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Method of Residues. Residue theorem. Evaluation of real definite integrals. Cauchy principal value. Summation of series.



ole.gif

Method of Residues. Let f(z) be analytic in a region R, except for a singular point at z = a, as shown in Fig. 1. Cauchy’s theorem tells us that the integral of f(z) around any simple closed curve that doesn’t enclose any singular points is zero. Thus for a curve such as C1 in the figure


             ole1.gif  


What is the value of the integral of f(z) around a curve such as C2 in the figure that does enclose a singular point? The answer:                                                

ole2.gif


where a-1 is the coefficient of 1/(z - a) in the Laurent expansion of f(z) about a. The Laurent expansion about z = a is given by


ole3.gif


where


ole4.gif



If we evaluate 3) for k = -1 we get


             ole5.gif


The same result can be obtained by taking the integral of f(z) in 2)


ole6.gif


and integrating term by term using the following theorem


Theorem 1. Let C be a simple closed curve containing point a in its interior. Then


             ole7.gif


Proof


whereby all terms except the a-1 term drop out.



Def. Residue of an analytic function at an isolated singular point. The residue of an analytic function f(z) at an isolated singular point z0 is


ole8.gif


ole9.gif

where a-1 is the coefficient of (z - a)-1 in the Laurent expansion of f(z) about z0 and C is a simple closed curve enclosing z0.



Residue theorem. Let f(z) be analytic inside and on a simple closed curve C except at the isolated singularities a, b, c, ... inside C which have residues given by ar, br, cr ... . Then


ole10.gif  


See Fig.2.

 

Proof




Calculation of residues


1. Residues at poles. The following theorem gives a simple procedure for the calculation of residues at poles.


Theorem 2. If f(z) has a pole of order m at z = a, then the residue of f(z) at z = a is given by


ole11.gif


if m =1, and by


ole12.gif


if m > 1.


Proof


Note. Formula 6) can be considered a special case of 7) if we define 0! = 1.


Example. Let


             ole13.gif


Then f(z) has two poles: z = -2, a pole of order 1, and z = 3, a pole of order 2.


The residue at z = -2 is given by


             ole14.gif



The residue at z = 3 is given by


             ole15.gif


                                     ole16.gif



Often the order of the pole will not be known in advance. In this case it is still possible to apply Theorem 2 by taking m = 1, 2, 3, ... , in turn, until the first time a finite limit is obtained for a-1. The value of m for which this occurs is the order of the pole and the value of a-1 thus computed is the residue.


2. Residues at essential points. Residues at essential singularities can sometimes be found by using known series expansions.


Example. In the function f(z) = e-1/z, z = 0 is an essential singularity. Using the known series expansion for eu


             ole17.gif


and setting u = -1/z we get the series expansion for e-1/z


             ole18.gif


The residue at z = 0 is the coefficient of 1/z and is -1.


Theorem 3. The power series expansion of a function about a point is unique.


The Laurent expansion about a point is unique. Thus if a series expansion of the Laurent type is found by any process, it must be the Laurent expansion.

 


3. Residues at removable singularities.


Theorem 4. The residue of a function at a removable singularity is zero.




Evaluation of real definite integrals. There are several large and important classes of real definite integrals that can be evaluated by the Method of Residues. We now treat the following types:


Type 1. The integral


ole19.gif   


where the integrand R1 is a finite-valued rational function of sin θ and cos θ for 0 ole20.gif θ ole21.gif 2π.

 


General procedure. Perform the substitution z = e. This substitution transforms integral 8) into the integral


ole22.gif

ole23.gif


where R2(z) is a rational function of z and C is the positively-sensed unit circle centered at z = 0 shown in Fig. 3. The residue theorem then gives the solution of 9) as


ole24.gif  


where Σ r is the sum of the residues of R2(z) at those singularities of R2(z) that lie inside C.


Details. We perform the substitution z = e as follows: Apply the substitution to


            ole25.gif


thus transforming them into


ole26.gif


and then substitute these expressions for sin θ and cos θ as expressed in terms of z and z-1 into R1(sin θ, cos θ). Let f(z) be the function obtained from R1(sin θ, cos θ) by the substitution. From z = e we get dθ = dz/iz. Then R2(z) = f(z)/iz. We can then calculate the residues of those singular points of R2(z) that lie within the unit circle by methods described above and the integral is evaluated as


             ole27.gif





Type 2. The integral

ole28.gif


where the integrand R(x) = P(x)/Q(x) is a rational function that has no poles on the real axis and is such that the degree of the polynomial Q(x) in the denominator is at least two greater than the degree of the polynomial P(x) in the numerator.


