Movatterモバイル変換

[0]ホーム
URL:

Jump to content
WikipediaThe Free Encyclopedia
Search

Residue theorem

From Wikipedia, the free encyclopedia
Concept of complex analysis
Mathematical analysis →Complex analysis
Complex analysis

Incomplex analysis, theresidue theorem, sometimes calledCauchy's residue theorem, is a powerful tool to evaluateline integrals ofanalytic functions over closed curves; it can often be used to compute real integrals andinfinite series as well. It generalizes theCauchy integral theorem andCauchy's integral formula. Theresidue theorem should not be confused with special cases of thegeneralized Stokes' theorem; however, the latter can be used as an ingredient of its proof.

Statement of Cauchy's residue theorem

[edit]
See also:Residue (complex analysis)

The statement is as follows:

Illustration of the setting

Residue theorem: LetU{\displaystyle U} be asimply connectedopen subset of thecomplex plane containing a finite list of pointsa1,…,an,{\displaystyle a_{1},\ldots ,a_{n},}U0=U∖{a1,…,an},{\displaystyle U_{0}=U\smallsetminus \{a_{1},\ldots ,a_{n}\},} and a functionf{\displaystyle f}holomorphic onU0.{\displaystyle U_{0}.} Lettingγ{\displaystyle \gamma } be a closedrectifiable curve inU0,{\displaystyle U_{0},} and denoting theresidue off{\displaystyle f} at each pointak{\displaystyle a_{k}} byRes⁡(f,ak){\displaystyle \operatorname {Res} (f,a_{k})} and thewinding number ofγ{\displaystyle \gamma } aroundak{\displaystyle a_{k}} byI⁡(γ,ak),{\displaystyle \operatorname {I} (\gamma ,a_{k}),} the line integral off{\displaystyle f} aroundγ{\displaystyle \gamma } is equal to2πi{\displaystyle 2\pi i} times the sum of residues, each counted as many times asγ{\displaystyle \gamma } winds around the respective point:

∮γf(z)dz=2πi∑k=1nI⁡(γ,ak)Res⁡(f,ak).{\displaystyle \oint _{\gamma }f(z)\,dz=2\pi i\sum _{k=1}^{n}\operatorname {I} (\gamma ,a_{k})\operatorname {Res} (f,a_{k}).}

Ifγ{\displaystyle \gamma } is apositively orientedsimple closed curve,I⁡(γ,ak){\displaystyle \operatorname {I} (\gamma ,a_{k})} is1{\displaystyle 1} ifak{\displaystyle a_{k}} is in the interior ofγ{\displaystyle \gamma } and0{\displaystyle 0} if not, therefore

∮γf(z)dz=2πi∑Res⁡(f,ak){\displaystyle \oint _{\gamma }f(z)\,dz=2\pi i\sum \operatorname {Res} (f,a_{k})}

with the sum over thoseak{\displaystyle a_{k}} insideγ.{\displaystyle \gamma .}[1]

The relationship of the residue theorem to Stokes' theorem is given by theJordan curve theorem. The generalplane curveγ must first be reduced to a set of simple closed curves{γi}{\displaystyle \{\gamma _{i}\}} whose total is equivalent toγ{\displaystyle \gamma } for integration purposes; this reduces the problem to finding the integral offdz{\displaystyle f\,dz} along a Jordan curveγi{\displaystyle \gamma _{i}} with interiorV.{\displaystyle V.} The requirement thatf{\displaystyle f} be holomorphic onU0=U∖{ak}{\displaystyle U_{0}=U\smallsetminus \{a_{k}\}} is equivalent to the statement that theexterior derivatived(fdz)=0{\displaystyle d(f\,dz)=0} onU0.{\displaystyle U_{0}.} Thus if two planar regionsV{\displaystyle V} andW{\displaystyle W} ofU{\displaystyle U} enclose the same subset{aj}{\displaystyle \{a_{j}\}} of{ak},{\displaystyle \{a_{k}\},} the regionsV∖W{\displaystyle V\smallsetminus W} andW∖V{\displaystyle W\smallsetminus V} lie entirely inU0,{\displaystyle U_{0},} hence

