Residue at the Point at Infinity
Cauchy Integral Formula Generalization
Let a point z_0 be an isolated singularity of a single-valued function f(z). In a punctured neighborhood of this point, f(z) admits a unique expansion into a Laurent series:
f(z) = \sum_{n=-\infty}^\infty c_n \left(z - z_0 \right)^n
The coefficients c_n are determined by the integral formula:
c_n = \frac{1}{2\pi i} \oint_{C} \frac{f(\zeta)}{(\zeta - z_0)^{n+1}} \mathrm d\zeta, \quad n \in \mathbb{Z}
where C is any simple closed contour in the annulus of convergence, oriented counter-clockwise, that encloses z_0. A coefficient of particular significance is c_{-1}, which is defined as the residue of the function f(z) at the singular point z_0.
The residue is formally defined by the integral:
\operatorname{Res} [f(z), z_0] = \frac{1}{2\pi i} \oint_C f(\zeta)\mathrm d\zeta
To see that this integral is indeed equal to c_{-1}, we can integrate the Laurent series term by term. The uniform convergence of the series on the compact contour C justifies the interchange of summation and integration:
\begin{aligned} \oint_C f(\zeta)\mathrm d\zeta &= \oint_C \sum_{n=-\infty}^\infty c_n \left(\zeta - z_0 \right)^n \mathrm d\zeta \\ &= \sum_{n=-\infty}^\infty c_n \oint_C \left(\zeta - z_0 \right)^n \mathrm d\zeta \end{aligned}
From the theory of complex integration, the contour integral of (\zeta - z_0)^n is zero for all integer values of n except for n=-1, for which its value is 2\pi i. Consequently, every term in the series vanishes except for the one corresponding to n=-1:
\oint_C f(\zeta)\mathrm d\zeta = c_{-1} (2\pi i)
This confirms the identity \operatorname{Res} [f(z), z_0] = c_{-1}. If z_0 is a regular point or a removable singularity, the principal part of the Laurent series is zero, meaning c_n=0 for all n < 0. In this situation, the residue is zero. The residue is a non-zero quantity only when z_0 is a pole or an essential singularity. The calculation of residues is a central operation in complex analysis, as it allows the evaluation of complex integrals to be reduced to an algebraic procedure of computing derivatives.
Suppose z_0 is a simple pole (a pole of the first order) of the function f(z). The Laurent series expansion in the vicinity of this point has the form:
f(z) = c_{-1} (z - z_0)^{-1} + \sum_{n=0}^\infty c_n (z-z_0)^n
To isolate the residue c_{-1}, we can multiply the series by the factor (z - z_0):
(z - z_0) f(z) = c_{-1} + \sum_{n=0}^\infty c_n (z-z_0)^{n+1}
Taking the limit as z approaches z_0, all terms in the summation vanish, yielding a direct formula for the residue:
\operatorname{Res} [f(z), z_0] = c_{-1} = \lim_{z \to z_0} (z - z_0) f(z)
A frequent scenario is when f(z) is expressed as a ratio of two analytic functions, f(z) = g(z)/h(z), where g(z_0) \neq 0 and z_0 is a simple zero of h(z), meaning h(z_0) = 0 and h^\prime(z_0) \neq 0. In this case, the calculation of the residue simplifies further. Using the limit formula:
\begin{aligned} \operatorname{Res} [f(z), z_0] &= \lim_{z \to z_0} (z - z_0) \frac{g(z)}{h(z)} \\ &= \lim_{z \to z_0} g(z) \cdot \lim_{z \to z_0} \frac{z - z_0}{h(z) - h(z_0)} \\ &= g(z_0) \cdot \left[ \lim_{z \to z_0} \frac{h(z) - h(z_0)}{z - z_0} \right]^{-1} \end{aligned}
The second limit is the reciprocal of the definition of the derivative of h(z) at z_0. This gives the formula:
\operatorname{Res} \left[\frac{g(z)}{h(z)}, z_0\right] = \frac{g(z_0)}{h^\prime(z_0)}
Now, let z_0 be a pole of order m > 1 for the function f(z). The Laurent series expansion is:
f(z) = \frac{c_{-m}}{(z - z_0)^m} + \dots + \frac{c_{-1}}{z - z_0} + \sum_{n=0}^\infty c_n (z-z_0)^n
To find c_{-1}, we first multiply f(z) by (z - z_0)^m. This operation removes the singularity at z_0 and results in a function, let’s call it g(z), which is analytic at z_0.
