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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?

Yair
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  • Well, $$\int_{a}^{b}\frac{x}{x-c-ie},\mathrm{d}x+\int_{a}^{b}\frac{-c-ie}{x-c-ie},\mathrm{d}x,\neq,\int_{a}^{b}1,\mathrm{d}x$$ is true, provided we interpret $\int_{a}^{b}\ldots,\mathrm{d}x$ as the integral along the line segment $[a,b]$ or any prescribed contour $a\to b$ avoiding $c+ie$. I guess your intended inequality is something like $$\int_{\gamma_1}\frac{z}{z-c-ie},\mathrm{d}z+\int_{\gamma_2}\frac{-c-ie}{z-c-ie},\mathrm{d}z,\neq,\int_{\gamma_3}1,\mathrm{d}z$$ in general, where $\gamma_1, \gamma_2, \gamma_3$ are any contours from $a$ to $b$ that avoids $c+ie$. – Sangchul Lee Apr 04 '23 at 22:04
  • Thanks for the answer Lee! Clearly, the pole itself is avoided under the assumption that I mentioned. However, how would you explain this difference to someone who cannot see that this can happen? – Yair Apr 04 '23 at 22:18

1 Answers1

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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*}

Sangchul Lee
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  • Do you agree that: $\int_{-3}^{3}\frac{-1-2i}{x-1-2i}=(-1-2i)\left(\log(-2+2i)-\log(4+2i)\right)

    =(-1-2i)\left(\log(2\sqrt{2}e^{\frac{3}{4}i\pi+2i\pi k})-\log(2\sqrt{5}e^{i0.46+2i\pi m})\right)

    =(-1-2i)\left(\log(\sqrt{2})-\log(\sqrt{5})+\frac{3}{4}i\pi-i0.46+2i\pi n\right)$

