Residue Theorem

The residue theorem converts many closed contour integrals into a finite sum of local coefficients at isolated singularities. It is the engine behind standard contour evaluations of Fourier integrals, Green functions, response functions, and scattering-amplitude pole contributions.

This page is the canonical home for the theorem and for practical residue computation. For contour deformation, large-arc estimates, and how to choose a contour, see Contour Integration.

Statement

Let $f$ be meromorphic on a region containing a positively oriented simple closed contour $C$ and its interior. Suppose $f$ has isolated poles $z_1,\ldots,z_N$ inside $C$ and no poles on $C$. Then

$$\oint_C f(z)\,dz = 2\pi i \sum_{j=1}^{N} \operatorname{Res}_{z=z_j} f.$$

The orientation matters. If $C$ is clockwise, the right-hand side acquires a minus sign.

The hypotheses matter too:

• poles must be isolated;
• singularities must not lie on the contour unless a prescription is supplied;
• branch cuts require additional contour pieces;
• large arcs in open-contour arguments must be shown to vanish.

What a Residue Is

If $f$ has an isolated singularity at $z_0$, its Laurent expansion has the form

$$f(z) = \sum_{n=-\infty}^{\infty} a_n(z-z_0)^n$$

in an annulus around $z_0$. The residue is the coefficient of $(z-z_0)^{-1}$:

$$\operatorname{Res}_{z=z_0}f = a_{-1}.$$

This coefficient is special because

$$\oint (z-z_0)^n\,dz =0 \quad \text{for } n\ne -1,$$

while

$$\oint \frac{dz}{z-z_0} = 2\pi i$$

for a small positively oriented circle around $z_0$.

Computing Residues

For a simple pole at $z_0$,

$$\operatorname{Res}_{z=z_0} f = \lim_{z\to z_0} (z-z_0)f(z).$$

If

$$f(z)=\frac{p(z)}{q(z)},$$

where $p$ and $q$ are analytic, $q(z_0)=0$, and $q'(z_0)\ne0$, then

$$\operatorname{Res}_{z=z_0}f = \frac{p(z_0)}{q'(z_0)}.$$

For a pole of order $m$,

$$\operatorname{Res}_{z=z_0}f = \frac{1}{(m-1)!} \lim_{z\to z_0} \frac{d^{m-1}}{dz^{m-1}} \left[ (z-z_0)^m f(z) \right].$$

In practice, simple-pole formulas and Laurent expansions handle most quantum-mechanics contour integrals. Higher-order pole formulas are useful but easy to misuse; check the order of the pole before differentiating.

Example: A Simple Pole

Let

$$f(z)=\frac{e^z}{z-a}.$$

At $z=a$, the residue is

$$\operatorname{Res}_{z=a}f = \lim_{z\to a} (z-a)\frac{e^z}{z-a} = e^a.$$

Therefore, for a positively oriented contour $C$ enclosing $a$ and no other singularities,

$$\oint_C \frac{e^z}{z-a}\,dz = 2\pi i\,e^a.$$

If $a$ lies outside $C$, the integral is zero because the integrand is analytic inside and on the contour.

Example: A Real Integral

For $a>0$, evaluate

$$I = \int_{-\infty}^{\infty} \frac{dx}{x^2+a^2}.$$

Consider

$$f(z)=\frac{1}{z^2+a^2} = \frac{1}{(z-ia)(z+ia)}.$$

Close the real line by a large semicircle in the upper half-plane. The large arc vanishes because $f(z)$ decays like $1/z^2$. The only pole inside is $z=ia$, and

$$\operatorname{Res}_{z=ia}f = \frac{1}{2ia}.$$

Thus

$$I = 2\pi i \frac{1}{2ia} = \frac{\pi}{a}.$$

This example is a model for more physical Fourier and Green-function integrals: choose the contour, justify the arc, identify enclosed singularities, then sum residues.

Fourier Integral Application

For

$$I(x) = \int_{-\infty}^{\infty} \frac{e^{ikx}}{k^2+\kappa^2}\,dk, \qquad \kappa>0,$$

the poles are at $k=i\kappa$ and $k=-i\kappa$. The exponential factor decides the contour:

• for $x>0$, close in the upper half-plane and use the pole $i\kappa$;
• for $x\lt0$, close in the lower half-plane and use the pole $-i\kappa$ with clockwise orientation.

The result is

$$I(x) = \frac{\pi}{\kappa} e^{-\kappa\lvert x\rvert}.$$

The residue theorem supplies the pole contribution, but contour integration supplies the permission to close the contour. Both steps are needed.

Green-Function Application

Resolvent and Green-function integrals often contain denominators such as

$$\frac{1}{E-\hbar^2k^2/(2m)\pm i0}.$$

Residues determine the contributions of isolated poles after the contour prescription is chosen. The $i0$ term tells the contour how to avoid real-axis singularities and selects a physical boundary condition, such as outgoing or incoming waves.