∫V∖Wd(fdz)−∫W∖Vd(fdz){\displaystyle \int _{V\smallsetminus W}d(f\,dz)-\int _{W\smallsetminus V}d(f\,dz)}

is well-defined and equal to zero. Consequently, the contour integral offdz{\displaystyle f\,dz} alongγj=∂V{\displaystyle \gamma _{j}=\partial V} is equal to the sum of a set of integrals along pathsγj,{\displaystyle \gamma _{j},} each enclosing an arbitrarily small region around a singleaj{\displaystyle a_{j}} — the residues off{\displaystyle f} (up to the conventional factor2πi{\displaystyle 2\pi i} at{aj}.{\displaystyle \{a_{j}\}.} Summing over{γj},{\displaystyle \{\gamma _{j}\},} we recover the final expression of the contour integral in terms of the winding numbers{I⁡(γ,ak)}.{\displaystyle \{\operatorname {I} (\gamma ,a_{k})\}.}

In order to evaluate real integrals, the residue theorem is used in the following manner: the integrand is extended to the complex plane and its residues are computed (which is usually easy), and a part of the real axis is extended to a closed curve by attaching a half-circle in the upper or lower half-plane, forming a semicircle. The integral over this curve can then be computed using the residue theorem. Often, the half-circle part of the integral will tend towards zero as the radius of the half-circle grows, leaving only the real-axis part of the integral, the one we were originally interested in.

Calculation of residues

[edit]
This section is an excerpt fromResidue (complex analysis) § Calculation of residues.[edit]

Suppose apunctured disk⁠D={z:0<|z−c|<R}{\displaystyle D=\{z:0<\vert z-c\vert <R\}}⁠ in the complex plane is given and⁠f{\displaystyle f}⁠ is aholomorphic function defined (at least) on⁠D{\displaystyle D}⁠. The residue⁠Res⁡(f,c){\displaystyle \operatorname {Res} (f,c)}⁠ of⁠f{\displaystyle f}⁠ at⁠c{\displaystyle c}⁠ is the coefficient⁠a−1{\displaystyle a_{-1}}⁠ of⁠(z−c)−1{\displaystyle (z-c)^{-1}}⁠ in theLaurent series expansion of⁠f{\displaystyle f}⁠ around⁠c{\displaystyle c}⁠. Various methods exist for calculating this value, and the choice of which method to use depends on the function in question, and on the nature of the singularity.

According to the residue theorem, we have:

Res⁡(f,c)=12πi∮γf(z)dz{\displaystyle \operatorname {Res} (f,c)={1 \over 2\pi i}\oint _{\gamma }f(z)\,dz}

where⁠γ{\displaystyle \gamma }⁠ traces out a circle around⁠c{\displaystyle c}⁠ in a counterclockwise manner and does not pass through or contain other singularities within it. We may choose the path⁠γ{\displaystyle \gamma }⁠ to be a circle of radius⁠ε{\displaystyle \varepsilon }⁠ around⁠c{\displaystyle c}⁠. Since⁠ε{\displaystyle \varepsilon }⁠ can be as small as we desire it can be made to contain only the singularity of⁠c{\displaystyle c}⁠ due to nature of isolated singularities. This may be used for calculation in cases where the integral can be calculated directly, but it is usually the case that residues are used to simplify calculation of integrals, and not the other way around.

Removable singularities

[edit]

If the function⁠f{\displaystyle f}⁠ can becontinued to aholomorphic function on the whole disk⁠|y−c|<R{\displaystyle \vert y-c\vert <R}⁠, then⁠Res⁡(f,c)=0{\displaystyle \operatorname {Res} (f,c)=0}⁠. The converse is not in general true.

Simple poles

[edit]

If⁠c{\displaystyle c}⁠ is asimple pole of⁠f{\displaystyle f}⁠, the residue of⁠f{\displaystyle f}⁠ is given by:

Res⁡(f,c)=limz→c(z−c)f(z).{\displaystyle \operatorname {Res} (f,c)=\lim _{z\to c}(z-c)f(z).}

If that limit does not exist, then⁠f{\displaystyle f}⁠ instead has an essential singularity at⁠c{\displaystyle c}⁠. If the limit is⁠0{\displaystyle 0}⁠, then⁠f{\displaystyle f}⁠ is either analytic at⁠c{\displaystyle c}⁠ or has a removable singularity there. If the limit is equal to infinity, then the order of the pole is higher than⁠1{\displaystyle 1}⁠.

