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Mathematics · 术语表

什么是 residue?

又称: meromorphic function

Definition 17.3 University Mathematics — Year 3 · 第 17 章 — Laurent Series and the Residue Theorem

If ff is holomorphic on a punctured disc D(a,R)∖{a}D(a, R)\setminus \{a\}, expand by Laurent (r=0r = 0). Three exclusive cases:

  • all cn=0c_n = 0 for n<0n < 0: removable singularity (the nonnegative series extends ff holomorphically to aa);
  • cn≠0c_n \neq 0 for finitely many, at least one, n<0n < 0: a pole of order m=−min⁡{n:cn≠0}m = -\min\{n : c_n \neq 0\}; equivalently f=g/(z−a)mf = g/(z-a)^m, gg holomorphic, g(a)≠0g(a) \neq 0; equivalently ∣f(z)∣→∞\abs{f(z)} \to \infty as z→az\to a;
  • infinitely many negative cn≠0c_n \neq 0: essential singularity.

The residue is Res⁡(f,a)=c−1\operatorname{Res}(f, a) = c_{-1}. A function holomorphic on Ω\Omega minus a set of poles is meromorphic on Ω\Omega.

示例

Example 17.7 (The four classical integral types)

(a) Rational over R\R: for ∫R ⁣dx1+x4\int_\R\frac{\dd x}{1 + x^4}, close with a large semicircle SRS_R in the upper half-plane: the integrand is O(R−4)O(R^{-4}) there, so ∫SR→0\int_{S_R} \to 0 (ML), and the residue theorem with the poles eiπ/4,e3iπ/4\eu^{\iu\pi/4}, \eu^{3\iu\pi/4} (simple, residues 14z3=z4z4=−z4\frac1{4z^3} = \frac{z}{4z^4} = -\frac z4 at a pole) gives

∫R ⁣dx1+x4=2iπ(−eiπ/44−e3iπ/44)=π2.\int_\R\frac{\dd x}{1 + x^4} = 2\iu\pi\Bigl(-\frac{\eu^{\iu\pi/4}}4 - \frac{\eu^{3\iu\pi/4}}4\Bigr) = \frac{\pi}{\sqrt2} .

(b) Fourier type: for t≥0t \geq 0, ∫Reitx1+x2 ⁣dx=2iπRes⁡(eitz1+z2,i)=2iπe−t2i=πe−t\int_\R\frac{\eu^{\iu tx}}{1 + x^2}\dd x = 2\iu\pi\operatorname{Res}\bigl(\tfrac{\eu^{\iu tz}}{1+z^2}, \iu\bigr) = 2\iu\pi\frac{\eu^{-t}}{2\iu} = \pi\eu^{-t} — the upper semicircle works because ∣eitz∣=e−tIm⁡z≤1\abs{\eu^{\iu tz}} = \eu^{-t\operatorname{Im}z} \leq 1 there; taking real parts: ∫Rcos⁡(tx)1+x2 ⁣dx=πe−∣t∣\int_\R\frac{\cos(tx)}{1+x^2}\dd x = \pi\eu^{-\abs t}, settling Exercise 10.10’s admitted formula. (c) Trigonometric over a period: substitute z=eitz = \eu^{\iu t}, cos⁡t=z+z−12\cos t = \frac{z + z^{-1}}2,  ⁣dt= ⁣dziz\dd t = \frac{\dd z}{\iu z}: ∫02π ⁣dta+cos⁡t\int_0^{2\pi}\frac{\dd t}{a + \cos t} (a>1a > 1) becomes a residue count inside the unit circle (Exercise 17.2). (d) Series: pair ff with πcot⁡(πz)\pi\cot(\pi z), whose poles are the integers with residue 11: the weekend problem sums ∑n−2\sum n^{-2} and ∑n−4\sum n^{-4} this way.

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