For bound states and resonances, poles of analytically continued Green functions or scattering amplitudes carry physical information. The residue may encode normalization, coupling strength, or transition amplitude, depending on the quantity being continued. The exact interpretation belongs to the relevant physics page; the residue theorem supplies the local complex-analysis mechanism.

Residue at Infinity

Sometimes it is easier to compute a contour integral by including the point at infinity. The residue at infinity is defined by

$$\operatorname{Res}_{z=\infty}f = - \operatorname{Res}_{w=0} \frac{1}{w^2}f(1/w).$$

For a meromorphic function on the extended complex plane,

$$\sum_{z_j\in\mathbb C} \operatorname{Res}_{z=z_j}f + \operatorname{Res}_{z=\infty}f =0.$$

This is often useful for rational functions. In physics contour calculations, however, infinity is also where growth and arc estimates live, so do not use this shortcut to avoid checking large-contour behavior in Fourier-type integrals.

Failure Modes

The residue theorem does not say that every complex integral is a sum over poles. It applies to meromorphic functions inside a closed contour. Several common quantum-mechanics situations require extra care:

• branch points and branch cuts are not isolated poles;
• essential singularities require Laurent coefficients but may be harder to control;
• real-axis poles require an indentation, principal value, or $i0$ prescription;
• continuous spectra often produce branch cuts rather than isolated poles;
• large arcs may fail to vanish if the exponential grows in the chosen half-plane.

When these issues appear, the residue theorem may still be part of the calculation, but it is not the whole calculation.

Common Mistakes

• Forgetting the contour orientation sign.
• Summing poles outside the chosen contour.
• Applying the theorem when a pole lies on the contour without specifying a prescription.
• Treating branch points as if they were poles with residues.
• Using the simple-pole formula for a higher-order pole.
• Assuming the large semicircle vanishes because the contour “looks standard.”
• Dropping contributions from small indentation arcs around real-axis poles.
• Forgetting that a physical pole’s residue depends on normalization and convention.

Cross-Links

• Analytic Functions
• Contour Integration
• Branch Cuts
• Complex Analysis Essentials
• Fourier Transform
• Principal Value Distributions
• Green Functions
• Asymptotic Analysis
• Scattering Amplitude
• Bound States and Scattering Poles
• Resonances

References

• L. V. Ahlfors, Complex Analysis, 3rd ed., McGraw-Hill, 1979.
• E. M. Stein and R. Shakarchi, Complex Analysis, Princeton University Press, 2003.
• J. W. Brown and R. V. Churchill, Complex Variables and Applications, 9th ed., McGraw-Hill, 2014.
• G. B. Arfken, H. J. Weber, and F. E. Harris, Mathematical Methods for Physicists, 7th ed., Academic Press, 2013.
• J. R. Taylor, Scattering Theory: The Quantum Theory of Nonrelativistic Collisions, Dover, 2006.

Exercises

1. Compute the residue of

$$f(z)=\frac{z^2+1}{z-2}$$

at $z=2$.

Solution

The pole is simple, so

$$\operatorname{Res}_{z=2}f = \lim_{z\to2} (z-2)\frac{z^2+1}{z-2} = 2^2+1 =5.$$

2. Compute the residue of $e^z/z^2$ at $z=0$.

Solution

Use

$$e^z = 1+z+\frac{z^2}{2}+\cdots.$$

Then

$$\frac{e^z}{z^2} = \frac{1}{z^2} +\frac{1}{z} +\frac12+\cdots.$$

The coefficient of $1/z$ is $1$, so

$$\operatorname{Res}_{z=0} \frac{e^z}{z^2} =1.$$

3. Use residues to evaluate

$$\int_{-\infty}^{\infty} \frac{dx}{x^2+4}.$$

Solution

Here $a=2$ in the standard example. Closing in the upper half-plane encloses the pole $z=2i$ with residue

$$\frac{1}{4i}.$$

Thus

$$\int_{-\infty}^{\infty} \frac{dx}{x^2+4} = 2\pi i \frac{1}{4i} = \frac{\pi}{2}.$$

4. A contour is closed in the lower half-plane around one simple pole with residue $r$. What is the closed-contour integral?

Solution

The orientation is clockwise, so it is negative relative to the usual positive orientation:

$$\oint_C f(z)\,dz = -2\pi i\,r.$$

5. Why can the residue theorem alone not evaluate an integral around a branch cut of $\log z$?

Solution

$\log z$ has a branch point, not an isolated pole. The residue theorem applies directly to meromorphic functions with isolated poles inside the contour. For a branch cut, the calculation must include the discontinuity between the two sides of the cut and any small circles around branch points. Residues may still appear from additional poles, but they do not account for the branch-cut contribution.