Complex varibles:contour integration examples

Transcription

1 omple varibles:ontour integration eamples 1 Problem 1 onsider the problem d If we take the substitution = tan θ then d = sec 2 θdθ, which leads to dθ = π sec 2 θ tan 2 θ + 1 dθ Net we consider the same problem using comple variables and a contour integration in the comple plane. We can recall from our earlier introduction to comple variables that if we have the given integral on a closed contour f(z)dz = 2πi Res(f(z n )) n where z n are the poles of f(z). Now, suppose we have, dz I 2 z where the contour is shown in figure 1 with the curved portion being a semi-circle with an infinite radius. The integrand has poles at Z = i. Then dz I 2 = z = dz z dz z 2 = 2πiRes(f(z = i)) = 2π i + 1 z + i z=i = π and provided which we have to prove. onsidering the integral at R = we substitute z = Re iθ and get I 2 = I dz z = dz z π ire iθ dθ π R R 2 e 2iθ + 1 ire iθ dθ R R 2 e 2iθ + 1 R where we have used the inequality π Z1 + Z2 Z 1 Z 2 And so we find that using contour integration and the residue theorem, d = π Rdθ R 2 1 = πr R R 2 1 1

2 8 Im R= Re Figure 1: ontour for problem 1. 2 Problem 2 onsider Again we replace this with I 2 = z 2 z dz = d d z 2 z dz = 2πi n Res(f(z = z n ) Here the integrand has the following poles, Z = e i π 4, e 3i π 4 and Z = e i π 4, e 3i π 4. The contour remains the same, ecept that now there are two poles within the contour as given above. Thus, the solution is (e i π 4 ) 2 I 2 = 2πi (e i π 4 e 3i π 4 )(e i π 4 e i π 4 )(e i π 4 e 3i π 4 ) + 2πi (e 3i (e 3i π 4 e i π 4 )(e 3i π 4 e i π 4 )(e 3i π 4 e 3i π 4 ) = π 4 ) 2 ( ) i 2πi (e i π 4 e 3i π 4 )(i 2)(e i π 4 e 3i π 4 ) + i (e 3i π 4 e i π 4 )(e 3i π 4 e i π 4 )(i = 2) ( ( )) i 2πi i (e i π 4 e 3i π 4 ) (e i π 4 e 3i π 4 ) + 1 (e 3i π 4 e i = π 4 ) ( 2πi (e 3i π 4 e i π 4 ) + (e i π 4 e 3i ) π 4 ) (e i π 4 e 3i = π 4 ) ( ) 2πi 2 2isin(3π/4) + 2isin(π/4) = 1 + i ( 1 + i) 4 2

3 iπ ( ) 2 2i = π Problem 3: sin/ Integration of sin / from to is an interesting problem 3.1 Method 1 In the first method let us consider e ia d = cos(a) d + i sin(a) d So when Z is real the integral is the imaginary part of the above integral. So if we select a contour as shown below, then on the real ais, Z = and if we can find the values of the other integrals we can find the answer as the imaginary part of the real line integral. Thus, we replace the integral with ɛ z dz = z ɛ dz + z dz + ɛ z dz + R z dz As ɛ tends to zero we find that the first and the third part give us the real line integral e ia d As for R integral, using Jordan s lemma, it goes to zero for a >. With regard to ɛ we have for z dz z = is a simple pole. Thus, one of the theorems says ɛ ɛ z dz = iφ 1 = iπ where 1 is the residue at the pole. φ = π because we go in the W direction. Thus we have z dz = e ia d iπ = since no pole is enclosed. Thus, Matching real to real etc., we have e ia d = iπ sina d = π 3

4 R 3 -R R z= Figure 2: ontour for problem Method 2 In the second method, we will include the pole within the contour. Thus, z dz = ɛ And the result follows. 3.3 Method 3 z dz = z dz + ɛ z dz + ɛ sin(a) d = z z dz + z ɛ dz + iπ = 2πi z dz = πi e ia e ia d 2i z dz + R z dz dz = 2πi(ResF (z = )) e iaz dz The logic here is this: The first integral (what we want) is eactly equal to the second representation. We make it into a comple integral (with real its) such that the comple integral will equal the two prior integrals when z =. Hence the comple integral (with real its) must be a branch of any contour we choose. Net, since the comple integral is a sum then it has to be split 4

