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The residue theorem states:

Suppose $U$ is a simply connected open subset of the complex plane, and $a_1,\ldots,a_n$ are finitely many points of $U$ and $f$ is a function which is defined and holomorphic on $U\setminus\{a_1,\ldots,a_n\}$. If $\gamma$ is a rectifiable curve in $U$ which does not meet any of the $a_k$, and whose start point equals its endpoint, then $$\oint_\gamma f(z) dz = 2\pi i \sum_{k=1}^n I(\gamma, a_k) \textrm{Res}(f, a_k)$$

But what if a pole $a_k$ lies on the curve $\gamma$? What could we say about the evaluation of our integral?

Take for instance,

$$\oint_C \frac{1}{z+1} dz$$

where $C$ is the unit circle. Is it fair to say that the value of this integral is 0 by the residue theorem?

Mateen Ulhaq
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2 Answers2

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Such an integral simply is not defined. You should change the curve and avoid the poles.

Even if we go down to the basics: $$\int_\gamma f dz = \int_a^b f(\gamma(t)) \gamma'(t) dt$$ But $f(\gamma(t))$ is not defined for the pole, thus the integral is undefined.

Here's a related question with answers: general-method-of-integration-when-poles-on-contour And here's a nice image on how to avoid poles: Residue theorem:When a singularity on the circle (not inside the circle)

Community
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Zabuzard
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No, the curve should not touch any of the poles themselves. When, for instance, you evaluate the contour integral of a function whose poles lie inconveniently on the path of integration, you must circumvent them via small indentations.

Exercise: by integrating around a semi-circular contour, show that this misconception would imply that $\int_{-\infty}^{\infty} \! \dfrac{\sin(x)}{x} \, \mathrm{d}x=0$.

Chris Sanders
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