# How integrating over a branch cut is made rigorous?

This is from Ch. 7 of the book Complex Variables by J Brown and R Churchill 8th ed.

In evaluation of the counter integration of $$f(z)=\dfrac{z^{-a}}{z+1}$$ the book first suggests the following :

where $$\mathbb{R^+} \cup {\{0}\}$$ is the branch cut. Then it claim that integration over the branch cut is not allowed however we will make it rigorousin the exercise 8. Here it is:

Well, the part (c) is where it's trying to make it 'rigorous' (part (a) and (b) are easy to understand).

My questions:

1. I can't understand part (c) about change of branches. In other words, how it integrates in other branches then sums them in another branch?! (esp. when still the legs from $$\rho$$ to $$R$$ both again relies over the branch cut $$\mathbb{R^+} \cup {\{0}\}$$).

2. When in the text I read that it is going to become a rigorous treating, I was thinking about letting the branch cut $$\mathbb{R^+} \cup {\{0}\}$$ be, and limiting the legs to that branch cut (ie. limiting $$\theta_1$$ and $$\theta_2$$ to zero) like this :

I tried it but I am not sure if the method is consistent with theorems. Also I don't know if I can enter the limit $$\theta_i \to 0$$ inside integral. So how can I accomplish this method if it is allowed or is there any better (more acceptable) method alternative for the one presented in the exercise above?

Edit: For question 2: How to prove that the following holds $$\lim_{\theta_1, \theta_2 \to 0} \int_{\rho}^{R} r^{-a} \Big( \dfrac{e^{-ai\theta_1}}{re^{\theta_1}+1} - \dfrac{e^{-2 \pi ai + ia\theta_2}}{re^{\theta_2}+1} \Big) dr = (1-e^{-2 \pi ai}) \int_{\rho}^{R} \dfrac{r^{-a}}{r+1} dr.$$

and can this be considered a rigorous-ization?

• When $f$ has multiple branches, in $\int_\gamma f(z)dz$ we mean choosing a branch of $f$ analytic around $\gamma(0)$ then continuing $f$ analytically along $\gamma(t),t : 0\to 1$. This way it is very possible that $\gamma(0) = \gamma(1)$ but $f(\gamma(0)) \ne f(\gamma(1))$ Oct 29, 2018 at 20:29
• @reuns, I read your comment several times, unfortunately I didn't understand that at all!
– user231343
Oct 29, 2018 at 22:26
• Let $\gamma_1(t) = e^{2i \pi t}, t \in (0,1)$ and $\gamma_2(t) = e^{4i \pi t}, t \in (0,1)$ then what do you get for $\int_{\gamma_1} z^{1/2}dz$ and $\int_{\gamma_2} z^{1/2}dz$ ? Oct 29, 2018 at 22:29
• Let $f(z) = z^{1/2}, f(1) = 1$ then $\int_{\gamma_2} z^{1/2}f(z)dz = \int_0^1 f(\gamma_2(t))\gamma_2'(t)dt$ $= \int_0^1 f(e^{4i \pi t}) 4i \pi e^{4i \pi t}dt = \int_0^1 (e^{4i \pi t})^{1/2} 4i \pi e^{4i \pi t}dt$ $=\int_0^1 e^{2i \pi t} 4i \pi e^{4i \pi t}dt = 0$. Do you see why I set $f(e^{4 i \pi t}) = (e^{4i \pi t})^{1/2} = e^{2i \pi t}$ and why it is not consistent with the idea that $f$ is function $\mathbb{C} \to \mathbb{C}$ ? Oct 29, 2018 at 23:39
• When rotating clockwise around $0$, then (again by analytic continuation) $\log z$ becomes $2i\pi+\log z$, right ? Here it is the same with $g_0(z) = \frac{z^a}{1+z}$ which becomes $g_1(z) = e^{2i \pi a} \frac{z^a}{1+z}$ so that $\int_0^\infty g_1(z)dz = e^{2i \pi a} \int_0^\infty g_0(z)dz$. Nov 2, 2018 at 15:44

We require to prove rigorously the equation (3) used in section 84 of the book, for which the author initially only gives an intuitive derivation, based on a limiting case as the branch cut is approached.

