6. Residue calculus. where C is any simple closed contour around z 0 and inside N ε.

Transcription

1 6. Residue calculus Let z 0 be an isolated singularity of f(z), then there exists a certain deleted neighborhood N ε = {z : 0 < z z 0 < ε} such that f is analytic everywhere inside N ε. We define Res(f, z 0 ) = 1 f(z) dz, 2πi C where C is any simple closed contour around z 0 and inside N ε. 1

2 Since f(z) admits a Laurent expansion inside N ε, where then f(z) = n=0 b 1 = 1 2πi a n (z z 0 ) n + n=1 b n (z z 0 ) n, C f(z) dz = Res(f, z 0). Example Res ( 1 (z z 0 ) k, z 0 ) = { 1 if k = 1 0 if k 1 Res(e 1/z,0) = 1 since e 1/z = ! z ! z 2 +, z > 0 Res ( 1 (z 1)(z 2),1 ) = by the Cauchy integral formula. 2

3 Cauchy residue theorem Let C be a simple closed contour inside which f(z) is analytic everywhere except at the isolated singularities z 1, z 2,, z n. C f(z) dz = 2πi[Res(f, z 1) + + Res(f, z n )]. This is a direct consequence of the Cauchy-Goursat Theorem. 3

4 Example Evaluate the integral using z =1 z + 1 z 2 dz (i) direct contour integration, (ii) the calculus of residues, (iii) the primitive function log z 1 z. Solution (i) On the unit circle, z = e iθ and dz = ie iθ dθ. We then have z =1 z + 1 2π z 2 dz = 0 (e iθ +e 2iθ )ie iθ 2π dθ = i 0 (1+e iθ ) dθ = 2πi. 4

5 (ii) The integrand (z + 1)/z 2 has a double pole at z = 0. The Laurent expansion in a deleted neighborhood of z = 0 is simply 1 2, where the coefficient of 1/z is seen to be 1. We have z + 1 z and so z =1 Res ( ) z + 1 z 2,0 z + 1 dz = 2πiRes z2 = 1, ( ) z + 1 z 2,0 = 2πi. (iii) When a closed contour moves around the origin (which is the branch point of the function log z) in the anticlockwise direction, the increase in the value of arg z equals 2π. Therefore, z =1 z + 1 z 2 dz = change in value of ln z + iarg z 1 z in traversing one complete loop around the origin = 2πi. 5

6 Computational formula Let z 0 be a pole of order k. In a deleted neighborhood of z 0, f(z) = Consider n=0 a n (z z 0 ) n + b 1 z z b k (z z 0 ) k, b k 0. g(z) = (z z 0 ) k f(z). the principal part of g(z) vanishes since g(z) = b k + b k 1 (z z 0 ) + b 1 (z z 0 ) k 1 + By differenting (k 1) times, we obtain b 1 = Res(f, z 0 ) = g (k 1) (z 0 ) (k 1)! d k 1 lim z z 0 dz k 1 [ (z z0 ) k f(z) (k 1)! ] n=0 a n (z z 0 ) n+k. if g (k 1) (z) is analytic at z 0 if z 0 is a removable singularity of g (k 1) (z) 6

7 Simple pole k = 1 : Res(f, z 0 ) = lim z z0 (z z 0 )f(z). Suppose f(z) = p(z) q(z) where p(z 0) 0 but q(z 0 ) = 0, q (z 0 ) 0. Res(f, z 0 ) = lim z z0 (z z 0 )f(z) = lim (z z z z0 0 ) = p(z 0) q (z 0 ). p(z 0 ) + p (z 0 )(z z 0 ) + q (z 0 )(z z 0 ) + q (z 0 ) 2! (z z 0 ) 2 + 7

8 Example Find the residue of at all isolated singularities. f(z) = e1/z 1 z Solution (i) There is a simple pole at z = 1. Obviously (ii) Since Res(f,1) = lim(z 1)f(z) = e 1/z = e. z 1 z=1 e 1/z = z + 1 2!z 2 + has an essential singularity at z = 0, so does f(z). Consider e 1/z ( 1 z = (1+z+z2 + ) z + 1 2!z 2 + the coefficient of 1/z is seen to be ! + 1 3! + = e 1 = Res(f,0). ), for 0 < z < 1, 8

9 Example Find the residue of f(z) = z1/2 z(z 2) 2 at all poles. Use the principal branch of the square root function z 1/2. Solution The point z = 0 is not a simple pole since z 1/2 has a branch point at this value of z and this in turn causes f(z) to have a branch point there. A branch point is not an isolated singularity. However, f(z) has a pole of order 2 at z = 2. Note that d Res(f,2) = lim z 2 dz ( ) z 1/2 z = lim z 2 ( z1/2 2z 2 ) = 1 4 2, where the principal branch of 2 1/2 has been chosen (which is 2). 9