5 into its components and then each integral has a singularity at z =. We must then calculate the penalty for integrating through the singularity. The first term is If this is integrated over the same contour The R integral is zero. The second term is dz = dz = dz = If this is integrated over the same contour e iaz dz = dz ɛ dz + w R dz + ɛ dz + w dz = iπ dz + 2i dz = π 2 e iaz dz = e iaz ɛ dz + w R dz + The R integral is infinity. So we take the bottom contour and include the pole. e iaz dz = e iaz dz + ɛ w dz = 2πiRes(F (z = )) Thus So e iaz dz = e iaz π dz + 2 e iaz dz = +π 2 π e iaz dz = +π 2 e iaz dz = π = = ( 1)2πiRes(F (z = )) sin(a) d 5

6 3.4 Method 4 It should be mentioned here that if an integral has to go through a singularity then there is no choice but to go through it using an ɛ contour and see if this is a integrable singularity. There is no avoiding it by deforming the contour. Avoiding the singularity by starting with an indented contour can be done only when the function is analytic in the domain. With a singularity it is not. However, in this case, there is no singularity in sin(a)/ to begin with at =. So when we break it into sin(a) d = e ia e ia d 2i e iaz dz the comple integral is analytic all the way along the real ais. Thus, we can deform the contour as we want. Hence, we choose an indented contour to begin with. The value of the integral must be the same. First let us observe that around the origin ɛ ɛ e iaz dz, withz = ɛ iθ, dz = iɛ iθ dθ we get e iaɛ(cosθ+isinθ) e iaɛ(cosθ+isinθ) ɛ ɛ 2iɛ iθ iɛ iθ dθ = Thus, the function is not singular and has no integral contributions from z =. So nothing new can come integrating through the z = point. We now choose to deform the already indented path. If we continuously deform the path now to + then one of the eponents becomes non-analytic at z =. (If the value of a function becomes infinity then it has a singularity at that point). The singularity can be avoided by deforming the contour, as long as the resultant integrands acquire finite values else they are considered to hit a singularity. We will now be forced to break the integrals into two parts. The first chosen path is indented not going through the z = point since there is no reason to. So one of the eponents goes to zero as we go to infinity in the upper half. The other eponent is not analytic and so has to go to. Here it will go through a singularity. So we deform it the other way. So for the first integral we have dz If this is integrated over the same contour dz = ɛ dz + w R dz + We will hold onto the end points, i.e., + and, and continuously deform the contour upwards, into R at infinity. The value of this integral by Jordan s Lemma is zero. 4 Problem 4 onsider cos cosa 2 a 2 d, a real 6

7 R R R Figure 3: ontour for problem 4. This integral is convergent and well defined at = ±a because the lhospitals rule shows cos cosa sin ±a 2 a 2 = = sina ±a 2 2a To convert the problem to comple domain we consider e iz cosa z 2 a 2 dz, a real where is the contour in figure 3. Here the original function has no singularity. However, once we define the comple function it has a singularity and finally we must take the real part of the answer. So the real argument carries throughout the derivation. Thus, we must stick with the original real line integral for the comple integrand as one branch of the contour. Because, only then after taking the real argument, the original integral is available on the real ais (z = ). Net, no poles are enclosed by and so = = e iz cosa z 2 a 2 dz ( a ɛ1 R + a ɛ2 R a+ɛ 1 a+ɛ 2 ɛ1 ɛ2 R ) e iz cosa z 2 a2 dz (1) From theorems and (Jordans lemma), the integral along R is zero. From theorem 4.3.1, the integrals around the small arcs become e iz ( ) cosa e iz cosa ɛ 1 ɛ1 z 2 a 2 dz = iπ z= a = πsina 2z 2a and ɛ 2 ɛ2 Thus as R (in the PV sense) e iz cosa z 2 a 2 dz = iπ and hence by taking the real part, ( ) e iz cosa 2z cos cosa 2 a 2 d = πsina a πsina a 7 z=a = πsina 2a