Equation (3) states :- $$\int_{\rho}^{R} \frac{r^{-a}}{r + 1} dr + \int_{C_{R}} f(z) dz - \int_{\rho}^{R} \frac{r^{-a}e^{-i2a\pi}}{r + 1} dr + \int_{C_{R}} f(z) dz = 2\pi i \: \mathrm{Res}(f, -1)$$

Firstly it is helpful to clarify the potentially confusing topic of the multi-valued logarithm and complex exponent functions and define some notation that might be helpful :-

Given $$\alpha \in \mathbb{R}$$ :-

• $$\arg_{\alpha} : \mathbb{C} \setminus \{0\} \rightarrow (\alpha - 2\pi, \alpha]$$ is the '$$\alpha$$-branch' of the argument function, $$\alpha = \pi$$ giving the Principal Argument function $$\mathrm{Arg} : \mathbb{C} \setminus \{0\} \rightarrow (-\pi, \pi]$$.

• $$\exp : S_{\alpha} \rightarrow \mathbb{C} \setminus \{0\}$$ is the bijective restriction of the exponential function to the strip $$S_{\alpha} = \{z \in \mathbb{C} : \alpha - 2\pi < \mathrm{Im}z \le \alpha \}$$.

• $$\log_{\alpha} : \mathbb{C} \setminus \{0\} \rightarrow S_{\alpha}$$ is the inverse of the latter bijective function. This is the '$$\alpha$$-branch' of the complex $$\log$$ function.

• $$R_{\alpha}$$ is the ray $$\{re^{i\alpha} : r \geq 0\}$$

• $$C_{\alpha}$$ is the 'cut-plane' $$\mathbb{C} \setminus R_{\alpha}$$.

We then have $$\log_{\alpha}z = \ln |z| + i \arg_{\alpha}z, \forall z \in \mathbb{C} \setminus \{0\}$$, with $$\log_{\alpha}$$ differentiable on the open region $$C_{\alpha}$$ (with derivative $$1/z$$), but discontinuous at every point of the cut $$R_{\alpha} \setminus \{0\}$$.

The general complex exponent function $$z^{a}$$ for $$a \in \mathbb{C}$$ and $$z \in \mathbb{C} \setminus \{0\}$$ is defined as $$z^{a} = \exp (a \log_{\alpha} z)$$ for every $$\alpha \in \mathbb{R}$$, analogously to the real case. This has $$\alpha$$-branches in the same way as the complex $$\log$$ function. For a given $$a$$ and $$z$$ there will generally be multiple values, $$\alpha = \pi$$ giving the Principle Value. This function reduces back to the familiar cases for exponential functions already covered, such as $$e^{z}$$, and integer and rational and real exponents.

The author uses a slightly different terminology than the above, calling the ray $$R_{\alpha}$$ a 'branch cut', and log function on the open region $$C_{\alpha}$$ a 'branch' - such a branch is then analytic. But the log function is defined on a larger domain than $$C_{\alpha}$$, ie $$\mathbb{C} \setminus \{0\}$$ but it has discontinuities in that larger domain, ie all along $$R_{\alpha}$$. Where the author says 'since $$f(z)$$ is not analytic, or even defined, on the branch cut involved' on pg 285 it is thus slightly confusing - but what he is refering to is the analytic restriction of $$f$$ (and of $$\log$$). However we do need to consider $$\log$$ on its full domain $$\mathbb{C} \setminus \{0\}$$ because the paths of integration $$C_{R}$$ and $$C_{\rho}$$ cross over the 'branch cut' (ie. the positive x-axis). Thus in integrating $$f$$ along these contours we are integrating a piecewise continuous function (which the author discusses on pg 127). Such functions have finitely many discontinuities, and these discontinuities do not prevent the integral from being well-defined - similarly to how a standard Reimann integral in $$\mathbb{R}$$ is unaffected by changing the value of the integrand arbitrarily at finitely many points of its domain.

The term 'multiple-valued function' can cause some confusion as by definition a function can only have one value for each point in its domain. It is perhaps better described as a 'family of functions'. So in practice we only deal with a single instance in this 'family' at a time - ie we select some value of $$\alpha$$ to work with. When we select $$\alpha = \pi$$ we are dealing with the Principle Value.

