The Calculus of Residues

Transcription

1 hapter 7 The alculus of Residues If fz) has a pole of order m at z = z, it can be written as Eq. 6.7), or fz) = φz) = a z z ) + a z z ) a m z z ) m, 7.) where φz) is analytic in the neighborhood of z = z. Now we have seen that if encircles z once in a positive sense, dz z z ) n = πiδ n,, 7.) where the Kronecker δ-symbol is defined by {, m n, δ m,n =, m = n.. 7.3) Proof: By auchy s theorem we may take to be a circle centered on z. On the circle, write z = z + re iθ. Then the integral in Eq. 7.) is i r n π dθ e i n)θ, 7.4) which evidently integrates to zero if n, but is πi if n =. QED. Thus if we integrate the function 7.) on a contour which encloses z, while φz) is analytic on and within, we find fz)dz = πia. 7.5) Because the coefficient of the z z ) power in the Laurent expansion of f plays a special role, we give it a name, the residue of fz) at the pole. If contains a number of poles of f, replace the contour by contours α, β, γ,... encircling the poles singly, as shown in Fig. 7.. The contour integral 63 Version of October 6,

2 64 Version of October 6, HAPTER 7. THE ALULUS OF RESIDUES α γ β Figure 7.: Integration of a function f around the contour which contains only poles of f may be reduced to the integrals around subcontours α, β, γ, etc., each of which contains but a single pole of f. around may be distorted to a sum of disjoint ones around α, β,..., so fz)dz = fz)dz + fz)dz +..., 7.6) α and since each small contour integral gives πi times the reside of the single pole interior to that contour, we have established the residue theorem: If f be analytic on and within a contour except for a number of poles within, fz)dz = πi residues, 7.7) β poles within where the sum is carried out over all the poles contained within. This result is very usefully employed in evaluating definite integrals, as the following examples show. 7. Example onsider the following integral over an angle: π Let us introduce a complex variable according to dθ, < p <. 7.8) p cosθ + p z = e iθ, dz = ie iθ dθ = iz dθ, 7.9) so that cosθ = z + ). 7.) z Therefore, we can rewrite the angular integral as an integral around a closed contour which is a unit circle about the origin: dz iz p z + z) + p

3 7.. A FORMULA FOR THE RESIDUE 65 Version of October 6, dz = i z pz + ) + p z = dz i pz)z p). 7.) The integrand exhibits two poles, one at z = /p > and one at z = p <. Only the latter is inside the contour, so since pz we have from the residue theorem z p = z p + πi i p ) pz p, 7.) p = π p. 7.3) Note that we could have obtained the residue without partial fractioning by evaluating the coefficient of /z p) at z = p: This observation is generalized in the following. pz = z=p p. 7.4) 7. A Formula for the Residue If fz) has a pole of order m at z = z, the residue of that pole is d m a = m )! dz m [z z ) m fz)]. 7.5) z=z The proof follows immediately from Eq. 7.). 7.3 Example This time we consider an integral along the real line, dx R x + ) 3 = lim dx R R x + ) 3, 7.6) where we have made explicit the meaning of the upper and lower limits. We relate this to a contour integral as sketched in Fig. 7.. Thus we have dz R z + ) 3 = R dx x + ) 3 + dz z + ) 3, 7.7)

4 66 Version of October 6, HAPTER 7. THE ALULUS OF RESIDUES R R i i R Figure 7.: The closed contour consists of the portion of the real axis between R and R, and the semicircle of radius R in the upper half plane. Also shown in the figure are the location of the poles of the integrand in Eq. 7.7). where we are to understand that the limit R is to be taken at the end of the calculation. It is easy to see that the integral over the large semicircle vanishes in this limit: dz z + ) 3 = π R i e iθ dθ R e iθ 3, R. 7.8) + ) Hence the integral desired is just the closed contour integral, dz z = πiresidue at i). 7.9) + ) 3 By the formula 7.5) the desired residue is a = d [ ]! dz z i) 3 z i) 3 z + i) z=i 3 = d! dz z + i) 3 z=i = 3) 4)! z + i) 5 so 7.4 Jordan s Lemma z=i = 3 6i, 7.) 3π 8. 7.) The evaluation of a class of integrals depends upon this lemma. If fz) uniformly with respect to argz as z for argz π, and fz) is analytic when z > c > and arg z π, then for α >, lim ρ ρ e iαz fz)dz =, 7.)

