5. Week 5: Estimating and Sometimes Computing Path Integrals
≪ 4. Week 4: Conformal Mappings | Table of Contents | 6. Week 6: The Fundamental Theorems of Complex Analysis ≫I’d like to preface this week’s discussion with a bit of context for where complex line integrals come into play in both the theory and the applications of complex analysis.
When we learned about harmonic functions, we characterised them as solutions to Laplace’s equation: \(u\) is harmonic whenever \(\frac{\partial ^2 u }{\partial x^2} + \frac{\partial ^2u }{\partial y^2} = 0\). It turns out that harmonic functions also enjoy a “mean value property”: if \(u\) is harmonic near a point \(z_0\) and \(\gamma \) is a circle centred at \(z_0\) of radius \(r\), then \[u\left( z_0 \right) = \frac{1}{2 \pi r} \oint _{\gamma } u(z) dz.\] In words, the value of \(u\) at \(z_0\) is its average value on a circle centred at \(z_0\).
Because the real and imaginary parts of holomorphic functions are harmonic, one may anticipate similar integral identities to hold for holomorphic functions, and indeed they do. In the coming weeks, we’ll be unraveling several integral formulae (most notably, Cauchy’s integral formula) that relate the integrals of functions related to holomorphic functions on closed loops to “special values” of the function and its derivatives inside the loop. In fact, one can even define what it means to be “holomorphic” using integrals alone (more or less)!
The key here is that these integral identities almost always involve closed loops. In both theory and applications, there are scenarios where we only have a regular line integral, often one over the real line. Thus, a very common technique is to create a closed loop where exactly one segment of the loop is a quantity of interest. These integral identities allow us te determine the value of the integral over the entire loop, and hopefully we will be able to neglect or cancel out the integrals over the remaining segments of the loop. As you can maybe guess, this is a core part of how contour integration works.
Today’s focus will thus be on bounding or, in some very special cases, computing path integrals.
Path Integrals
When directly computing a path integral \(\int _{\gamma }f(z) dz\) by hand, you have two options:
- Parameterise the curve \(\gamma : \left[ a, b \right]\to \mathbb{C}\) and compute \(\int _{a}^{b} f\left( \gamma (t) \right) \gamma ‘(t) dt\) by hand. Note that the parameterisation doesn’t matter!
- Find a holomorphic antiderivative (or primitive) \(F(z)\) of \(f(z)\), if it exists. Then \(\int _{\gamma } f(z) dz = F\left( z_1 \right) - F\left( z_0 \right)\), where \(z_0\) and \(z_1\) are the endpoints of \(\gamma \) (in order).
The first point is more or less the same process that you use to compute line integrals in vector calculus! The second point is a consquence of Stokes’ theorem from vector calculus. However, the condition that \(F\) is holomorphic is critical to the computation, as we will see momentarily.
Example 1. Textbook Exercise IV.2.2
Using an appropriate primitive, evaluate \(\int _{\gamma }\frac{dz}{z}\) for a path \(\gamma \) that travels from \(-\pi i\) to \(\pi i\) in the right half-plane, and also for a path \(\gamma \) that travels from \(- \pi i \) to \(\pi i\) in the left half-plane. For each path give a precise definition of the primitive used to evaluate the integral.
Solution
When \(\gamma \) is confined to the right half-plane, we can take our antiderivative to be the standard branch of \(\operatorname{Log}\). This is holomorphic on \(\mathbb{C}\setminus (-\infty, 0]\), which includes the right-half plane. Thus, we get \[\int _{\gamma } \frac{dz}{z} = \operatorname{Log}(\pi i) - \operatorname{Log}(-\pi i).\] We get that \(\operatorname{Log} \left( \pi i \right) = \ln \pi + \frac{i \pi }{2}\) and that \(\operatorname{Log} \left( - \pi i \right) = \ln \pi - \frac{i \pi }{2}\) (draw a picture!), hence the integral simplifies to \(i \pi\).
The standard branch of the complex logarithm no longer works when \(\gamma \) is confined to the left half-plane: \(\gamma \) would have to jump across the branch cut, which is impossible. Instead, let’s define another branch of the complex logarithm, given by \[\operatorname{Log}\left( z \right) = \ln \left\lvert z \right\rvert + i \theta, \] with \(\theta \) chosen to be the argument of \(z\) between \(0\) and \(2 \pi \). Here, we are making a branch cut along the positive real axis instead. This time, we still get \[\int _{\gamma } \frac{dz}{z} = \operatorname{Log}(\pi i) - \operatorname{Log}(-\pi i),\] and \(\operatorname{Log}(\pi i) = \ln \pi + \frac{i \pi }{2}\) as before. However, this time, we get \(\operatorname{Log}(-\pi i) = \ln \pi + \frac{i 3 \pi }{2}\)! Hence, we get \(\int _{\gamma } \frac{dz}{z} = -\pi i\) instead.
A critical part of this solution was selecting the right branch cut of the complex logarithm to ensure holomorphicity at every point on the path! If you are feeling ambitious, you should also verify this solution by computing these integrals via an explicit parameterisation of the curves \(\gamma \) (I suggest taking a semicircular arc).
Problem 2. Textbook Problem IV.2.4
Let \(D=\mathbb{C}\setminus (-\infty, 1]\). Given that \(\operatorname{Log}\left( z+\sqrt{z^2-1} \right)\) is an antiderivative of \(\frac{1}{\sqrt{z^2-1}}\) on \(D\) (with the standard branch of \(\operatorname{Log}\) and \(\sqrt\cdot\)), compute \[\oint _{\left\lvert z \right\rvert = 2} \frac{dz}{\sqrt{z^2-1}}.\]
Hint
The ML Bound
Theorem 3. The ML Bound
Let \(\gamma \) be a piecewise smooth curve in \(\mathbb{C} \), and let \(f\) be an integrable function defined around the image of \(\gamma \). Then, \[\left\lvert \int _{\gamma } f(z) dz \right\rvert \leq \left\lvert \gamma \right\rvert \cdot \sup \left\lvert f(z) \right\rvert, \] where \(\left\lvert \gamma \right\rvert\) is the arc length of \(\gamma \).
In words, the integrand can cancel itself out via “destructive interference” over the course of the path \(\gamma \): try integrating \(1\) around a circle. Heuristically, the \(dz\)’s on diametrically opposed points of the circle cancel each other out. This works the other way too: the integrand can also perform “constructive interference”: an example from lecture is the integral of \(\frac{1}{z}\) around a circle. The theorem is saying that the absolute maximum amount of “constructive interference” that can ever happen is the above quantity.
While arc length computations in vector calculus are rather tedious and dense, in complex analysis the vast majority of paths we will be integrating consist of simple geometric shapes whose arc lengths are well-known (arcs on a circle and line segments, for the most part). The primary difficult typically rests upon bounding the integrand.
Problem 4.
Let \(\gamma_R \) be a semicircular arc of radius \(R>0\) centred at the origin, starting from the point \(+R\) and traveling counterclockwise until it meets the point \(-R\). Show that \[\lim _{R\to\infty} \int _{\gamma _R} \frac{dz}{z^4+1} = 0.\]
Hint
Problem 5.
Let \(\gamma _R\) be a circular arc, oriented counterclockwise, from the point \(R\) to the point \(R e ^{\frac{i \pi }{4}}\). Show that \[\lim _{R\to\infty} \int _{\gamma _R} \frac{1}{z \operatorname{Log}(z)} dz = 0.\]