Proof (1) of Equation (3) - Using the Author's Method in Ex. 8

In the case of 8(a), $$f_{1}$$ is defined using the branch of $$z^{-a}$$ with $$\alpha = 3\pi/2$$ (this $$\alpha$$ is marked with a dotted line in Fig 104), so that this $$f_{1}$$ is analytic on the open region $$C_{3\pi/2}$$ (apart from the pole at $$z = -1$$). The closed contour on the LHS of Fig 104 lies wholly within $$C_{3\pi/2}$$, so the Residue Theorem applies.

With 8(b) we have $$\alpha = 5\pi/2$$ (again marked with a dotted line in Fig 104) and $$f_{2}$$ is analytic on the open region $$C_{5\pi/2}$$ (= $$C_{\pi/2}$$) (apart from the pole at $$z = -1$$). The closed contour on the RHS of Fig 104 lies wholly within $$C_{5\pi/2}$$, and does not wind around any point outside of it, and it does not wind around the pole - and we can easily exclude the area where the pole is located (and thus have a fully analytic function - rather than just a Meromorphic function as in 8(a)), so the Cauchy-Goursat Theorem applies.

The function $$f$$ of 8(c) is defined using the branch of $$z^{-a}$$ with $$\alpha = 2\pi$$, so that this $$f$$ is analytic on the open region $$C_{2\pi}$$. We thus cannot apply either of the above two theorems to $$f$$, since the above contours are not confined to this region $$C_{2\pi}$$ where $$f$$ is analytic apart from its pole.

To summarize :- $$\begin{eqnarray*} f_{1}(z) & = & \mathrm{exp}(-a \log_{3\pi/2}z) / (z + 1) \\ f_{2}(z) & = & \mathrm{exp}(-a \log_{5\pi/2}z) / (z + 1) \\ f(z) & = & \mathrm{exp}(-a \log_{2\pi}z) / (z + 1) \\ \end{eqnarray*}$$

To answer the question 8(c) firstly consider part 8(a). On $$\Gamma_{R}$$ we have $$\arg_{3\pi/2} = \arg_{2\pi}$$ (trace around $$\Gamma_{R}$$ and compare these two arguments at each point) - with the exception of the point $$(R, 0)$$, where $$f$$ has its discontinuity, and the two arguments are $$0, 2\pi$$ respectively. This means $$f$$ and $$f_{1}$$ are equal on $$\Gamma_{R}$$ apart from at this one point. Because the value of a contour integral remains the same if the integrand function is changed arbitrarily at finitely many points (as with a standard Riemann integral in $$\mathbb{R}$$), we then have $$\int_{\Gamma_{R}} f_{1} = \int_{\Gamma_{R}} f$$. By similar argument we also have $$\int_{\Gamma_{\rho}} f_{1} = \int_{\Gamma_{\rho}} f$$. In the case of the segment $$L$$ it is wholly in the region where $$\arg_{3\pi/2} = \arg_{2\pi}$$ so $$f$$ and $$f_{1}$$ are equal all along it, and so $$\int_{L} f_{1} = \int_{L} f$$. And finally in a neighborhood of $$z = -1$$ we again have $$\arg_{3\pi/2} = \arg_{2\pi}$$, so $$f = f_{1}$$ in that neighborhood, and therefore $$\mathrm{Res}(f, -1) = \mathrm{Res}(f_{1}, -1)$$. It follows then that the entire equation in 8(a) remains valid if we replace every instance of $$f_{1}$$ with $$f$$.

Now consider 8(b). On each of $$\gamma_{\rho}$$, $$\gamma_{R}$$, and $$-L$$ we can check that $$\arg_{5\pi/2} = \arg_{2\pi}$$. This time there are no exception points and this equality holds all along these three curves, and hence $$f$$ and $$f_{2}$$ are equal along all of them. (Also note neither $$f$$ nor $$f_{2}$$ have any discontinuities along these curves). Thus the equation 8(b) remains valid if we replace every instance of $$f_{2}$$ with $$f$$.

We can then proceed to add 8(a) and 8(b) to arrive at equation (3).

Proof (2) of Equation (3) - Using Proposed Method in OP's Qu 2

Let $$f(z) = z^{-a}/(z + 1)$$ be defined by taking the $$\alpha = 2\pi$$ branch of $$\log_{\alpha}$$, so $$f(z) = \mathrm{exp}(-a \log_{2\pi}z) / (z + 1)$$, with $$\log_{2\pi}z = \ln |z| + i \arg_{2\pi} z$$, and $$\arg_{2\pi} z \in (0, 2\pi]$$. Again, $$z^{-a}$$ is defined on the whole of $$\mathbb{C} \setminus \{0\}$$, is differentiable on cut plane $$C_{2\pi}$$, and discontinuous at every point of $$R_{2\pi}$$ (= positive $$x$$-axis). $$f$$ has similar properties but with the exclusion of the point $$z_{0} = -1$$, where it has a simple pole with residue $$e^{-ia\pi}$$.