5 7.5. EXAMPLE 3 67 Version of October 6, where ρ is a semicircle of radius ρ above the real axis with center at the origin. f. Fig. 7..) Proof: Putting in polar coordinates, π e iαz fz)dz = e iαρ cos θ+iρsin θ) f ρe iθ) ρe iθ i dθ. 7.3) ρ If we take the absolute value of this equation, we obtain the inequality π e iαz fz)dz ρ e αρsin θ f ρe iθ ) ρ dθ < ε π e αρ sin θ ρ dθ, 7.4) if f ρe iθ ) < ε for all θ when ρ is sufficiently large. This is what we mean by going to zero uniformly for large ρ.) Now when θ π θ, sin θ π, 7.5) which is easily verified geometrically. Therefore, the integral on the right-hand side of Eq. 7.4) is bounded as follows, Hence π e αρsin θ ρ dθ < ρ = π α π/ e αρθ/π dθ e αρ ). 7.6) e iαz fz)dz ρ < επ e αρ ) 7.7) α may be made as small as we like by merely choosing ρ large enough so ε ). QED. 7.5 Example 3 onsider the integral The associated contour integral is e iz R z + a dz = R cosx x dx. 7.8) + a e ix x + a dx + e iz z dz, 7.9) + a where the contour is a large semicircle of radius R centered on the origin in the upper half plane, as in Fig. 7.. The only difference here is that the

6 68 Version of October 6, HAPTER 7. THE ALULUS OF RESIDUES pole inside the contour is at ia.) The second integral on the right-hand side vanishes as R by Jordan s lemma. Note carefully that this would not be true if we replace e iz by cosz in the above.) Because only the even part of e ix survives symmetric integration, e ix x + a dx = e iz z + a dz = πi ia eiia) = π a e a. 7.3) Note that if were closed in the lower half plane, the contribution from the infinite semicircle would not vanish. Why?) 7.6 auchy Principal Value To this point we have assumed that the path of integration never encounters any singularities of the integrated function. On the contrary, however, let us now suppose that fx) has simple poles on the real axis, and try to attach meaning to fx)dx. 7.3) For simplicity, suppose fz) has a simple pole at only one point on the real axis, fz) = φz) + a z x, 7.3) where φz) is analytic on the entire real axis. Then we define the auchy) principal value of the integral as x δ ) P fx)dx = lim fx)dx + fx)dx, 7.33) δ + x +δ which means that the immediate neighborhood of the singularity is to be omitted symmetrically. The limit exists because fx) a /x x ) near x = x, which is an odd function. We can apply the residue theorem to such integrals by considering a deformed indented) contour, as shown in Fig For simplicity, suppose the function falls off rapidly enough in the upper half plane so that fz)dz =, 7.34) where is the infinite semicircle in the upper half plane. Then the integral around the closed contour shown in the figure is fz)dz = P fx)dx iπa, 7.35)

7 7.6. AUHY PRINIPAL VALUE 69 Version of October 6, x Figure 7.3: ontour which avoids the singularity along the real axis by passing above the pole. x Figure 7.4: ontour which avoids the singularity along the real axis by passing below the pole. where the second term comes from an explicit calculation in which the simple pole is half encircled in a negative sense giving / the result if the pole were fully encircled in the positive sense). On the other hand, from the residue theorem, fz)dz = πi residues), 7.36) poles UHP where UHP stands for upper half plane. Alternatively, we could consider a differently deformed contour, shown in Fig Now we have fz)dz = P fx)dx + iπa = πi residues) + a, 7.37) so in either case P fx)dx = πi poles UHP poles UHP residues) + πia, 7.38) where the sum is over the residues of the poles above the real axis, and a is the residue of the simple pole on the real axis.

