Residue theorem - Alchetron, The Free Social Encyclopedia

In complex analysis, the residue theorem, sometimes called Cauchy's residue theorem, is a powerful tool to evaluate line integrals of analytic functions over closed curves; it can often be used to compute real integrals as well. It generalizes the Cauchy integral theorem and Cauchy's integral formula. From a geometrical perspective, it is a special case of the generalized Stokes' theorem.

Let U be a simply connected open subset of the complex plane containing a finite list of points a₁, ..., a_n, and f a function defined and holomorphic on U {a₁,...,a_n}. Let γ be a closed rectifiable curve in U which does not meet any of the a_k, and denote the winding number of γ around a_k by I(γ, a_k). The line integral of f around γ is equal to 2πi times the sum of residues of f at the points, each counted as many times as γ winds around the point:

∮ γ f ( z ) d z = 2 π i ∑ k = 1 n I ⁡ ( γ , a k ) Res ⁡ ( f , a k ) .

If γ is a positively oriented simple closed curve, I(γ, a_k) = 1 if a_k is in the interior of γ, and 0 if not, so

∮ γ f ( z ) d z = 2 π i ∑ Res ⁡ ( f , a k )

with the sum over those a_k inside γ.

The relationship of the residue theorem to Stokes' theorem is given by the Jordan curve theorem. The general plane curve γ must first be reduced to a set of simple closed curves {γ_i} whose total is equivalent to γ for integration purposes; this reduces the problem to finding the integral of f dz along a Jordan curve γ_i with interior V. The requirement that f be holomorphic on U₀ = U {a_k} is equivalent to the statement that the exterior derivative d(f dz) = 0 on U₀. Thus if two planar regions V and W of U enclose the same subset {a_j} of {a_k}, the regions V W and W V lie entirely in U₀, and hence

∫ V ∖ W d ( f d z ) − ∫ W ∖ V d ( f d z )

is well-defined and equal to zero. Consequently, the contour integral of f dz along γ_j = ∂V is equal to the sum of a set of integrals along paths λ_j, each enclosing an arbitrarily small region around a single a_j — the residues of f (up to the conventional factor 2πi) at {a_j}. Summing over {γ_j}, we recover the final expression of the contour integral in terms of the winding numbers {I(γ, 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.

Example

The integral

∫ − ∞ ∞ e i t x x 2 + 1 d x

arises in probability theory when calculating the characteristic function of the Cauchy distribution. It resists the techniques of elementary calculus but can be evaluated by expressing it as a limit of contour integrals.

Suppose t > 0 and define the contour C that goes along the real line from −a to a and then counterclockwise along a semicircle centered at 0 from a to −a. Take a to be greater than 1, so that the imaginary unit i is enclosed within the curve. Now consider the contour integral

∫ C f ( z ) d z = ∫ C e i t z z 2 + 1 d z .

Since e^itz is an entire function (having no singularities at any point in the complex plane), this function has singularities only where the denominator z² + 1 is zero. Since z² + 1 = (z + i)(z − i), that happens only where z = i or z = −i. Only one of those points is in the region bounded by this contour. Because f(z) is

e i t z z 2 + 1 = e i t z 2 i ( 1 z − i − 1 z + i ) = e i t z 2 i ( z − i ) − e i t z 2 i ( z + i ) ,

the residue of f(z) at z = i is

Res z = i ⁡ f ( z ) = e − t 2 i .

According to the residue theorem, then, we have

∫ C f ( z ) d z = 2 π i ⋅ Res z = i ⁡ f ( z ) = 2 π i e − t 2 i = π e − t .

The contour C may be split into a straight part and a curved arc, so that

∫ s t r a i g h t f ( z ) d z + ∫ a r c f ( z ) d z = π e − t

and thus

∫ − a a f ( z ) d z = π e − t − ∫ a r c f ( z ) d z .

Using some estimations, we have

and

lim a → ∞ π a a 2 − 1 = 0.

Note that, since t > 0 and for complex numbers in the upper halfplane the argument lies between 0 and π, one can estimate

| e i t z | = | e i t | z | ( cos ⁡ ϕ + i sin ⁡ ϕ ) | = | e − t | z | sin ⁡ ϕ + i t | z | cos ⁡ ϕ | = e − t | z | sin ⁡ ϕ ≤ 1.

Therefore

∫ − ∞ ∞ e i t z z 2 + 1 d z = π e − t .

If t < 0 then a similar argument with an arc C′ that winds around −i rather than i shows that

∫ − ∞ ∞ e i t z z 2 + 1 d z = π e t ,

and finally we have

∫ − ∞ ∞ e i t z z 2 + 1 d z = π e − | t | .

(If t = 0 then the integral yields immediately to elementary calculus methods and its value is π.)

Example 2

The fact that π cot(πz) has simple poles with residue one at each integer can be used to compute the sum

∑ n = − ∞ ∞ f ( n ) .

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

1 2 π i ∫ Γ N f ( z ) π cot ⁡ ( π z ) d z = Res z = 0 + ∑ n = − N n ≠ 0 N n − 2 .

The left-hand side goes to zero as N → ∞ since the integrand has order O(N⁻²). On the other hand,

z 2 cot ⁡ ( z 2 ) = 1 − B 2 z 2 2 ! + ⋯ , B 2 = 1 6 .

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

∑ n = 1 ∞ 1 n 2 = π 2 6

which is a proof of the Basel problem.

The same trick can be used to establish

π cot ⁡ ( π z ) = lim N → ∞ ∑ n = − N N ( z − n ) − 1

that is, the Eisenstein series.

We take f(z) = (w − z)⁻¹ with w a non-integer and we shall show the above for w. The difficulty in this case is to show the vanishing of the contour integral at infinity. We have:

∫ Γ N π cot ⁡ ( π z ) z d z = 0

since the integrand is an even function and so the contributions from the contour in the left-half plane and the contour in the right cancel each other out. Thus,

∫ Γ N f ( z ) π cot ⁡ ( π z ) d z = ∫ Γ N ( 1 w − z + 1 z ) π cot ⁡ ( π z ) d z

goes to zero as N → ∞.

See the corresponding article in French Wikipedia for further examples.

References

Residue theorem Wikipedia

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Contents

Example

Example 2

References