Theorem 5. If U(z) is a function which is analytic in the upper half of the z plane except at a finite number of poles, none of which are on the real axis, and if zU(z) converges uniformly to zero when z ole29.gif through values for which 0 ole30.gif arg z ole31.gif π, then ole32.gif is equal to 2πi times the sum of the residues at the poles of U(z) which lie in the upper half plane.


Proof


A function R(x) = P(x)/Q(x) automatically satisfies all the requirements of Theorem 5 if the degree of the denominator exceeds the degree of the numerator by at least two. Thus we have the following corollary.


Corollary 1. If P(x) and Q(x) are real polynomials such that the degree of Q(x) is at least two more than the degree of P(x), and if Q(x) has no real roots, then


             ole33.gif



General procedure. We determine the poles from the zeros of Q(x) and then compute the residues at the poles in the upper half plane by the method of Theorem 2 above.



Example. Evaluate the integral


             ole34.gif


Solution. The integral meets the requirements of Corollary 1. The only poles are at z = ole35.gif ai, ole36.gif bi. Only the poles ai and bi lie in the upper half plane. At z = ai the residue is


              ole37.gif


From symmetry it can be seen that the residue at z = bi must be b/2i(b2 - a2). Thus the value of the integral is

 

             ole38.gif  





Type 3. The integral


             ole39.gif



where the function R(x) = P(x)/Q(x) is a rational function that has no poles on the real axis and the degree of the polynomial in the denominator is at least one greater than that of the polynomial in the numerator.


Solution. Let  


ole40.gif


and consider the function R(z)eimz . Let Σ r be the sum of the residues of R(z)eimz in the upper half plane. Then


             ole41.gif


and I1 and I2 are, respectively, the real and imaginary parts of I. In other words,

 

             ole42.gif


             ole43.gif



Example. Evaluate


             ole44.gif


Solution. Consider the related function


             ole45.gif


Its only pole in the upper half plane is z = i, and its residue there is


             ole46.gif


So


             ole47.gif




ole48.gif

Before proceeding to the next type we need to define the term Cauchy principal value.



The Cauchy principal value of integrals. Let f(x) be a function which is finite at all points of the closed interval [a, b] except at the point x = c, where it becomes infinite. See Fig. 4. Suppose that f(x) is integrable on the intervals [a, c - ε1] and [c + ε2, b] for any positive values of ε1 and ε2. Then we define


             ole49.gif


In some cases the above limit does not exist for ε1 ole50.gif ε2 but does exist if we take ε1 = ε2 = ε. In such a case we define 


             ole51.gif


and call it the Cauchy principal value, or simply principal value, of integral ole52.gif and write


             ole53.gif



Example. The following integral


             ole54.gif


does not exist, however the Cauchy principal value with ε1 = ε2 = ε does exist and equals zero.




Type 4. The integral


             ole55.gif


ole56.gif

where the associated complex function f(z) is a meromorphic function which may have simple poles on the real axis and which approaches zero uniformly on any circular arc centered at z = 0 as the radius of the arc approaches infinity. See Fig. 5.


Let a function f(z) satisfy the inequality |f(z)| < Kρ when z is on a circular arc Γρ of radius ρ, and let Kρ depend only on ρ so that the inequality holds for all z on Γρ, regardless of the argument of z. If Kρ → 0 as ρ → ∞, then f(z) approaches zero uniformly on Γρ as ρ → ∞.


If ρ is allowed to become sufficiently large all poles in the upper half plane will fall within the contour shown in Fig. 5.


Solution. Consider the associated function f(z)eimz = f(z) cos mx + f(z) sin mx. Let Σ r be the sum of the residues of f(z)eimz at all poles lying in the upper half plane (not including those on the real axis). Let Σ r' be the sum of the residues of f(z)eimz at all simple poles lying on the real axis. Then


             ole57.gif



             ole58.gif



             ole59.gif





Type 5. The integral


             ole60.gif


where Q(z) is analytic everywhere in the z plane except at a finite number of poles, none of which lies on the positive half of the real axis.



