Lesson 12: Branch Cuts & the Keyhole Contour

Lesson 7 evaluated $\int_{-\infty}^\infty\frac{dx}{1+x^2}$ with a semicircular contour, and it worked. It worked because of three assumptions that were never stated out loud:

1. The integrand decayed fast enough that a plain ML-estimate killed the arc.
2. The integrand was single-valued.
3. No singularity sat on the contour.

Each of those can fail independently, and each failure has its own repair. Slow decay is handled by Jordan's lemma; a pole sitting on the path is handled by an indented contour. This lesson takes the middle one, multivaluedness, because that is the case standing between Lesson 9 and its reflection formula.

The diagnostic worth internalizing first: the question is never "where are the poles." Poles exist in every residue problem. The actual checks are whether a non-integer power or a logarithm has made the integrand multivalued, whether the decay is fast enough when you bound $|f|$ directly on a large arc rather than inferring it from pole locations, and whether any pole's location is a real number and therefore sits on the path you wanted to use.

What a branch cut actually is

Take the smallest instance that shows the problem: $\sqrt z$ on the unit circle, $z=e^{i\theta}$, so $\sqrt z=e^{i\theta/2}$.

Start at $z=1$, walk one full loop, arrive back at $z=1$ — and $\sqrt z$ has gone from $1$ to $-1$. The input returned; the output did not.

The conflict is a three-way one, and something has to give. You cannot simultaneously have a function that is (i) single-valued, (ii) continuous, and (iii) defined on a loop enclosing the origin. The piece that gives is (iii). Erect a barrier, the branch cut, running from $0$ out to $\infty$, which no path is permitted to cross. Then no path can ever complete a loop around the origin, and the contradiction never arises. Note what did not change: the function is the same function. Only the domain of allowed paths shrank.

Three things a branch cut is not, each easy to assume and each wrong:

The cut is not a set of singularities. $\sqrt z$ is perfectly well-behaved at every point on the cut ray, and everywhere else off it. The only genuine singularity is the branch point at $z=0$. The cut is a chosen no-crossing barrier, not a locus of trouble.

The cut is a choice, not a property of the function. The principal convention cuts along the negative reals, $-\pi<\theta<\pi$. This lesson cuts along the positive reals instead, $0<\theta<2\pi$. At $z=e^{-3i\pi/4}$ the principal convention gives $\sqrt z=e^{-3i\pi/8}$ while this one gives $e^{5i\pi/8}$: reached by adding a full turn, $-\frac{3\pi}4+2\pi=\frac{5\pi}4$, not by reflecting. Those two values differ by exactly $-1$. At $z=e^{3i\pi/4}$, whose angle already sits inside both ranges, the conventions agree exactly.

The jump across the cut is the mechanism, not a defect. Crossing from one side of the cut to the other multiplies the value by a specific factor. The next section extracts precisely that factor as the answer to a real integral.

Key takeaway: a branch cut restricts the domain so that no path can loop the branch point, which is what makes a single-valued continuous choice possible. The cut is chosen, the branch point is not.

The keyhole contour

Target: $\displaystyle\int_0^\infty\frac{x^{a-1}}{1+x}\,dx=\frac{\pi}{\sin(\pi a)}$ for $0<a<1$. Everything below is checked concurrently at $a=\frac12$, where the answer should come out to $\pi$.

The contour has four pieces: an outer circle of radius $R$, an inner circle of radius $\varepsilon$, and two straight segments hugging the positive real axis, one from above and one from below. The cut is placed along the positive reals — deliberately directly on top of where the integral lives.

The contrast with Lesson 7 is the key move. The crosscut construction in the Residue Theorem's proof has the same geometry: an outer curve, an inner curve, and two connecting segments traversed in opposite directions. There, the two segments cancelled. They cancelled for two reasons, not one: they were the same path reversed, and the integrand took the same value on both of them. Here the first reason still holds and the second does not, because of the branch jump. The cancellation in Lesson 7 was earned by single-valuedness, and this integrand does not have it.