8 Discussion: The original integral is not singular at the two denominator zeros. Thus, we could choose the indented contour for the comple integral provided we choose the comple integrand as cos(z) cosa z 2 a 2 dz, a real which also has no singularity at (z = ±a). In this case we can start with an already indented contour and proceed like we did for the earlier problem. However, the function could also be e iz cosa z 2 a 2 dz, a real where we have to take the real value at every step. However, there is now a singularity on the real ais and if we must obtain the original integral we must stick to the real ais its. Thus, the singularity is unavoidable. We cannot choose an indented contour. 5 Problem 5 onsider We write it as d d zdz z = d For finding the singularities of the integrand we have z = R zdz z zdz z z 3 = e (iπ+i2πk) (2) z = e (iπ/3+i2πk/3), k =, 1, 2 (3) z = e (iπ/3), e (iπ), e (i5π/3) (4) Below we take a closed contour which encloses only the first pole. integral goes to zero. Thus, The residue is d Res(e (iπ/3) ) = p q z=e (iπ/3) = zdz z = 2πi Res(e(iπ/3) ) e(iπ/3) 3e (2iπ/3) Thus d zdz z = 2πi Let us look at the integral along the straight line contour. zdz z Set z = re iθ, dz = dre iθ 3 e( iπ/3) One can see that the R z =, r = (5) z =, r = (6) 8

9 R Figure 4: ontour for problem 5. since θ = 2π/3. So zdz z = re iθ dre iθ r 3 e 3iθ + 1 = I(1 e i4π/3 ) = 2πi 3 e( iπ/3) 2π 3 3 rdre 2iθ r = Iei4π/3 6 Problem 5a Let us do the same problem using the method of contour deformation. We consider the original integral I as an integral of a comple integrand. zdz z where the integral is still on the real ais. Now we deform this straight line contour as shown in the figure below. Since we cannot cut across a pole, the contour encircles the pole with two R Figure 5: ontour for problem 5a. opposite going segments. The function is well defined on these segments. Thus we get I c + I R + I two opposites + I round the pole I R and I two opposites are zero so we are left with I c + I round the pole 9

10 I c from the previous method is (mind the sign). Thus, Ie i4π/3 This gives the same answer as before. 7 Problem 6 I(1 e i4π/3 ) = I round the pole = 2πi Res(e (iπ/3) ) onsider We replace it with As before the poles are 2 d z 2 dz z z = z 6 = e (iπ+i2πk) (7) z = e (iπ/6+i2πk/6), k =, 1, 2 (8) z = e (iπ/6), e (iπ/2)... (9) We can do this in more than one way. We take the following contour where is the location of R I Figure 6: ontour for problem 6. the first pole. Only one pole is enclosed. We get z 2 dz z = I + I R + I c We can see that I R is zero. Thus, I + I c = 2πi Res(e iπ/6 ) For I c, we do the following I + z = re iπ/3 dz = dre iπ/3 r 2 dre iπ r eiπ/3 = 2πi = π/3 6e5iπ/6 2 π/3 or π/6 1

11 8 Problem 7 onsider The poles are at See the contour below I + R+i2π R+2iπ e a d 1 + e, a < 1 e az dz = I + [etra terms] 1 + ez R+i2π R+i2π + + R R I + term2 + term3 + term4 z = e iπ+i2πk, k =, 1, 2.. X -R+i 2π X i 3π i 2π X i π R+i 2π -R R Figure 7: ontour for problem term 2 R+i2π e az dz R R+2iπ 1 + e z z = t + i2π dz = dt (1) z = R + i2π t = R (11) z = R + i2π t = R (12) The integral becomes R+i2π R R+2iπ e az dz R 1 + e z = R R e ai2π e at dt R e ai2π e at dt = 1 + et+i2π R R 1 + e t = Ie ia2π 11