10 Example Evaluate Res(g(z)f (z)/f(z), α) if α is a pole of order n of f(z), g(z) is analytic at α and g(α) 0. Solution Since α is a pole of order n of f(z), there exists a deleted neighborhood {z : 0 < z α < ε} such that f(z) admits the Laurent expansion: f(z) = b n (z α) n+ b n 1 (z α) n 1+ + b 1 (z α) + n=0 a n (z α) n, b n 0. Within the annulus of convergence, we can perform termwise differentiation of the above series 10

11 f (z) = nb n (z α) n+1 (n 1)b n (z α) n b 1 (z α) 2 + Provided that g(α) 0, it is seen that (z α) = lim g(z) z α = ng(α) 0, [ n=0 na n (z α) n 1. nb n (z α) n+1 (n 1)b n (z α) n b 1 (z α) 2 + n=0 na n (z α) n 1 b n (z α) n + b n 1 (z α) n b 1 z α + n=0 a n (z α) n so that α is a simple pole of g(z)f (z)/f(z). Furthermore, Remark Res ( g f f, α ) = ng(α). When g(α) = 0, α becomes a removable singularity of gf /f. ] 11

12 Example Suppose an even function f(z) has a pole of order n at α. Within the deleted neighborhood {z : 0 < z α < ε}, f(z) admits the Laurent expansion f(z) = b n (z α) n + + b 1 (z α) + n=0 Since f(z) is even, f(z) = f( z) so that f(z) = f( z) = b n ( z α) n + + b 1 ( z α) + a n (z α) n, b n 0. n=0 a n ( z α) n, which is valid within the deleted neighborhood {z : 0 < z + α < ε}. Hence, α is a pole of order n of f( z). Note that Res(f(z), α) = b 1 and Res(f(z), α) = b 1 so that Res(f(z), α) = Res(f(z), α). For an even function, if z = 0 happens to be a pole, then Res(f,0) = 0. 12

13 Example z =2 tan z z dz = 2πi [ Res ( tan z, π z 2 ) + Res ( tan z, π z 2 )] since the singularity at z = 0 is removable. Observe that π 2 is a ( simple pole and cos z = sin z π ), we have 2 Res ( tan z, π z 2 ) = lim z π 2 ( z π 2 ) tan z z ( z π 2 ) sin z = lim z π 2 = 1 π 2 z [ ( z π 2 = 2 π. ] ) (z π + 2 )

14 ( tan z As tan z/z is even, we deduce that Res, π z 2 result from the previous example. We then have Remark z =2 tan z dz = 0. z Let p(z) = sin z/z, q(z) = cos z, and observe that p 0 and q ( π 2 ) = 1 0, then ) ( ) π 2 = 2 using the π = 2 π, q ( π 2 ) = Res ( tan z, π z 2 ) = p ( π ) / q ( π 2 2 ) = 2 π. 14

15 Example Evaluate Solution C z 2 (z 2 + π 2 ) 2 sin z dz. lim z 0 z sin z z (z 2 + π 2 ) 2 = ( lim z 0 z sin z ) ( lim z 0 so that z = 0 is a removable singularity. ) z (z 2 + π 2 ) 2 = 0 15

16 It is easily seen that z = iπ is a pole of order 2. d Res(f, iπ) = lim z iπ dz [(z iπ)2 f(z)] d = lim z iπ dz [ z 2 (z + iπ) 2 sin z ] 2z(z + iπ)sin z z 2 [(z + iπ)cos z + 2sin z] = lim z iπ (z + iπ) 3 sin 2 z 2sinh π + ( π cosh π sinh π) 1 = 4πsinh 2 = π 4π sinh π + cosh π 4π sinh 2 π. Recall that sin iπ = isinh π and cos iπ = cosh π. Hence, C z 2 (z 2 + π 2 ) 2 sin z dz = 2πiRes(f, iπ) = i 2 ( 1 sinh π + cosh π ) sinh 2. π 16

17 Theorem If a function f is analytic everywhere in the finite plane except for a finite number of singularities interior to a positively oriented simple closed contour C, then f(z) dz = 2πiRes C ( 1 z 2f ( ) 1 z ),0. 17

18 We construct a circle z = R 1 which is large enough so that C is interior to it. If C 0 denotes a positively oriented circle z = R 0, where R 0 > R 1, then where In particular, f(z) = c n = 1 2πi n= c n z n, R 1 < z <, (A) f(z) dz n = 0, ±1, ±2,. C 0 zn+1 2πic 1 = C 0 f(z) dz. How to find c 1? First, we replace z by 1/z in Eq. (A) such that the domain of validity is a deleted neighborhood of z = 0. 18

