Explaining the concept of complex integration

Question

I need some help to explain a colleague that complex integration has different properties than real integration. As an example, I gave him the following integration: \begin{equation} \int_{-3}^{3}\frac{x}{x-1-2i}\,dx+\int_{-3}^{3}\frac{-1-2i}{x-1-2i}\,dx\,\neq\,6 \end{equation} Since the LHS yields a multi-value result, unlike the RHS which is fixed on a single value. The integration in Mathematica clearly shows a difference, but he insists that the software simply fails to make additional necessary simplifications to establish the equivalence. Is there any advice how to make it easier explain?

Answer 1

1. You have to be clear as to what $\int_{a}^{b} \ldots \,\mathrm{d}z$ means, because this expression often refers to the contour integral along the line segment $\overline{ab}$ from $a$ to $b$. Adopting this convention, the integral $\int_{a}^{b}\ldots\,\mathrm{d}z$ is single-valued, and we clearly have linearity:

$$ \alpha \int_{a}^{b}f(z)\,\mathrm{d}z + \beta \int_{a}^{b}g(z)\,\mathrm{d}z = \int_{a}^{b} (\alpha f(z) + \beta g(z)) \,\mathrm{d}z. $$

2. However, even with this convention, we can come up with an inequality that can be a surprise in view of real-analytic integrals:

$$ \int_{1}^{i} \frac{\mathrm{d}z}{z} + \int_{i}^{-1} \frac{\mathrm{d}z}{z} + \int_{-1}^{-i} \frac{\mathrm{d}z}{z} + \int_{-i}^{1} \frac{\mathrm{d}z}{z} = 2\pi i \neq 0 = \int_{1}^{1} \frac{\mathrm{d}z}{z} \tag{*}$$

Of course, this is an immediate consequence of the residue computation. But the left-hand side can be verified by a direct computation only using the Riemann-sum definition of contour integral and real-analytic methods. To show this, note:

Observation. Let$a, b \in \mathbb{C} \setminus\{0\}$ be such that the line segment $\overline{ab}$ does not pass through the origin. Then

1. We have $$ \int_{a}^{b} \frac{\mathrm{d}z}{z} = \lim_{n\to\infty} \sum_{k=1}^{n} \frac{(b-a)\frac{1}{n}}{a+(b-a)\frac{k}{n}} = \int_{0}^{1} \frac{(b - a) \, \mathrm{d}t}{a + (b - a)t}. $$
2. For any $c \in \mathbb{C}\setminus\{0\}$, $$ \int_{ca}^{cb} \frac{\mathrm{d}z}{z} = \int_{0}^{1} \frac{(cb - ca) \, \mathrm{d}t}{ca + (cb - ca)t} = \int_{0}^{1} \frac{(b - a) \, \mathrm{d}t}{a + (b - a)t} = \int_{a}^{b} \frac{\mathrm{d}z}{z}. $$

Using this, the left-hand side of $\text{(*)}$ reduces to

\begin{align*} \text{[LHS of (*)]} = 4 \int_{1}^{i} \frac{\mathrm{d}z}{z} = 4 \int_{0}^{1} \frac{(i-1)\,\mathrm{d}t}{1 + (i-1)t}. \end{align*}

Then by noting that

$$ \frac{i-1}{1 + (i-1)t} = 2 \cdot \frac{(2t-1) + i}{(2t-1)^2 + 1}, $$

we get

\begin{align*} \text{[LHS of (*)]} &= 8 \int_{0}^{1} \frac{(2t-1) + i}{(2t-1)^2 + 1}\,\mathrm{d}t \\ &= \left[ 2 \log((2t-1)^2 + 1) + 4i \arctan(2t - 1) \right]_{0}^{1} \\ &= 2\pi i. \end{align*}