g(z) = (z - z_0)^m f(z) = c_{-m} + c_{-m+1}(z-z_0) + \dots + c_{-1}(z-z_0)^{m-1} + \sum_{n=0}^\infty c_n (z-z_0)^{n+m}
The expression on the right is the Taylor series expansion of the analytic function g(z) around the point z_0. The coefficient c_{-1} is the coefficient of the (z-z_0)^{m-1} term in this Taylor series. The standard formula for the coefficients of a Taylor series states that the coefficient of (z-z_0)^k for a function g(z) is given by \frac{1}{k!} \frac{\mathrm d^k g}{\mathrm d z^k}\bigg|_{z=z_0}.
Applying this formula for k=m-1, we identify c_{-1}:
c_{-1} = \frac{1}{(m-1)!} \frac{\mathrm d^{m-1}g(z)}{\mathrm d z^{m-1}}\bigg|_{z=z_0}
Substituting the expression for g(z) back yields the general formula for the residue at a pole of order m:
\operatorname{Res} [f(z), z_0] = \frac{1}{(m-1)!}\lim_{z \to z_0} \frac{\mathrm d^{m-1}}{\mathrm d z^{m-1}} \left[ (z - z_0)^m f(z)\right]
For the case of a simple pole (m=1), it reduces to the previous result.
Theorem: Let a function f(z) be analytic inside and on a simple closed contour C, except for a finite number of isolated singular points z_1, z_2, \dots, z_n located inside C. Then, the integral of f(z) taken counter-clockwise along C is proportional to the sum of the residues of the function at these singular points:
\oint_C f(\zeta) \, \mathrm d \zeta = 2\pi i \sum_{k=1}^{n} \operatorname{Res}[f(z), z_k]
The proof of this theorem can be found here. The utility of this result is that it transforms the often-difficult task of direct integration into the algebraic problem of calculating residues.
The concept of a residue can be extended to the point at infinity. Let z=\infty be an isolated singular point for an analytic function f(z). This implies that f(z) is analytic in an annulus R < |z| < \infty for some sufficiently large R.
The residue of f(z) at the point z = \infty is defined by the integral:
\operatorname{Res}[f(z), \infty] = \frac{1}{2\pi i}\oint_{C^-} f(\zeta) \, \mathrm d \zeta = - \frac{1}{2\pi i}\oint_{C^+} f(\zeta) \, \mathrm d \zeta
The integration contour C is any simple closed curve, such as a circle |z| = \rho > R, outside of which f(z) has no finite singularities. The notation C^- signifies that the contour is traversed in the clockwise (negative) direction. This orientation is chosen so that the unbounded region containing z=\infty is kept to the left, analogous to how the interior is kept to the left for a standard counter-clockwise (C^+) contour.
To relate this definition to the Laurent series of f(z) at infinity, we perform a change of variables z = 1/\zeta. The clockwise path C^- for z becomes a counter-clockwise path \gamma^+ for \zeta.