19 Now 1 z 2f ( ) 1 z = n= c n z n+2 = n= c n 2 z n, 0 < z < 1 R 1, so that c 1 = Res ( 1 z 2f ( ) 1 z ),0. Remark By convention, we may define the residue at infinity by Res(f, ) = 1 f(z) dz = Res 2πi C z 2f,0, z where all singularities in the finite plane are included inside C. With the choice of the negative sign, we have all ( 1 Res(f, z i ) + Res(f, ) = 0. ( ) 1 ) 19

20 Example Evaluate z =2 5z 2 z(z 1) dz. Solution Write f(z) = 5z 2. For 0 < z < 1, z(z 1) so that 5z 2 z(z 1) = 5z 2 1 z 1 z = ( 5 2 ) ( 1 z z 2 ) z Res(f,0) = 2. 20

21 For 0 < z 1 < 1, 5z 2 z(z 1) so that 5(z 1) = z (z 1) ( = ) [1 (z 1) + (z 1) 2 (z 1) 3 + ] z 1 Res(f,1) = 3. Hence, z =2 5z 2 z(z 1) dz = 2πi[Res(f,0) + Res(f,1)] = 10πi. 21

22 On the other hand, consider 1 z 2f ( ) 1 z = 5 2z z(1 z) = 5 2z 1 z 1 z ( ) 5 = z 2 (1 + z + z 2 + ) = 5 z z, 0 < z < 1, so that z =2 5z 2 z(z 1) dz = 2πiRes(f, ) = 2πiRes ( 1 z 2f ( ) 1 z ),0 = 10πi. 22

23 Evaluation of integrals using residue methods A wide variety of real definite integrals can be evaluated effectively by the calculus of residues. Integrals of trigonometric functions over [0,2π] We consider a real integral involving trigonometric functions of the form 2π R(cos θ,sin θ) dθ, 0 where R(x, y) is a rational function defined inside the unit circle z = 1, z = x+iy. The real integral can be converted into a contour integral around the unit circle by the following substitutions: 23

24 z = e iθ, dz = ie iθ dθ = iz dθ, cos θ = eiθ + e iθ = 1 ( z + 1 ), 2 2 z sin θ = eiθ e iθ = 1 ( z 1 ). 2i 2i z The above integral can then be transformed into 2π R(cos θ,sin θ) dθ = 0 z =1 [ = 2πi ( 1 z + z 1 iz R 2, z z 1 2i ) dz sum of residues of 1 iz R ( z + z 1 2, z ) ] z 1 inside z = 1. 2i 24

25 Example Compute I = 2π 0 cos2θ 2 + cos θ dθ. Solution 2π 0 cos2θ 2 + cos θ dθ = i z =1 = i z =1 1 2 ( z z 2 ) ( z + 1 z ) dz z z z 2 (z 2 + 4z + 1) dz. The integrand has a pole of order two at z = 0. Also, the roots of z 2 +4z+1 = 0, namely, z 1 = 2 3 and z 2 = 2+ 3, are simple poles of the integrand. 25

26 Write f(z) = z = 1. z z 2 (z 2 + 4z + 1). Note that z 1 is inside but z 2 is outside 26

27 d Res(f,0) = lim z 0 dz z z 2 + 4z + 1 = lim z 0 3z 3 (z 2 + 4z + 1) (z 4 + 1)(2z + 4) (z 2 + 4z + 1) 2 = 4 Res(f, 2 + 3) = z4 + 1 z 2 / d z= 2+ 3 = ( 2 + 3) ( 2 + 3) 2 dz (z2 + 4z + 1) z= ( 2 + 3) + 4 = 7 3. I = ( i)2πi [ Res(f,0) + Res(f, 2 + 3) ] ( ) = 2π. 27

28 Example Evaluate the integral Solution I = π 0 1 a bcos θ dθ, a > b > 0. Since the integrand is symmetric about θ = π, we have I = 1 2 2π 0 1 2π a bcos θ dθ = 0 2ae iθ b(e 2iθ + 1) dθ. The real integral can be transformed into the contour integral I = i z =1 e iθ 1 bz 2 2az + b dz. The integrand has two simple poles, which are given by the zeros of the denominator. 28

29 Let α denote the pole that is inside the unit circle, then the other pole will be α 1. The two poles are found to be α = a a 2 b 2 1 and b α = a + a 2 b 2. b Since a > b > 0, the two roots are distinct, and α is inside but α 1 is outside the closed contour of integration. We then have I = 1 ib z =1 = 2πi ib Res 1 (z α) (z α 1 ) dz = 2πi ib ( α α 1 ) = 1 (z α) (z α 1 ), α π a 2 b 2. 29

30 Integral of rational functions where f(x) dx, 1. f(z) is a rational function with no singularity on the real axis, 2. lim z zf(z) = 0. It can be shown that f(x) dx = 2πi [sum of residues at the poles of f in the upper half-plane]. 30