It may be that the function⁠f{\displaystyle f}⁠ can be expressed as a quotient of two functions,⁠f(z)=g(z)/h(z){\displaystyle f(z)={g(z)}/{h(z)}}⁠, where⁠g{\displaystyle g}⁠ and⁠h{\displaystyle h}⁠ areholomorphic functions in aneighbourhood of⁠c{\displaystyle c}⁠, with⁠h(c)=0{\displaystyle h(c)=0}⁠ and ⁠h′(c)≠0{\displaystyle h'(c)\neq 0}⁠. In such a case,L'Hôpital's rule can be used to simplify the above formula to:

Res⁡(f,c)=limz→c(z−c)f(z)=limz→czg(z)−cg(z)h(z)=limz→cg(z)+zg′(z)−cg′(z)h′(z)=g(c)h′(c).{\displaystyle {\begin{aligned}\operatorname {Res} (f,c)&=\lim _{z\to c}(z-c)f(z)=\lim _{z\to c}{\frac {zg(z)-cg(z)}{h(z)}}\\[4pt]&=\lim _{z\to c}{\frac {g(z)+zg'(z)-cg'(z)}{h'(z)}}={\frac {g(c)}{h'(c)}}.\end{aligned}}}

Limit formula for higher-order poles

[edit]

More generally, if⁠c{\displaystyle c}⁠ is apole of order⁠p{\displaystyle p}⁠, then the residue of⁠f{\displaystyle f}⁠ around⁠z=c{\displaystyle z=c}⁠ can be found by the formula:

Res⁡(f,c)=1(p−1)!limz→cdp−1dzp−1((z−c)pf(z)).{\displaystyle \operatorname {Res} (f,c)={\frac {1}{(p-1)!}}\lim _{z\to c}{\frac {d^{p-1}}{dz^{p-1}}}\left((z-c)^{p}f(z)\right).}

This formula can be very useful in determining the residues for low-order poles. For higher-order poles, the calculations can become unmanageable, andseries expansion is usually easier. Foressential singularities, no such simple formula exists, and residues must usually be taken directly from series expansions.

Residue at infinity

[edit]

In general, theresidue at infinity is defined as:

Res⁡(f(z),∞)=−Res(1z2f(1z),0).{\displaystyle \operatorname {Res} (f(z),\infty )=-{\operatorname {Res} }\left({\frac {1}{z^{2}}}f\left({\frac {1}{z}}\right),0\right).}

If the following condition is met:

lim|z|→∞f(z)=0,{\displaystyle \lim _{|z|\to \infty }f(z)=0,}

then theresidue at infinity can be computed using the following formula:

Res⁡(f,∞)=−lim|z|→∞zf(z).{\displaystyle \operatorname {Res} (f,\infty )=-\lim _{|z|\to \infty }zf(z).}

If instead

lim|z|→∞f(z)=c≠0,{\displaystyle \lim _{|z|\to \infty }f(z)=c\neq 0,}

then theresidue at infinity is

Res⁡(f,∞)=lim|z|→∞z2f′(z).{\displaystyle \operatorname {Res} (f,\infty )=\lim _{|z|\to \infty }z^{2}f'(z).}

For functions that are meromorphic on the entire complex plane with finitely many singularities, the sum of the residues at the (necessarily) isolated singularities plus the residue at infinity is zero, which gives:

Res⁡(f(z),∞)=−∑kRes⁡(f(z),ak).{\displaystyle \operatorname {Res} (f(z),\infty )=-\sum _{k}\operatorname {Res} (f(z),a_{k}).}

Series methods

[edit]

If parts or all of a function can be expanded into aTaylor series orLaurent series, which may be possible if the parts or the whole of the function has a standard series expansion, then calculating the residue is significantly simpler than by other methods. The residue of the function is simply given by the coefficient of⁠(z−c)−1{\displaystyle (z-c)^{-1}}⁠ in theLaurent series expansion of the function.

Examples

[edit]

An integral along the real axis

[edit]

The integral∫−∞∞eitxx2+1dx{\displaystyle \int _{-\infty }^{\infty }{\frac {e^{itx}}{x^{2}+1}}\,dx}

The contourC.

arises inprobability theory when calculating thecharacteristic function of theCauchy distribution. It resists the techniques of elementarycalculus but can be evaluated by expressing it as a limit ofcontour integrals.

