dx=\int_a^e f(x) dx+\int_e^b f(x) dx=F(e)-F(a)+F(b)-F(e),$$ with $a\leq e< b$.

What is necessary, that this is true for comlex limits?

I got so far that I need Cauchy's Integral Theorem. There I found that:

"One important consequence of the theorem is that path integrals of holomorphic functions on simply connected domains can be computed in a manner familiar from the fundamental theorem of real calculus".

Because I'm specially interested in the definite integral of $\text{li}(a+ib)=\int_0^{a+ib}\frac{1}{\ln u}du$, I continue with $f(u)=\frac{1}{\ln u}$.

I set $z=re^{i\phi}$ and get $$f(u)=\frac{1}{\ln r +i\phi}=\frac{\ln r -i\phi}{(\ln r)^2 -\phi^2}=u(r,\phi)+iv(r,\phi)$$ To check for holomorphy in polar coordinates if have to check: $${ \partial u \over \partial r }=\frac{(\ln r)^2+\phi^2}{r((\ln r)^2-\phi^2)^2}={1 \over r}{ \partial v \over \partial \theta}, \quad{ \partial v \over \partial r }=-\frac{2\phi \ln(r)}{r((\ln r)^2-\phi^2)^2} =-{1 \over r}{ \partial u \over \partial \theta}.$$ I used WolframAlpha to check that (links didn't work, sorry).

So everythings seems fine and if there's nobody out there screaming "STOP, don't do this because of ..." I split it!

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What "we all know from school" depends on us all intuiting a certain relation between $f$ and $F$, which relation would be better stated explicitly. –  Gerry Myerson Jan 17 '12 at 23:22

Caution: integrals in the complex plane are not "from $a$ to $b$", they are "on a particular path from $a$ to $b$". If the domain where the function is analytic is not simply connected, the answer will depend on which path you choose.

In this case, $f(z) = \frac{1}{\ln z}$ has a branch cut, which you can take to be on the negative real axis, but also a simple pole at $z=1$. An antiderivative $F(z)$ of this function will not be analytic on the whole region where $f(z)$ is analytic: it will have another branch cut, which may be taken to be the interval $(0,1)$ on the real axis. So you can define $\text{li}(x)$ to be analytic on the complement of the ray $(-\infty, 1]$, and it will satisfy $\int_C \frac{dz}{\ln(z)} = \text{li}(b) - \text{li}(a)$ for any directed contour $C$ that starts at $a$ and ends at $b$ and avoids $(-\infty, 1]$.

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Thx a lot. For me this was an eye opener!!! –  draks ... Jan 18 '12 at 13:16

I'll add to Robert's excellent advise this one : avoid evaluating expressions like $\operatorname{li}(e^{x+iy})$ or $\operatorname{li}(z^{\rho})$ with complex ${\rho}$ because the exact phase will be lost!
Computer evaluation of the exponential will lose the phase information since $e^z=e^{z+2 k\pi i}$ and the logarithm will return the 'principal branch' of the result so that $\ln(e^{x+iy})$ becomes $x+iy'$ with $y'-y = 2 k\pi$.
This is a classical problem when using $\zeta$ zeros in the Riemann's explicit formula for example.

To avoid losing $2 k\pi i$ terms it may be convenient to replace your $\operatorname{li}$ function by the exponential integral function $\operatorname{Ei}$ using the relation $\operatorname{li}(z)= \operatorname{Ei}(\ln z)$ or more specifically $\operatorname{li}(e^{x+iy})= \operatorname{Ei}(x+iy)$.

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There is an analog. Suppose that $\alpha, \beta:[0,1]\rightarrow \mathbb{C}$ are paths with $\alpha(1) = \beta(0)$. Define the path $$(\alpha + \beta)(t) = \left\{ \begin{array}{lc} \alpha(2t) & 0 \le t\le 1/2\\ \beta(2t - 1) & 1/2\le t \le 1. \end{array} \right.$$ Then if $f$ is holomorphic on a neighborhood of these paths, $$\int_{\alpha + \beta} f = \int_\alpha f + \int_\beta f.$$ This is just a "concatenation" of the two paths.

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