\operatorname{Res}[f(z), \infty] = -\frac{1}{2\pi i} \oint_{C^+} f(z) \mathrm dz = -\frac{1}{2\pi i} \oint_{\gamma^+} f(1/\zeta) \left(-\frac{1}{\zeta^2}\right) \mathrm d\zeta
The Laurent series for f(1/\zeta) around \zeta=0 is:
f(1/\zeta) = \sum_{n=-\infty}^\infty c_n \zeta^{-n} = \dots + c_{-1}\zeta + c_0 + c_1\zeta^{-1} + \dots
The integral becomes:
\operatorname{Res}[f(z), \infty] = \frac{1}{2\pi i} \oint_{\gamma^+} \frac{1}{\zeta^2} \sum_{n=-\infty}^\infty c_n \zeta^{-n} \mathrm d\zeta = \frac{1}{2\pi i} \oint_{\gamma^+} \sum_{n=-\infty}^\infty c_n \zeta^{-n-2} \mathrm d\zeta
By the term-by-term integration of the series, the only non-zero contribution comes from the term where the power of \zeta is -1. This occurs when -n-2 = -1, which implies n=-1. The coefficient of this term is c_{-1}. Therefore, the integral evaluates to c_{-1}. The calculation shows:
\operatorname{Res}[f(z), \infty] = -c_{-1}
The negative sign is a direct result of the definition. This leads to a distinction: while the residue at a finite removable singularity is always zero (since c_{-1}=0), the residue at a removable singularity at infinity can be non-zero. For instance, for f(z)=1/z, the point z=\infty is a removable singularity. The Laurent series at infinity is f(z) = z^{-1}, so c_{-1}=1. The residue at infinity is \operatorname{Res}[f(z), \infty]=-c_{-1}=-1.
A consequence of this theory is that the sum of all residues of a function over the entire extended complex plane is zero.
Theorem: If a function f(z) is analytic on the extended complex plane except for a finite number of isolated singular points \{z_1, \dots, z_{N-1}\} and z_N = \infty, then the sum of all its residues is zero:
\sum_{k=1}^N \operatorname{Res} [f(z), z_k] = 0
Proof: Let C be a closed contour, oriented counter-clockwise, that encloses all the finite singular points z_1, \dots, z_{N-1}. According to the residue theorem:
\frac{1}{2\pi i}\oint_{C^+} f(\zeta) \, \mathrm d \zeta = \sum_{k=1}^{N-1} \operatorname{Res}[f(z), z_k]
From the definition of the residue at infinity, the same integral is related to \operatorname{Res}[f(z), \infty]:
\operatorname{Res} [f(z), \infty] = - \frac{1}{2\pi i}\oint_{C^+} f(\zeta) \, \mathrm d \zeta
Combining these two equations gives:
\sum_{k=1}^{N-1} \operatorname{Res}[f(z), z_k] = -\operatorname{Res} [f(z), \infty]
Rearranging the terms completes the proof.
The sum of residues theorem provides a method to simplify the evaluation of certain contour integrals. Suppose we need to compute the integral of f(z) along a closed contour C, and there are numerous singularities inside C, making a direct application of the residue theorem laborious. If, however, there are only a few singularities outside C, we can use the sum theorem.
Let \{z_k^{\text{in}}\} be the singularities inside C and \{z_k^{\text{out}}\} be those outside. The theorem states:
\sum \operatorname{Res}[f, z_k^{\text{in}}] + \sum \operatorname{Res}[f, z_k^{\text{out}}] + \operatorname{Res}[f, \infty] = 0
Since \oint_{C^+} f(\zeta) \, \mathrm d \zeta = 2\pi i \sum \operatorname{Res}[f, z_k^{\text{in}}], we can write:
\oint_{C^+} f(\zeta) \, \mathrm d \zeta = -2\pi i \left( \textstyle \sum_{k} \operatorname{Res}[f, z_k^{\text{out}}] + \operatorname{Res} [f(z), \infty] \right)
This formula can be used in generalizing Cauchy’s integral formula to unbounded domains. Let f(z) be analytic everywhere outside a simple closed contour C. Assume z=\infty is a removable singularity of f(z), so that \lim_{z\to \infty} f(z) = f(\infty) exists and is finite. We wish to evaluate the integral of \frac{f(\zeta)}{\zeta-z_0} around C.