The jump, computed. Write $z^{a-1}=|z|^{a-1}e^{i(a-1)\theta}$. On the top segment, $\theta\to0^+$, giving $z^{a-1}=x^{a-1}$. On the bottom segment, $\theta\to2\pi^-$, giving $z^{a-1}=x^{a-1}e^{2\pi i(a-1)}=x^{a-1}e^{2\pi ia}$. At $a=\frac12$ and $x=4$ that reads $\frac12$ on top and $-\frac12$ on the bottom, a ratio of $-1=e^{i\pi}$.

Reversing the bottom segment into a forward integral and combining the two:

$$\left(1-e^{2\pi ia}\right)\int_0^\infty\frac{x^{a-1}}{1+x}\,dx$$

At $a=\frac12$ the prefactor is $1-(-1)=2$. The two sides add. They do not cancel.

Both circles vanish, for different reasons. The two halves of the condition $0<a<1$ turn out to be doing one job each:

• Outer circle, $R\to\infty$: the bound goes like $R^{a-1}\to0$, which needs $a<1$. A power vanishes as its base grows only if the exponent is negative.
• Inner circle, $\varepsilon\to0$: the bound goes like $\varepsilon^{a}\to0$, which needs $a>0$. A power vanishes as its base shrinks only if the exponent is positive; a negative exponent on a shrinking number blows up instead.

The residue, at the single enclosed pole $z=-1$. There is a trap here worth naming: within the range $\theta\in(0,2\pi)$, the point $-1$ must be written $e^{i\pi}$. Writing it $e^{-i\pi}$ corresponds to $\theta=-\pi$, which is outside the chosen range and therefore a different branch.

$$\mathrm{Res}=(-1)^{a-1}=\left(e^{i\pi}\right)^{a-1}=e^{i\pi(a-1)}=e^{i\pi a}e^{-i\pi}=-e^{i\pi a}$$

At $a=\frac12$ this gives $-e^{i\pi/2}=-i$.

Assembly.

$$\left(1-e^{2\pi ia}\right)I=2\pi i\left(-e^{i\pi a}\right) \;\Longrightarrow\; I=\frac{-2\pi ie^{i\pi a}}{1-e^{2\pi ia}}=\frac{-2\pi i}{e^{-i\pi a}-e^{i\pi a}}=\frac{-2\pi i}{-2i\sin(\pi a)}=\frac{\pi}{\sin(\pi a)}$$

Checking against the running instance: at $a=\frac12$, the right side of the residue equation is $2\pi i(-i)=2\pi$, and dividing by the prefactor $2$ gives $I=\pi$; independently, $\pi/\sin(\pi/2)=\pi$. Both routes agree, so

$$\int_0^\infty\frac{x^{-1/2}}{1+x}\,dx=\pi$$

For general $a\in(0,1)$ this is a family of real integrals that real-variable calculus alone does not reach. The $a=\frac12$ instance shown here is an exception rather than the evidence: $x=u^2$ turns it into $2\int_0^\infty\frac{du}{1+u^2}$, which is elementary. The keyhole earns its keep on the rest of the family, which is what the reflection formula needs.

Key takeaway: the keyhole contour puts the cut where the integral lives, and the branch jump (the very thing that broke single-valuedness) becomes the factor $\left(1-e^{2\pi ia}\right)$ that makes the two segments add instead of cancel.

The reflection formula

The answer just computed, $\frac{\pi}{\sin\pi a}$, is already the right-hand side of the Gamma function's reflection formula, $\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin(\pi s)}$. That is not a coincidence, and it is the reason this contour was worth building.

The bridge from this integral to a product of two Gammas runs through the Beta function, and Lesson 9 walks it in full: the substitution $t=\frac{x}{1+x}$ turns $B(p,1-p)$ into exactly the keyhole integral above, the Beta–Gamma relation converts that into $\Gamma(p)\Gamma(1-p)$, and the identity theorem carries the result off the strip $0<p<1$ to all of $\mathbb{C}\setminus\mathbb{Z}$.

The one ingredient Lesson 9 could not produce for itself is the keyhole integral. That is this lesson.

Key takeaway: the keyhole integral is the right-hand side of the reflection formula. Lesson 9 builds the Beta bridge between the two; the keyhole contour is what makes the far end of that bridge computable.