12 8.2 term 3 R+i2π e az dz R R 1 + e z z = R + ip dz = idp (13) z = R p = (14) z = R + i2π p = 2π (15) The integral becomes 2π R e ar e aip idp 2π 1 + e R = eip R e (a 1)R eaip idp e R + e ip = since a < 1. Similarly term 4 is also zero. We have the residue at z = iπ left. The residue is e iaπ e iπ Finally considering I and term 2 along with the residue we get I(1 e ia2π ) = 2iπe iaπ I( 2i)sin(aπ)e iaπ = 2iπe iaπ π sin(aπ) 9 Problem 7a We will do the same problem by the method of deformation of contours. onsider e a d 1 + e = e az dz 1 + e z Now it is an integral on the comple domain. If we continuously deform the contour the value of the integral remains the same as long as we do not cut across a pole. Thus, we continuously deform the contour upward and as we hit a pole we deform the contour round it. The length of the contour keeps stretching. There will be vertical parts of the contour going in opposite directions and the poles get encircled in the anti-clockwise direction. We go till infinity upwards and in R. Thus the R integral goes to. Thus, we are left with integrations around the poles, i.e., residues only ( ) e iaπ 2πi e iπ + e3iaπ e iπ + e5iaπ +... eiπ ( ) 2πi e iaπ + e 3iaπ + e 5iaπ +... Using the geometric series formula with r = e ia2π ( ) 2πie iaπ 1 + e 2iaπ + e 4iaπ

13 X i 5π X i 3π R X i π -R R Figure 8: ontour for problem 7a. e iaπ 2πi 1 e 2iaπ e iaπ 2πi e iaπ (e iaπ e iaπ ) π sin(aπ) 1 Problem 8 onsider on a key hole contour given below d (1 + 2 ) R ε Real ais We replace the integral by Figure 9: ontour for problem 8. dz z(1 + z 2 ) The square root function has a branch cut along the positive ais. So we get ɛ + + The R integral is zero. For the ɛ we have ɛ = ɛ 2π ɛ R + ɛ z = ɛe iθ dz = ɛie iθ dθ (16) ɛe iθ idθ ɛe iθ/2 (1 + ɛ 2...) = (17) 13

14 Thus, we are left with ɛ ɛ + = upper + Lower Branch r ( r The residues at i, i. dre iθ re iθ/2 (1 + r 2 e 2iθ ) θ= + r ( r r r 2 dr r(1 + r 2 ) θ= + r dre iθ re iθ/2 (1 + r 2 e 2iθ ) θ=2π r ) dr re iπ (1 + r 2 ) dr r(1 + r 2 = 2πiRes(i, i) ) 1 1 z=i + z= i z(z + i) z(z i) 1 1 z(z + i) z=e iπ/2 + z(z i) z=e i3π/2 1 e iπ/4 2i + 1 e i3π/4 ( 2i) 2 1 ( 1 2i e iπ/4 + 1 ) e iπ/4 2πi 2 1 2i 2cos(π/4)2πi z = re iθ dz = dre iθ (18) ) (19) cos(π/4)π = π 2 11 Problem 9 onsider the integral We replace this with the contour integral z z 2 Using polar coordinates (z 2 1) 1/2 = ρ 1 ρ 2 e i(φ 1+φ 2 )/2, φ 1, φ 2 < 2π where (z 1) 1/2 = ρ 1 e iφ1/2, ρ 1 = z 1 (z + 1) 1/2 = ρ 2 e iφ2/2, ρ 2 = z + 1 (2) 14