Consider the auxiliary function g(z) = \frac{f(z)}{z - z_0}. The singularities of g(z) outside of C can only be at z=z_0 and at z=\infty.
The residue of g(z) at infinity is:
\operatorname{Res}[g(z), \infty] = -c_{-1}^g
where c_{-1}^g is the coefficient of z^{-1} in the Laurent expansion of g(z) for large z. For large z:
\begin{aligned} g(z) &= \frac{1}{z-z_0}f(z) = \frac{1}{z(1-z_0/z)} \left(f(\infty) + \frac{c_{-1}}{z} + \dots \right) \\ &= \frac{1}{z}\left(1+\frac{z_0}{z}+\dots\right)\left(f(\infty) + \frac{c_{-1}}{z} + \dots \right) \\ &= \frac{f(\infty)}{z} + O(z^{-2}) \end{aligned}
The coefficient c_{-1}^g is f(\infty), so \operatorname{Res}[g(z), \infty] = -f(\infty).
Now we analyze two cases based on the position of z_0.
Case 1: z_0 is in the bounded domain interior to C. The function g(z) is analytic everywhere outside C. The only exterior singularity is at infinity. Applying our formula:
\oint_{C^+} \frac{f(\zeta)}{\zeta - z_0} \, \mathrm d \zeta = -2\pi i \operatorname{Res}[g(z), \infty] = -2\pi i(-f(\infty)) = 2\pi i f(\infty)
Case 2: z_0 is in the unbounded domain exterior to C. In this case, g(z) has two singularities outside C: a simple pole at z=z_0 and the point at infinity. The residue at the pole is:
\operatorname{Res}[g(z), z_0] = \lim_{z \to z_0} (z-z_0) \frac{f(z)}{z-z_0} = f(z_0)
The sum of the exterior residues is \operatorname{Res}[g(z), z_0] + \operatorname{Res}[g(z), \infty] = f(z_0) - f(\infty). The integral is:
\oint_{C^+} \frac{f(\zeta)}{\zeta - z_0} \, \mathrm d \zeta = -2\pi i \left( f(z_0) - f(\infty) \right) = 2\pi i \left( f(\infty) - f(z_0) \right)
Combining these results gives a generalized Cauchy integral formula for an exterior domain, where C is traversed counter-clockwise:
\frac{1}{2\pi i}\oint_{C^+} \frac{f(\zeta)}{\zeta - z_0} \, \mathrm d \zeta = \begin{cases} f(\infty) & \text{if } z_0 \text{ is inside } C \\ f(\infty) - f(z_0) & \text{if } z_0 \text{ is outside } C \end{cases}
Certain classes of integrals, which can be challenging to compute with standard real analysis techniques, become quite manageable when transformed into contour integrals in the complex plane.
A common class of definite integrals for which these methods can be used involves rational functions of trigonometric expressions, integrated over a full period.
Consider an integral of the general form:
I = \int_0^{2\pi} f(\cos(\theta), \sin(\theta)) \, \mathrm d\theta
where f is a rational function of its two arguments. The evaluation strategy involves a change of variables that maps this real integral into a contour integral over the unit circle in the complex plane.
Let the complex variable z be defined as z = e^{i \theta}. As the real variable \theta ranges from 0 to 2\pi, the complex variable z traverses the unit circle, |z|=1, in the counter-clockwise direction.
