The Residue Theorem & the Evaluation of Integrals

the dogbone contour

Sometimes a multivalued integrand has not one branch point but two, joined by a finite branch cut — a segment, say from -1 to 1, across which a square root like the square root of (1 - x^2) jumps sign. The dogbone contour is the curve tailored for this: it wraps tightly around the whole finite cut, looking like a dog's bone or a dumbbell, with a small loop at each branch point joined by two straight runs along the cut.

The shape: start just above one end of the cut, run along the top of the cut to the other end, loop around that branch point, return along the bottom of the cut, loop around the first branch point, and close. Because the integrand changes sign (or picks up a phase) when you cross from the top edge to the bottom edge, the two straight runs reinforce rather than cancel, just as in the keyhole. The trick for evaluating it is usually to view the dogbone from the outside: the function is single-valued in the region outside the cut (the two branch points are 'tied together' by the cut, so a large loop around both sees no net change), so by deforming the dogbone outward to a huge circle, the dogbone integral equals minus 2 pi i times the residue at infinity (plus any residues from poles outside the cut).

This is the right tool for integrals like the integral from -1 to 1 of dx over (1 + x^2) times the square root of (1 - x^2), and more generally rational functions times a square root of a quadratic with two real roots. The key distinction from the keyhole: the keyhole handles one branch point with an infinite cut to infinity; the dogbone handles two branch points joined by a finite cut, and the evaluation often routes through the residue at infinity rather than residues inside.

For the integral from -1 to 1 of dx over the square root of (1 - x^2) times (a + x), a dogbone around [-1, 1] reduces the problem to the residue of the integrand at infinity and at z = -a, giving a clean closed form.

Wrap the finite cut, then deform outward and read off the residue at infinity.

The branch of the root must be chosen so the function is single-valued on the dogbone's path and outside the cut; on the two edges of the cut the root takes opposite signs, which is what makes the straight pieces add. Mixing up that sign convention flips the answer.

Also called
dumbbell contourthe dog-bone狗骨圍道啞鈴形圍道