15 with this choice of the branch we find < < (z 2 1) 1/2 = 2 1 < < 1 i < < 1, y + i < < 1, y Using the epressions in the contour integral J, it follows that 1 ɛ2 i ɛ1 1+ɛ d + i 1 2 ( ) (z 2 1) 1/2 1 ɛ d ɛ1 ɛ2 R 1 + z 2 dz [( (z 2 1) 1/2 ) ( (z 2 1) 1/2 ) ] = i2π 2z e iπ/2 + 2z e i3π/2 (21) We note that the crosscut integrals vanish ( + L L i ) (z 2 1) 1/2 1 + z 2 dz = e1 z=i e2 R z=-1 z=1 z=-i Figure 1: ontour for problem 9. r2= z+1 r1= z-1 t 1 t 2 z=-1 z=1 Figure 11: Square root function of problem 9. because L and L i are chosen in a region where (z 2 1) 1/2 is continuous and single valued, and L and L i are arbitrarily close to each other. Theorem 4.3.1a shows that ɛi as ɛ, i.e., (z ɛi 2 1) 1/2 1 + z 2 dz < 2π 2π ( ɛ 2 i e2iθ 1 1/2 ) 1 + ɛ 2 ɛ i dθ using the inequalities at the end i e2iθ ɛ 2 i + 1 ɛ 2 i 1 ɛ 2 ɛ i dθ = + 1 2π i 1 ɛ 2 ɛ i dθ as ɛ i (22) i 15

16 The contribution from R is calculated as follows R (z 2 1) 1/2 1 + z 2 dz = 2π (R 2 e 2iθ 1) 1/2 1 + R 2 e 2iθ ire iθ dθ We note that (R 2 e 2iθ 1) 1/2 Re iθ as R because the chosen branch implies z (z 2 1) 1/2 = z. Hence, (z R R 2 1) 1/2 1 + z 2 dz = 2πi alculations of the residues requires computing the correct branch of (z 2 1) 1/2. z 2 1 2e i(3π/4+π/4)/2 2z z=e iπ/2 = = 1 2i 2 z 2 1 2e i(5π/4+7π/4)/2 2z z=e i3π/2 = = 1 (23) 2i 2 Thus Taking ɛ i, R and substituting the above results in the epression for J, we find i d = 2iπ( 2 1) 1 12 Some useful theorems π( 2 1) Theorem Let f(z) = N(z)/D(z) be a rational function such that the degree of D(z) eceeds the degree of N(z) by at least two. Then R f(z)dz = R Theorem 4.2.2: Jordan Lemma Suppose on a circular arc R that is in the upper half of the comple plane, f(z) uniformly as R then R f(z)e ikz dz =, R k > Alternatively Suppose on a circular arc R that is in the lower half of the comple plane, f(z) uniformly as R then R f(z)e ikz dz =, R k > Theorem 4.3.1a Suppose on a contour ɛ, as shown in the figure below, (z z )f(z) uniformly as ɛ, then ɛ f(z)dz = ɛ Theorem 4.3.1b Suppose f(z) has a simple pole at z = z with a residue= 1 then ɛ f(z)dz = iφ 1 ɛ where φ is the angle subtended by the arc and is taken positive in the anti-clockwise sense. 16

17 e f Z 13 Some useful inequalities Figure 12: Figure for a theorem. If z 1 and z 2 are any two comple numbers z 1 + z 2 z 1 + z 2 z 1 + z 2 z 1 z 2 z 1 z 2 z 1 + z 2 z 1 z 2 z 1 z 2 (24) 14 Finding residues at singularities If we have a function that is a ratio of two polynomials, for eample, z, singular at z = 1, and z = 2 (z 1)(z 2) then the residue at z = 1 is found as follows: Res(z = 1) = the residue at z = 2 is found as follows: Res(z = 2) = z (z 1)(z 2) (z 1) z=1 = z (z 1)(z 2) (z 2) z=2 = z (z 2) z=1 = 1 z (z 1) z=2 = 2 If f(z) is a ratio of two polynomials p(z) and q(z) then the residue at z = z o is also found as p(z) Res z=zo q(z) = p(z o) q (z o ) If provided q (z o ). If q(z o ) =, then this is the only way. If m = 1 and if m 2 then f(z) = φ(z) (z z o ) m Res z=zo f(z) = φ(z = z o ) Res z=zo f(z) = 1 d m 1 φ(z = z o ) (m 1)! dz m 1 17