From this substitution, we derive the following expressions for the differential and the trigonometric functions:
\begin{aligned} &\mathrm dz = i e^{i \theta} \mathrm d\theta = i z \mathrm d\theta \\ &\mathrm d\theta = \frac{\mathrm dz}{iz} \end{aligned}
The trigonometric functions \cos(\theta) and \sin(\theta) can be expressed in terms of z using Euler’s formula:
\begin{aligned} & \cos(\theta) = \frac{1}{2}\left(e^{i \theta} + e^{-i \theta}\right) = \frac{1}{2}\left(z + \frac{1}{z}\right) \\ & \sin(\theta) = \frac{1}{2i}\left(e^{i \theta} - e^{-i \theta}\right) = \frac{1}{2i}\left(z - \frac{1}{z}\right) \end{aligned}
Substituting these expressions transforms the original real integral into a complex contour integral:
I = \frac{1}{i} \oint_{|z|=1} F\left(z + \frac{1}{z}, z - \frac{1}{z} \right) \frac{\mathrm dz}{z}
The integrand is now a new function, let’s call it G(z), which is a rational function of z.
Assuming G(z) is analytic within the contour |z|=1 except for a finite number of poles, the integral can be evaluated using the residue theorem:
I = 2\pi i \sum_{k} \operatorname{Res}[G(z), z_k]
where the sum is taken over all poles z_k of G(z) that are located inside the unit circle, |z_k| < 1. If a pole z_k is of order \beta_k, its residue is calculated by the general formula:
\operatorname{Res}[G(z), z_k] = \frac{1}{(\beta_k-1)!} \lim_{z \to z_k} \frac{\mathrm d^{\beta_k-1}}{\mathrm d z^{\beta_k-1}} \left[ (z - z_k)^{\beta_k} G(z) \right]
For example, let’s evaluate the integral:
I = \int_0^{2\pi} \frac{\mathrm d\theta}{1 + a \, \cos(\theta)}, \quad |a|< 1
Applying the substitution z = e^{i\theta}:
\begin{aligned} I &= \oint_{|z|=1} \frac{1}{1 + a \left(\frac{z + z^{-1}}{2}\right)} \frac{\mathrm dz}{iz} \\ &= \oint_{|z|=1} \frac{1}{1 + \frac{a}{2z}(z^2 + 1)} \frac{\mathrm dz}{iz} \\ &= \frac{1}{i} \oint_{|z|=1} \frac{2z}{2z + a(z^2 + 1)} \frac{\mathrm dz}{z} \\ &= \frac{2}{i} \oint_{|z|=1} \frac{1}{az^2 + 2z + a} \mathrm dz \end{aligned}
The integrand has singularities where the denominator is zero. The roots of the quadratic equation az^2 + 2z + a = 0 are:
z_{1,2} = \frac{-2 \pm \sqrt{4 - 4a^2}}{2a} = \frac{-1 \pm \sqrt{1 - a^2}}{a}
These are two simple poles. Let’s analyze their location. The product of the roots is z_1 z_2 = a/a = 1. Since |z_1||z_2| = 1 and the roots are distinct for |a|<1, one root must lie inside the unit circle and the other outside. Let us identify the root with modulus less than one:
z_1 = \frac{-1 + \sqrt{1 - a^2}}{a}
For |a|<1, the magnitude of z_1 is less than 1. The integral’s value is determined by the residue at this pole. Since it is a simple pole of the form P(z)/Q(z), the residue is P(z_1)/Q^\prime(z_1). Here, P(z) = 1 and Q(z) = az^2 + 2z + a, so Q^\prime(z) = 2az + 2.
Applying the residue theorem, we obtain the final result for the integral:
\begin{aligned} I &= \frac{2}{i} 2\pi i \operatorname{Res}\left[\frac{1}{az^2+ 2z + a}, z_1 \right] =4\pi \frac{1}{2az_1 + 2} \\ & = 4\pi\frac{1}{2a\left(\frac{-1 + \sqrt{1 - a^2}}{a}\right) + 2} \\ & = 4\pi\frac{1}{-2 + 2\sqrt{1-a^2} + 2} = 4\pi\frac{1}{2\sqrt{1-a^2}} \\ & = \frac{2\pi}{\sqrt{1-a^2}} \end{aligned}