Supposet > 0 and define the contourC that goes along thereal line from−a toa and then counterclockwise along a semicircle centered at 0 froma to−a. Takea to be greater than 1, so that theimaginary uniti is enclosed within the curve. Now consider the contour integral∫Cf(z)dz=∫Ceitzz2+1dz.{\displaystyle \int _{C}{f(z)}\,dz=\int _{C}{\frac {e^{itz}}{z^{2}+1}}\,dz.}

Sinceeitz is anentire function (having nosingularities at any point in the complex plane), this function has singularities only where the denominatorz2 + 1 is zero. Sincez2 + 1 = (z +i)(z −i), that happens only wherez =i orz = −i. Only one of those points is in the region bounded by this contour. Becausef(z) iseitzz2+1=eitz2i(1z−i−1z+i)=eitz2i(z−i)−eitz2i(z+i),{\displaystyle {\begin{aligned}{\frac {e^{itz}}{z^{2}+1}}&={\frac {e^{itz}}{2i}}\left({\frac {1}{z-i}}-{\frac {1}{z+i}}\right)\\&={\frac {e^{itz}}{2i(z-i)}}-{\frac {e^{itz}}{2i(z+i)}},\end{aligned}}}theresidue off(z) atz =i isResz=i⁡f(z)=e−t2i.{\displaystyle \operatorname {Res} _{z=i}f(z)={\frac {e^{-t}}{2i}}.}

According to the residue theorem, then, we have∫Cf(z)dz=2πi⋅Resz=i⁡f(z)=2πie−t2i=πe−t.{\displaystyle \int _{C}f(z)\,dz=2\pi i\cdot \operatorname {Res} \limits _{z=i}f(z)=2\pi i{\frac {e^{-t}}{2i}}=\pi e^{-t}.}

The contourC may be split into a straight part and a curved arc, so that∫straightf(z)dz+∫arcf(z)dz=πe−t{\displaystyle \int _{\mathrm {straight} }f(z)\,dz+\int _{\mathrm {arc} }f(z)\,dz=\pi e^{-t}}and thus∫−aaf(z)dz=πe−t−∫arcf(z)dz.{\displaystyle \int _{-a}^{a}f(z)\,dz=\pi e^{-t}-\int _{\mathrm {arc} }f(z)\,dz.}

Using someestimations, we have|∫arceitzz2+1dz|≤πa⋅suparc|eitzz2+1|≤πa⋅suparc1|z2+1|≤πaa2−1,{\displaystyle \left|\int _{\mathrm {arc} }{\frac {e^{itz}}{z^{2}+1}}\,dz\right|\leq \pi a\cdot \sup _{\text{arc}}\left|{\frac {e^{itz}}{z^{2}+1}}\right|\leq \pi a\cdot \sup _{\text{arc}}{\frac {1}{|z^{2}+1|}}\leq {\frac {\pi a}{a^{2}-1}},}andlima→∞πaa2−1=0.{\displaystyle \lim _{a\to \infty }{\frac {\pi a}{a^{2}-1}}=0.}

The estimate on the numerator follows sincet > 0, and forcomplex numbersz along the arc (which lies in the upper half-plane), the argumentφ ofz lies between 0 andπ. So,|eitz|=|eit|z|(cos⁡φ+isin⁡φ)|=|e−t|z|sin⁡φ+it|z|cos⁡φ|=e−t|z|sin⁡φ≤1.{\displaystyle \left|e^{itz}\right|=\left|e^{it|z|(\cos \varphi +i\sin \varphi )}\right|=\left|e^{-t|z|\sin \varphi +it|z|\cos \varphi }\right|=e^{-t|z|\sin \varphi }\leq 1.}

Therefore,∫−∞∞eitzz2+1dz=πe−t.{\displaystyle \int _{-\infty }^{\infty }{\frac {e^{itz}}{z^{2}+1}}\,dz=\pi e^{-t}.}

Ift < 0 then a similar argument with an arcC′ that winds around−i rather thani shows that

The contourC′.

∫−∞∞eitzz2+1dz=πet,{\displaystyle \int _{-\infty }^{\infty }{\frac {e^{itz}}{z^{2}+1}}\,dz=\pi e^{t},}

and finally we have∫−∞∞eitzz2+1dz=πe−|t|.{\displaystyle \int _{-\infty }^{\infty }{\frac {e^{itz}}{z^{2}+1}}\,dz=\pi e^{-\left|t\right|}.}

(Ift = 0 then the integral yields immediately to elementary calculus methods and its value isπ.)

Evaluating zeta functions

[edit]

The fact thatπ cot(πz) has simple poles with residue 1 at each integer can be used to compute the sum∑n=−∞∞f(n).{\displaystyle \sum _{n=-\infty }^{\infty }f(n).}

Consider, for example,f(z) =z−2. LetΓN be the rectangle that is the boundary of[−N −⁠1/2⁠,N +⁠1/2⁠]2 with positive orientation, with an integerN. By the residue formula,

12πi∫ΓNf(z)πcot⁡(πz)dz=Resz=0+∑n=−Nn≠0Nn−2.{\displaystyle {\frac {1}{2\pi i}}\int _{\Gamma _{N}}f(z)\pi \cot(\pi z)\,dz=\operatorname {Res} \limits _{z=0}+\sum _{n=-N \atop n\neq 0}^{N}n^{-2}.}

