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 :

enter image description here

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:

enter image description here

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 :

enter image description here

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?

  • 2
    $\begingroup$ 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 $f(\gamma(0)) \ne f(\gamma(1))$ $\endgroup$
    – reuns
    Oct 29, 2018 at 20:29
  • $\begingroup$ @reuns, I read your comment several times, unfortunately I didn't understand that at all! $\endgroup$
    – user231343
    Oct 29, 2018 at 22:26
  • 1
    $\begingroup$ 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 $\int_{\gamma_2} z^{1/2}dz$ ? $\endgroup$
    – reuns
    Oct 29, 2018 at 22:29
  • 1
    $\begingroup$ 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 $\mathbb{C} \to \mathbb{C}$ ? $\endgroup$
    – reuns
    Oct 29, 2018 at 23:39
  • 1
    $\begingroup$ 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 $\int_0^\infty g_1(z)dz = e^{2i \pi a} \int_0^\infty g_0(z)dz$. $\endgroup$
    – reuns
    Nov 2, 2018 at 15:44

1 Answer 1


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 $\phi \in (0, \pi)$.

Diagram for Proof (2) Diagram for Proof (2)

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).

  • 2
    $\begingroup$ Quite a detailed answer. Thank you for the time you took to write it all up :) $\endgroup$ Nov 3, 2018 at 7:31
  • 1
    $\begingroup$ 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). $\endgroup$
    – user231343
    Nov 3, 2018 at 15:55
  • $\begingroup$ First part is not complete at all. For 2nd one just enough to show uniform continuity. $\endgroup$
    – user200918
    Nov 4, 2018 at 1:25
  • $\begingroup$ @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.) $\endgroup$ Nov 4, 2018 at 16:05
  • $\begingroup$ 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.) $\endgroup$ Nov 4, 2018 at 16:05

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