8 7 Version of October 6, HAPTER 7. THE ALULUS OF RESIDUES R k + iǫ k iǫ R Figure 7.5: The closed contour for the integral in Eq. 7.4). Equivalently, instead of deforming the contour to avoid the singularity, one can displace the singularity, x x ± iǫ. Then gx) dx x x iǫ = P dx gx) x x ± iπgx ), 7.39) if g is a regular function on the real axis. [Proof: Homework.] 7.7 Example 4 onsider the integral e iqx q k dq, x >, 7.4) + iǫ which is important in quantum mechanics. We can replace this integral by the contour integral e iqx q k dq, x >, 7.4) + iǫ where the closed contour is shown in Fig The integral over the infinite semicircle is zero according to Jordan s lemma. By redefining ǫ, but not changing its sign, we write the integral as k = + k ) [ ] dq e iqx e iqx = πi q k iǫ) q + k iǫ) q k iǫ) in the end taking ǫ. 7.8 Example 5 We will consider two ways of evaluating q= k iǫ) = πi k e ikx, 7.4) dx + x )

9 7.8. EXAMPLE 5 7 Version of October 6, Figure 7.6: ontour used in the evaluation of the integral 7.44). Shown also is the branch line of the logarithm along the +z axis, and the poles of the integrand. The integrand is not even, so we cannot extend the lower limit to. How can contour methods be applied? 7.8. Method onsider the related integral log z dz, 7.44) + z3 over the contour shown in Fig Here we have chosen the branch line of the logarithm to lie along the +z axis; the discontinuity across it is disc log z = log ρ log ρe iπ = iπ. 7.45) The integral over the large circle is zero, as is the integral over the little circle: Therefore, lim ρ, π log ρe iθ + ρ e 3iθ ρeiθ i dθ =. 7.46) log z πi + z 3 dz = residues). 7.47) poles inside

10 7 Version of October 6, HAPTER 7. THE ALULUS OF RESIDUES To find the sum of the residues, we note that the poles occur at the three cube roots of, namely, e iπ/3, e iπ, and e 5iπ/3, so ) residues) = log e iπ/3 e iπ/3 e iπ e iπ/3 e i5π/3 ) + log e iπ e iπ e iπ/3 e iπ e i5π/3 ) + log e 5iπ/3 e 5iπ/3 e iπ/3 e 5iπ/3 e iπ = i π 3 + ) 3i + + ) 3i 3i or 7.8. Method + iπ + 3i ) 3i + i5π 3 ) 3i + 3i ) 3i + = iπ 3 3i 3 + 3i + iπ i5π 3 3i 3 3i = π [ i) + 4i ] 3i) = π 3 3, 7.48) π ) An alternative method which is simpler algebraically is the following. onsider dz z 3 +, 7.5) where the contour is shown in Fig The integral over the arc of the circle at infinity,, evidently vanishes as the radius of that circle goes to infinity. The integral over is the integral I. The integral over 3 is 3 since e iπ/3) 3 =. Thus dz z 3 + = dxe iπ/3 ) xe iπ/3 ) 3 + = eiπ/3 I, 7.5) dz z 3 + = I e πi/3) = Ie iπ/3 i sin π ) )

11 7.9. EXAMPLE 6 73 Version of October 6, 3 π/3 Figure 7.7: ontour used in the evaluation of Eq. 7.5). The only pole of /z 3 + ) contained within is at z = e iπ/3, the residue of which is so e iπ/3 e iπ e iπ/3 e i5π/3 = e πi/3 e iπ/3 e iπ/3 πie 6πi/3 i) 3 sin π 3 ) sin π 3 e 3iπ/3, 7.53) e iπ/3 eiπ/3, 7.54) or since sin π 3 = sin π 3 = 3, π ) 3 = π , 7.55) the same result 7.49) as found by method. 7.9 Example 6 onsider We may use the contour integral z) µ dz = e iπµ ) xµ dx + z + x x µ dx, < µ <. 7.56) + x e iπµ ) xµ dx + x, 7.57) where is the same contour shown in Fig. 7.6, and because µ is between zero and one it is easily seen that the large circle at infinity and the small circle about the origin both give vanishing contributions. The pole now is at z =, so z) µ dz = πi, 7.58) + z where the phase is measured from the negative real z axis. Thus πi = e iπµ ) e iπµ )) ii sinπµ, 7.59)