     – Yair Apr 04 '23 at 23:17
  • @Yair, It depends on how you define $\int_{-3}^{3}\ldots,\mathrm{d}x$. Contour integrals are single-valued, because its definition requires us to explicate the contour being used in the integration, like: $$\int_{\gamma}f(z),\mathrm{d}z=\int_{t_0}^{t_1}f(\gamma(t))\gamma'(t),\mathrm{d}t$$ for a path $\gamma:[t_0,t_1]\to\mathbb{C}$. And what people usually do with the expression $\int_a^b f(x),\mathrm{d}x$ is either spare it for integrals over real-axis only or interpret it as a contour integral along the line segment $\overline{ab}$. Either way, they are usually defined as single-valued. – Sangchul Lee Apr 04 '23 at 23:24
  • @Yair, So, in that sense, I disagree with what you have written. Of course, you have freedom to define your own idiosyncratic version of the expression $\int_{a}^{b}f(z),\mathrm{d}z$ as something like $$\int_{a}^{b}f(z),\mathrm{d}z=\left{\int_{\gamma}f(z),\mathrm{d}z:\gamma\text{ is any contour from $a$ to $b$ for which the integral is defined}\right}$$ and then with this very definition I would agree with your formula. – Sangchul Lee Apr 04 '23 at 23:28
  • There is no contour in the integration that we have to perform, it involves only integration over the real axis alone, and not a closed shape. The point that I try to make is that the value of each complex log is determined only up to a factor of $2i\pi$. This integration is over a Riemann surface (branch cut), so I cannot see how it can yield a single value result. – Yair Apr 04 '23 at 23:34
  • @Yair, I don't understand what you mean by "This integral is over a Riemann surface", especially considering that you first mentioned "it involves only integration over the real axis". I am pretty sure anyone who read this will envision a contour integral $\int_{\gamma}f(z),\mathrm{d}z$ where $\gamma$ is a contour in a Riemann surface (such as the universal cover $\tilde{X}$ of $X=\mathbb{C}\setminus{0}$ on which the complex-logarithm is defined). Even in this context, the contour integral is single-valued. – Sangchul Lee Apr 04 '23 at 23:50
  • It is true that the value of $\log(z)$ is determined only up to $2\pi i\mathbb{Z}$, reflecting the fact that $2\pi i\mathbb{Z}$ is the monodromy group of the universal covering of $\mathbb{C}\setminus{0}$. However, this ambiguity does not enter when it comes to defining and/or computing the integral $$\int_{a}^{b}f(x),\mathrm{d}x,$$ whether it be a Riemann integral along the interval $[a, b]$ or a contour integral along the line segment $\overline{ab}$. This expression is defined via a very specific limit that, of course, only admits at most one possible value. – Sangchul Lee Apr 04 '23 at 23:54
  • Arguing differently, any branch $\operatorname{Log}:U\to\mathbb{C}$ of the complex logarithm, where $U$ is a simply-connected domain in $\mathbb{C}\setminus{0}$ that contains both $-2+2i$ and $4+2i$, can be used to compute your integral: $$\int_{-3}^{3}\frac{-1-2i}{x-1-2i},\mathrm{d}x=[(-1-2i)\operatorname{Log}(x-1-2i)]_{-3}^{3}$$ Furthermore, this value does not depend on the choice of $\operatorname{Log}$ (essentially because the ambiguity of $2k\pi i$ part cancels out each other). – Sangchul Lee Apr 04 '23 at 23:59
  • But why do you assume that the two complex logs from the upper and lower bounds belong to the same branch? in my case it is more natural to assume that each one of them belong to a different branch and then the ambiguous part is generated. – Yair Apr 05 '23 at 00:04
  • @Yair, Actually there is a slight mistake in my last comment. For the computation above, you need a branch $\text{Log}:U\to\mathbb{C}$ such that $U$ contains the line segment from $a=-4-2i$ to $b=2-2i$, namely the path $$\overline{ab}:x\mapsto(x-1-2i),\quad-3\le x\le3$$ (and not just $U$ satisfying $a,b\in U$ as in my previous comment). The reason is that you are essentially integrating $\frac{1}{z}$ over $\overline{ab}$, hence $\text{Log}$ must also be analytic along $\overline{ab}$. This prohibits certain branches from being used in conjunction with the fundamental theorem of calculus. – Sangchul Lee Apr 05 '23 at 00:15
  • To give a sense of the above explanation, consider the following example: $-\arctan(1/z)$ is a branch of the antidervative of $1/(1+z^2)$. However, you can't use it, in conjunction with FoC, to compute the integral of $1/(1+x^2)$ for $-1\leq x\leq 1$: $$\frac{\pi}{2}=\int_{-1}^{1}\frac{\mathrm{d}x}{1+x^2}\neq\left[-\arctan(1/x)\right]_{-1}^{1}=-\frac{\pi}{2}$$ This is because the line $[-1,1]$ crosses the branch cut of $-\arctan(1/z)$. – Sangchul Lee Apr 05 '23 at 00:37
  • And if you still have questions, please clarify and provide your definition of $\int_{a}^{b}f(x),\mathrm{d}x$ for a general holomorphic function $f$, so that we can analyze it together and see if your claim holds true or not. (And again, I would like to remind you that what you have claimed is not true, if you are using the standard definition of Riemann integral and/or contour integral.) – Sangchul Lee Apr 05 '23 at 00:37
  • Thanks for all the comments and clarifications. However, I find it hard to accept that. Let me explain by a slightly better example. According to what you said, $\int_{0}^{1}\frac{e^{i\pi x}}{e^{i\pi x}-C}dx-\int_{0}^{1}\frac{C}{e^{i\pi x}-C}dx,=,\int_{0}^{1}1,dx$. However, for this to be true for any given complex $C$, the following identity must hold: $\log(-1+C)-\log(C(-1+C))-\log(1+C)-\log(C(1+C))=0$. By taking non-symmetric branch definition with respect to the argument of the logs, one can obtain a non-vanishing value. – Yair Apr 05 '23 at 02:51
  • @Yair, Again, what is your definition of the definite integral $\int_{0}^{1}\frac{e^{i\pi x}}{e^{i\pi x}-C},\mathrm{d}x$? Is it different from the Riemann integral, defined as the limit of Riemann sums? $$\int_{a}^{b}f(x),\mathrm{d}x=\lim_{|\mathcal{P}|\to0} \sum_{k} \frac{e^{i\pi x^_k}}{e^{i\pi x^k}-C}(x_k - x{k-1})$$ or something like this? – Sangchul Lee Apr 05 '23 at 03:18
  • Let us continue this discussion in chat. – Yair Apr 05 '23 at 03:19