The Residue Theorem then tells us that for the closed contour $$\gamma$$ in the diagram below, which winds once ($$+1$$) around the pole $$z_{0} = -1$$, we have $$\int_{\gamma} f = 2 \pi i \: \mathrm{Res}(f, z_{0})$$, and this is applicable for every $$\phi \in (0, \pi)$$.

Thus taking the limit as $$\phi \rightarrow 0^{+}$$ we have :-

$$\lim_{\phi \rightarrow 0^{+}} \int_{\gamma_{1}} \! f + \lim_{\phi \rightarrow 0^{+}} \int_{\gamma_{2}} \! f + \lim_{\phi \rightarrow 0^{+}} \int_{\gamma_{3}} \! f + \lim_{\phi \rightarrow 0^{+}} \int_{\gamma_{4}} \! f = 2 \pi i \: \mathrm{Res}(f, z_{0})$$

provided each of these four limits exist (which we show below). In the following we make use of these two theorems :-

(A) (The Estimation Lemma) If $$f : D \rightarrow \mathbb{C}$$ ($$D \subseteq \mathbb{C}$$) is continuous and $$\gamma$$ a contour in $$D$$ of length $$L$$ and $$M$$ an upper bound for $$|f|$$ on $$\gamma$$ then $$|\int_{\gamma} f| \le ML$$. [eg see Complex Analysis by Stewart & Tall, 1983, pg 111].

(B) (Integrals Depending On A Parameter). If $$g : [a, b] \times [c, d] \rightarrow \mathbb{R}$$ is continuous then $$\int_{a}^{b} g(x, t) dx$$ is continuous as a function of $$t \in [c, d]$$. [eg see Elements of Real Analysis by Bartle, 1964, pg 306].

Theorem (B) readily generalizes to functions $$g$$ taking values in $$\mathbb{R}^{n}$$ and is therefore applicable to the present case of $$\mathbb{C}$$. This is the theorem that allows us to 'crash' the limit through the integral sign.

Because $$C_{R}$$ is compact (ie closed and bounded) any continuous function on it is bounded. But note $$f$$ is discontinuous at the point $$(R, 0)$$ of $$C_{R}$$ because that point lies on the branch cut $$R_{2\pi}$$. However this is the only point of discontinuity of $$f$$ on $$C_{R}$$ and we can temporarily re-define its value there and split it into two separate continuous functions on the upper and lower half, which are then bounded. Thus there is a bound $$M_{1}$$ for $$|f|$$ on $$C_{R}$$ - and likewise a bound $$M_{2}$$ for $$|f|$$ on $$C_{\rho}$$. So applying theorem (A) we have :-

$$\left|\int_{\gamma_{2}} f - \int_{C_{R}} f \right| = \left|\int_{\gamma_{6}} f \right| \le M_{1} \cdot 2\phi R \;\; \rightarrow 0 \mbox{ as } \phi \rightarrow 0^{+},$$

$$\left|\int_{\gamma_{4}} f - \int_{C_{\rho}} f \right| = \left|\int_{\gamma_{5}} f \right| \le M_{2} \cdot 2\phi \rho \;\; \rightarrow 0 \mbox{ as } \phi \rightarrow 0^{+}.$$

Thus $$lim_{\phi \rightarrow 0^{+}}\int_{\gamma_{2}}f = \int_{C_{R}}f$$ and $$lim_{\phi \rightarrow 0^{+}}\int_{\gamma_{4}}f = \int_{C_{\rho}}f$$.

