cn=0 for finitely many, at least one, n<0: a pole of order m=−min{n:cn=0}; equivalently f=g/(z−a)m, gholomorphic, g(a)=0; equivalently ∣f(z)∣→∞ as z→a;
infinitely many negative cn=0: essential singularity.
The residue is Res(f,a)=c−1. A function holomorphic on Ω minus a set of poles is meromorphic on Ω.
Examples
Example 17.7(The four classical integral types)
(a) Rational over R: for ∫R1+x4dx, close with a large semicircle SR in the upper half-plane: the integrand is O(R−4) there, so ∫SR→0 (ML), and the residue theorem with the poles eiπ/4,e3iπ/4 (simple, residues4z31=4z4z=−4z at a pole) gives
∫R1+x4dx=2iπ(−4eiπ/4−4e3iπ/4)=2π.
(b) Fourier type: for t≥0, ∫R1+x2eitxdx=2iπRes(1+z2eitz,i)=2iπ2ie−t=πe−t — the upper semicircle works because eitz=e−tImz≤1 there; taking real parts: ∫R1+x2cos(tx)dx=πe−∣t∣, settling Exercise 10.10’s admitted formula. (c) Trigonometric over a period: substitute z=eit, cost=2z+z−1, dt=izdz: ∫02πa+costdt (a>1) becomes a residue count inside the unit circle (Exercise 17.2). (d) Series: pair f with πcot(πz), whose poles are the integers with residue1: the weekend problem sums ∑n−2 and ∑n−4 this way.