Solution. The solution is given by the following theorem:


Theorem. Let Q(z) be analytic everywhere in the z plane except at a finite number of poles, none of which lies on the positive half of the real axis. If |zaQ(z)| converges uniformly to zero when z → 0 and when z → ∞, then


             ole61.gif


provided arg z is taken in the interval (-π, π).


It should be noted that unless a is an integer, (-z)a-1 is a multiple-valued function which, using the defining formula az = e z ln a, is given by

 

            (-z)a-1 = e (a-1) ln (-z) = e (a-1)[ln |z| + i arg (-z)}          -π < arg z ole62.gif π




Type 6. The integral


             ole63.gif


where R(z) is a rational function of z which has no poles at z = 0 nor on the positive part of the real axis and k is not an integer.



Solution. To insure convergence of this integral it is necessary that it have the proper behavior at zero and infinity. It is sufficient that ole64.gif . In evaluating the integral we employ the related function z-kR(z) which is a multiple-valued function. The branch of this function that is used is z-k = e -k(ln |z| + i arg z).


Let Σ r be the sum of the residues of z-kR(z) at the poles of R(z). The integral is evaluated by the formula


             ole65.gif


provided z-k = e -k(ln |z| + i arg z).



For types of integrals not covered above, evaluation by the method of residues, when possible at all, usually requires considerable ingenuity in selecting the appropriate contour and in eliminating the integrals over all but the selected portion of the contour.



Special theorems used in evaluating definite integrals. In evaluating definite integrals by the method of residues the following theorems are often useful.

ole66.gif

Theorem 1. Let Γρ be a semicircular arc of radius ρ, in the upper half plane, centered at the origin. See Fig. 6. If |f(z)| ole67.gif M/ρk for z = ρe where k > 1 and M are constants then

                                                                        

             ole68.gif


Proof

 

Theorem 2. Let Γρ be a semicircular arc of radius ρ, in the upper half plane, centered at the origin. See Fig. 6. If |f(z)| ole69.gif M/ρk for z = ρe where k > 1 and M are constants then


             ole70.gif  



Theorem 3. Let Γρ be any circular arc of radius ρ centered at the origin. Let R(z) = P(z)/Q(z) be a rational function in which P(z) and Q(z) are polynomials and the degree of Q(z) is at least two greater than that of P(z). Then


             ole71.gif



Jordan’s lemma. Let Γρ be a semicircular arc of radius ρ, in the upper half plane, centered at the origin. See Fig. 6. Let R(z) = P(z)/Q(z) be a rational function in which P(z) and Q(z) are polynomials and the degree of Q(z) is at least one greater than that of P(z). Then


             ole72.gif



Leibnitz’s rule for differentiation under the integral sign. A method sometimes useful for evaluating integrals utilizes Leibnitz’s rule for differentiation under the integral sign. The rule states that


             ole73.gif


The rule is valid if a and b are constants, α is a real parameter such that α1 ole74.gif α ole75.gif α2 where α1 and α1 are constants, and f(x, α) is continuous and has a continuous partial derivative with respect to α for a ole76.gif x ole77.gif b, α1 ole78.gif α ole79.gif α2. It can be extended to cases where the limits a and b are infinite or dependent on α.

 






Residue theorem used to sum series. . The residue theorem can often be used to sum various types of series. Let us denote an infinite series such as, for example,


             ole80.gif


by the notation


             ole81.gif



The following results are valid under very mild restrictions on f(z) which are usually satisfied whenever the series converge. Note that we replace n by the complex number z in the formula, viewing f(z) as complex.


 


ole82.gif


ole83.gif


ole84.gif


ole85.gif


 




Example. Evaluate


             ole86.gif


where a > 0.


Solution. Let

             ole87.gif


which has simple poles at z = ±ai.


The residue of


             ole88.gif


at z = ai is


             ole89.gif



Similarly the residue at z = -ai is


             ole90.gif


The sum of the two residues is then


             ole91.gif


Consequently,


             ole92.gif





References

  Spiegel. Complex Variables (Schaum)

  Wylie. Advanced Engineering Mathematics

  Hauser. Complex Variables with Physical Applications.


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