The left-hand side goes to zero asN → ∞ since|cot⁡(πz)|{\displaystyle |\cot(\pi z)|} is uniformly bounded on the contour, thanks to usingx=±(12+N){\displaystyle x=\pm \left({\frac {1}{2}}+N\right)} on the left and right side of the contour, and so the integrand has orderO(N−2){\displaystyle O(N^{-2})} over the entire contour. On the other hand,[2]

z2cot⁡(z2)=1−B2z22!+⋯{\displaystyle {\frac {z}{2}}\cot \left({\frac {z}{2}}\right)=1-B_{2}{\frac {z^{2}}{2!}}+\cdots } where theBernoulli numberB2=16.{\displaystyle B_{2}={\frac {1}{6}}.}

(In fact,⁠z/2⁠ cot(⁠z/2⁠) =⁠iz/1 −e−iz⁠ −⁠iz/2⁠.) Thus, the residueResz=0 is−⁠π2/3⁠. We conclude:

∑n=1∞1n2=π26{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{6}}}which is a proof of theBasel problem.

The same argument works for allf(x)=x−2n{\displaystyle f(x)=x^{-2n}} wheren{\displaystyle n} is a positive integer,giving usζ(2n)=(−1)n+1B2n(2π)2n2(2n)!.{\displaystyle \zeta (2n)={\frac {(-1)^{n+1}B_{2n}(2\pi )^{2n}}{2(2n)!}}.}The trick does not work whenf(x)=x−2n−1{\displaystyle f(x)=x^{-2n-1}}, since in this case, the residue at zero vanishes, and we obtain the useless identity0+ζ(2n+1)−ζ(2n+1)=0{\displaystyle 0+\zeta (2n+1)-\zeta (2n+1)=0}.

Evaluating Eisenstein series

[edit]

The same trick can be used to establish the sum of theEisenstein series:πcot⁡(πz)=limN→∞∑n=−NN(z−n)−1.{\displaystyle \pi \cot(\pi z)=\lim _{N\to \infty }\sum _{n=-N}^{N}(z-n)^{-1}.}

Proof

Pick an arbitraryw∈C∖Z{\displaystyle w\in \mathbb {C} \setminus \mathbb {Z} }. As above, define

g(z):=1w−zπcot⁡(πz){\displaystyle g(z):={\frac {1}{w-z}}\pi \cot(\pi z)}

By the Cauchy residue theorem, for allN{\displaystyle N} large enough such thatΓN{\displaystyle \Gamma _{N}} encirclesw{\displaystyle w},

12πi∮ΓNg(z)dz=−πcot⁡(πz)+∑n=−NN1z−n{\displaystyle {\frac {1}{2\pi i}}\oint _{\Gamma _{N}}g(z)dz=-\pi \cot(\pi z)+\sum _{n=-N}^{N}{\frac {1}{z-n}}}

It remains to prove the integral converges to zero. Sinceπcot⁡(πz)/z{\displaystyle \pi \cot(\pi z)/z} is an even function, andΓN{\displaystyle \Gamma _{N}} is symmetric about the origin, we have∮ΓNπcot⁡(πz)/zdz=0{\displaystyle \oint _{\Gamma _{N}}\pi \cot(\pi z)/zdz=0}, and so

∮ΓNg(z)dz=∮ΓN(1z+1w−z)πcot⁡(πz)dz=−w∮ΓN1z(z−w)πcot⁡(πz)dz=O(1/N){\displaystyle \oint _{\Gamma _{N}}g(z)dz=\oint _{\Gamma _{N}}\left({\frac {1}{z}}+{\frac {1}{w-z}}\right)\pi \cot(\pi z)dz=-w\oint _{\Gamma _{N}}{\frac {1}{z(z-w)}}\pi \cot(\pi z)dz=O(1/N)}

See also

[edit]

Notes

[edit]
  1. ^Whittaker & Watson 1920, p. 112, §6.1.
  2. ^Whittaker & Watson 1920, p. 125, §7.2. Note that the Bernoulli numberB2n{\displaystyle B_{2n}} is denoted byBn{\displaystyle B_{n}} in Whittaker & Watson's book.

References

[edit]

External links

[edit]
Retrieved from "https://en.wikipedia.org/w/index.php?title=Residue_theorem&oldid=1272574429"
Categories:
Hidden categories:
[8]ページ先頭
©2009-2026 Movatter.jp