How Do Dogbone Contours, Branch Cuts, and Residues at Infinity Work?

A dogbone contour is a path that follows both sides of a branch cut, usually joined by small arcs around the cut’s endpoints. It turns a contour integral into a relation between the two boundary values of a multivalued function. The main bookkeeping is to distinguish the function’s value on each bank from the direction in which that bank is traversed.

A dogbone contour does not by itself mean that you should use the residue at infinity. Apply the residue theorem to the poles enclosed in the chosen branch domain. A residue at infinity is valid only when the function is single-valued and meromorphic in a neighborhood of infinity.

Set the branch and contour first

For a multivalued expression such as $z^\alpha$ or $\log z$, choose a branch before integrating. For example, the choice

$$0<\arg z<2\pi$$

places the branch cut along the positive real axis. The values just above and below that axis are then different: for $x>0$,

$$(x^\alpha)_{\text{above}}=x^\alpha, \qquad (x^\alpha)_{\text{below}}=e^{2\pi i\alpha}x^\alpha.$$

A dogbone boundary follows one bank in one direction and the other bank in the opposite direction. The opposite orientations are part of the contour integral; they are not optional sign changes added afterward. The small arcs at the cut’s endpoints must also be considered, unless estimates show their contributions vanish.

Worked example: a keyhole around the positive axis

Evaluate the real integral

$$I=\int_0^\infty \frac{x^{a-1}}{1+x}\,dx, \qquad 0<a<1.$$

Use $f(z)=z^{a-1}/(1+z)$ with $0<\arg z<2\pi$, so the cut is the positive real axis. Take a keyhole contour with an outer circle of radius $R$, an inner circle of radius $\varepsilon$, and straight sections just above and below the cut. In the limit, the two straight sections give:

• Upper bank: traversed from $\varepsilon$ to $R$, where $z^{a-1}=x^{a-1}$.
• Lower bank: traversed from $R$ to $\varepsilon$, where $z^{a-1}=e^{2\pi i(a-1)}x^{a-1}=e^{2\pi ia}x^{a-1}$.

Together their contribution tends to

$$(1-e^{2\pi ia})I.$$

The only pole inside the contour is at $z=-1$. On the chosen branch, $\arg(-1)=\pi$, so

$$\operatorname{Res}_{z=-1} f(z)=(-1)^{a-1}=e^{i\pi(a-1)}.$$

The residue theorem therefore gives

$$(1-e^{2\pi ia})I=2\pi i e^{i\pi(a-1)}.$$

Using $1-e^{2\pi ia}=-2i e^{i\pi a}\sin(\pi a)$ and $e^{i\pi(a-1)}=-e^{i\pi a}$ yields

$$I=\frac{\pi}{\sin(\pi a)}.$$

The omitted circular arcs do vanish for $0<a<1$. Near zero, $f(z)$ is of order $|z|^{a-1}$ and the arc length is proportional to $\varepsilon$, so the small-arc integral is of order $\varepsilon^a$. At large radius, $f(z)$ is of order $R^{a-2}$ and the arc length is proportional to $R$, so the large-arc integral is of order $R^{a-1}$. Both tend to zero in the stated range.

When the residue at infinity applies

For a function meromorphic near infinity, the residue at infinity is defined by

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

Equivalently, for a Laurent expansion valid near infinity, it is the negative of the coefficient of $1/z$. With the usual counterclockwise orientation of a large circle,

$$\oint_{|z|=R} f(z)\,dz=-2\pi i\operatorname{Res}_{z=\infty}f(z),$$

provided $f$ is meromorphic throughout the exterior of the circle and has no finite poles on or outside it. If the function is meromorphic on the extended complex plane, then $\operatorname{Res}_{\infty} f=-\sum \operatorname{Res} f$ over its finite poles.

The example above does not have a residue at infinity in this sense: $z^{a-1}$ is multivalued, and infinity is a branch point for a general noninteger $a$. Its contour calculation uses the residue at $-1$ and the jump between the two banks of the cut. Do not replace that jump with a residue at infinity.

Common mistakes

• Forgetting the lower-bank phase. A circuit around the branch point changes the argument by $2\pi$, so fractional powers acquire a factor such as $e^{2\pi i\alpha}$.
• Giving both banks the same direction. A positively oriented boundary traverses the two sides of a slit in opposite directions.
• Dropping endpoint arcs without checking. Their integrals vanish only when the local growth of the integrand makes them tend to zero.
• Applying the residue theorem across an unresolved cut. First define a branch on the region bounded by the contour. Then identify the isolated poles inside that region.
• Treating every large contour as a residue at infinity. A branch point at infinity prevents the usual meromorphic residue-at-infinity argument.

For a different branch cut or a different contour orientation, the phase factors and signs may change. The reliable method is to write the value of the integrand and the direction of traversal on each segment before simplifying the contour integral.