On $$\gamma_{1}$$, $$z = t e^{i\phi}$$, with $$t \in [\rho, R]$$, so $$\log_{2\pi}z = \ln t + i\phi$$, thus $$z^{-a} = \mathrm{exp}(-a \ln t - ai\phi) = t^{-a}e^{-ai\phi}$$. Also, $$\gamma_{1}'(t) = e^{i\phi}$$, so $$\begin{eqnarray*} \int_{\gamma_{1}} f & = & \int_{\rho}^{R} \frac{ t^{-a} e^{-ai\phi} }{ te^{i\phi} + 1 } e^{i\phi} dt \\ & = & e^{i\phi (1 - a)} \int_{\rho}^{R} \frac{ t^{-a} }{1 + te^{i\phi}} dt. \\ \end{eqnarray*}$$

Although we have only defined $$\gamma$$ for $$\phi \in (0, \pi)$$ this last expression is well defined for $$\phi = 0$$ also, and the integrand is continuous as a function of $$(t, \phi)$$, and thus we can apply theorem (B) above to obtain :-

$$\lim_{\phi \rightarrow 0^{+}} \int_{\gamma_{1}} f = \int_{\rho}^{R} \frac{t^{-a}}{1 + t} dt.$$

Similarly on $$\gamma_{3}$$, $$z = te^{i(2\pi - \phi)}$$, and $$\gamma_{3}'(t) = e^{i(2\pi - \phi)}$$, and $$\log_{2\pi}z = \ln t + i(2\pi - \phi)$$, so $$z^{-a} = \mathrm{exp}( -a \ln t - a i (2\pi - \phi) ) = t^{-a}e^{-ai (2\pi - \phi)}$$, thus :-

$$\begin{eqnarray*} \int_{\gamma_{3}} f & = & -\int_{\rho}^{R} \frac{ t^{-a} e^{-ai(2\pi - \phi)} }{ te^{i(2\pi - \phi)} + 1 } e^{i(2\pi - \phi)} dt \\ & = & e^{i(2\pi - \phi) (1 - a)} \int_{\rho}^{R} \frac{ t^{-a} }{1 + te^{i(2\pi - \phi)}} dt \\ & \rightarrow & -e^{-2\pi ia} \int_{\rho}^{R}\frac{t^{-a}}{1 + t} dt \mbox{ as } \phi \rightarrow 0^{+}, \mbox{ by theorem (B) }. \\ \end{eqnarray*}$$

Then adding these 4 limits over $$\gamma_{1}, \gamma_{2}, \gamma_{3}, \gamma_{4}$$ we obtain the required equation (3).

• Quite a detailed answer. Thank you for the time you took to write it all up :) Nov 3, 2018 at 7:31
• I am reading through your answer... 1) about the proof by book's method: You showed $f=f_i$ for all but the line joining ρ and R where is the source of the problem! ; 2) about the alternative proof in Q.2: I took the angles $\phi_1$ and $\phi_2$ different before limiting; can we choose both to be equal? (in evaluating infinite integrals the book introduced P.V. and showed that choosing both upper and lower bounds results differently than non-equal case unless f is even so when there are two limits involved I choose them non-equal, though it's in a different topic).
– user231343
Nov 3, 2018 at 15:55
• First part is not complete at all. For 2nd one just enough to show uniform continuity.
– user200918
Nov 4, 2018 at 1:25
• @Edi - For Proof (1) you don't need to show $f = f_{1}$ along $[\rho, R]$ for 8(a) because the term in 8(a) that integrates along $[\rho, R]$ does not involve $f_{1}$. All of the other 4 terms in 8(a) we do consider though. We find there is equality of $f$ and $f_{1}$ everywhere required except at $(\rho, 0)$ and $(R, 0)$ - but this is no problem as these single points do not affect the value of the integral. Likewise you do not need to show $f = f_{2}$ along $[\rho, R]$ for 8(b) because the term in 8(b) that integrates along $[\rho, R]$ does not involve $f_{2}$. I think perhaps there (cont.) Nov 4, 2018 at 16:05
• is some confusion between the diagrams in Fig 104 and the equations 8(a) and 8(b) which come from those diagrams. For Proof (2) you can just use a single $\phi$. The kind of problem with improper integrals and the PV that you mention does not occur here so long as when you write $\lim_{\phi \rightarrow 0^{+}} \int_{\gamma_{1}} \! f + \lim_{\phi \rightarrow 0^{+}} \int_{\gamma_{2}} \! f + \lim_{\phi \rightarrow 0^{+}} \int_{\gamma_{3}} \! f + \lim_{\phi \rightarrow 0^{+}} \int_{\gamma_{4}} \! f = \lim_{\phi \rightarrow 0^{+}} \int_{\gamma} f$ you are sure each (cont.) Nov 4, 2018 at 16:05