12 74 Version of October 6, HAPTER 7. THE ALULUS OF RESIDUES y x R R Figure 7.8: ontour used in integral K, Eq. 7.6). Here the two lines making an angle of π/4 with respect to the real axis are closed with vertical lines at x = ±R, where we will take the limit R. or π sin πµ. 7.6) 7. Example 7 Here we demonstrate a method of evaluating the Gaussian integral, J = onsider the contour integral K = e x dx. 7.6) e iπz cscπz dz, 7.6) where is the contour shown in Fig The equation for the two lines making angles of π/4 with respect to the real axis are z = ± + ρeiπ/4, 7.63) so z = 4 ± ρeiπ/4 + iρ. 7.64)

13 7.. EXAMPLE 8 75 Version of October 6, Within the contour the only pole of cscπz is at z =, which has residue /π, so by the residue theorem Directly, however, K = e iπ/4 dρ exp K = πi = i. 7.65) π [ iπ iρ + ρe iπ/4 + )] csc π ρe iπ/4 + ) 4 [ e iπ/4 dρ exp iπ iρ ρe iπ/4 + )] csc π ρe iπ/4 ),7.66) 4 since the vertical segments give exponentially vanishing contributions as R. ombining these two integrals, we encounter [ exp iπρe iπ/4] cscπ ρe iπ/4 + ) [ exp iπρe iπ/4] cscπ ρe iπ/4 ) = exp iπρe iπ/4) + exp iπρe iπ/4) exp iπρe iπ/4) + exp iπρeiπ/4) =, 7.67) since e ±iπ/ = ±i. Hence K = e iπ/4 e iπ/4 dρ e πρ = i π dxe x, 7.68) so comparing with Eq. 7.65) we have for the Gaussian integral 7.6) 7. Example 8 Our final example is the integral J = π. 7.69) If we make the substitution e x = t, this is the same as log t dt t t = xdx ex. 7.7) dt log t [ t + ]. 7.7) t If we make the further substitution in the first form of Eq. 7.7) we have u = t, log u u du u = dt t, 7.7) du u = log u du, 7.73) u

14 76 Version of October 6, HAPTER 7. THE ALULUS OF RESIDUES If we average the two forms 7.7) and 7.73) we have dt log t t + dt log t t. 7.74) The two integrals here separately are divergent, but the sum is finite. We regulate the two integrals by putting in a large t cutoff: [ Λ lim dt log t ] Λ + dt log t. 7.75) Λ t t The first integral here is elementary, Λ dt log t = t log t Λ = log Λ, 7.76) while the second is evaluated by considering K = dz log z z, 7.77) where again is the contour shown in Fig Now, however, the sole pole is on the positive real axis, so no singularities are contained within, and hence by auchy s theorem K =. This time the contribution of the large circle is not zero: π π Λe iθ idθ log Λe iθ = i dθ [log Λ + iθ] Λeiθ = i [π log Λ + i π) log Λ 3 ] π)3.7.78) The discontinuity of the log across the branch line is log x log xe iπ) = log x log x + iπ) = 4iπ log x + 4π. 7.79) Finally, notice that there is a contribution from the pole at z = below the real axis see Fig. 7.9): Explicitly, the contribution from the small semicircle below the pole is π dθ iρeiθ [ 4iπ log + ρe iθ ) ρe iθ 4π ] = 4iπ 3, 7.8) π as ρ. The desired integral is obtained by taking the imaginary part, Λ Im K = 4π dt log t ] [π t log Λ 8π3 4π 3 =, 7.8) 3 so Λ dt log t t = log Λ π )

15 7.. EXAMPLE 8 77 Version of October 6, Figure 7.9: Portion of integral K, Eq. 7.77), corresponding to the integration below the cut on the real axis. The pole of the integrand at z = contributes here because log iǫ) = iπ. Thus the contribution of the small semicircle to K is +iπiπ) = 4iπ 3, in agreement with Eq. 7.8). Thus averaging this with Eq. 7.76) we obtain π ) A slight check of this procedure comes from computing the real part of K: ReK = 4π P Λ dt t + 4π log Λ =. 7.84)