31 Integrate f(z) around a closed contour C that consists of the upper semi-circle C R and the diameter from R to R. By the Residue Theorem C f(z) dz = R C R f(z) dz = 2πi [sum of residues at the poles of f inside C]. R f(x) dx + 31

32 As R, all the poles of f in the upper half-plane will be enclosed inside C. To establish the claim, it suffices to show that as R, lim R C R f(z) dz = 0. The modulus of the above integral is estimated by the modulus inequality as follows: C R f(z) dz π 0 f(reiθ ) R dθ max 0 θ π f(reiθ ) R = max zf(z) π, z C R π 0 dθ which goes to zero as R, since lim z zf(z) = 0. 32

33 Example Evaluate the real integral by the residue method. x x 6 dx Solution The complex function f(z) = z4 3 + i has simple poles at i, 1 + z6 2 and 3 + i in the upper half-plane, and it has no singularity on 2 the real axis. The integrand observes the property lim zf(z) = 0. z We obtain [ ( ) ( 3 + i f(x) dx = 2πi Res(f, i) + Res f, + Res f, 3 + i 2 2 )]. 33

34 The residue value at the simple poles are found to be Res(f, i) = 1 = i 6z 6, z=i and so that Res Res x 4 ( ( f, 3 + i 2 f, 3 + i 2 dx = 2πi 1 + x6 ) ) ( = 1 6z = 1 6z z= 3+i 2 z= 3+i 2 i i 12 = = 3 i 12, 3 + i 12, 3 + i 12 ) = 2π 3. 34

35 Integrals involving multi-valued functions Consider a real integral involving a fractional power function 0 f(x) dx, 0 < α < 1, xα 1. f(z) is a rational function with no singularity on the positive real axis, including the origin. 2. lim z f(z) = 0. We integrate φ(z) = f(z) z α along the closed contour as shown. 35

36 The closed contour C consists of an infinitely large circle and an infinitesimal circle joined by line segments along the positive x-axis. 36

37 (i) line segment from ε to R along the upper side of the positive real axis: z = x, ε x R; (ii) the outer large circle C R : z = Re iθ, 0 < θ < 2π; (iii) line segment from R to ε along the lower side of the positive real axis z = xe 2πi, ε x R; (iv) the inner infinitesimal circle C ε in the clockwise direction z = εe iθ, 0 < θ < 2π. 37

38 Establish: lim R C R φ(z) dz C ε φ(z) dz C R φ(z) = 0 and 2π 0 2π 0 lim φ(z) = 0. ε 0 C ε φ(re iθ )Re iθ dθ 2π max z C R zφ(z) φ(εe iθ ) ε dθ 2π max z C ǫ zφ(z). It suffices to show that zφ(z) 0 as either z or z Since lim z f(z) = 0 and f(z) is a rational function, deg (denominator of f(z)) 1 + deg (numerator of f(z)). Further, 1 α < 1, zφ(z) = z 1 α f(z) 0 as z. 2. Since f(z) is continuous at z = 0 and f(z) has no singularity at the origin, zφ(z) = z 1 α f(z) 0 f(0) = 0 as z 0. 38

39 The argument of the principal branch of z α is chosen to be 0 θ < 2π, as dictated by the contour. C φ(z) dz = φ(z) dz + φ(z) dz C R C ǫ R f(x) ǫ + ǫ x α dx + f(xe 2πi ) R x α dx e2απi = 2πi [sum of residues at all the isolated singularities of f enclosed inside the closed contour C]. By taking the limits ǫ 0 and R, the first two integrals vanish. The last integral can be expressed as Combining the results, 0 0 f(x) x α dx = e 2απi e2απi 0 f(x) x α f(x) x α dx = 2πi [sum of residues at all the isolated 1 e 2απi singularities of f in the finite complex plane]. dx. 39

40 Example Evaluate Solution 0 1 dx, 0 < α < 1. (1 + x)xα 1 f(z) = is multi-valued and has an isolated singularity at (1 + z)zα z = 1. By the Residue Theorem, C 1 (1 + z)z α dz = (1 e 2απi ) = 2πi Res ( R ǫ dx (1 + x)x α + 1 (1 + z)zα, 1 ) C R = 2πi e απi. dz (1 + z)z α + C ǫ dz (1 + z)z α 40

41 The moduli of the third and fourth integrals are bounded by 1 C R (1 + z)z α dz 2πR (R 1)R α R α 0 as R, 1 C ǫ (1 + z)z α dz 2πǫ (1 ǫ)ǫ α ǫ1 α 0 as ǫ 0. On taking the limits R and ǫ 0, we obtain so 0 (1 e 2απi ) 0 1 2πi dx = (1 + x)xα e απi; 1 (1 + x)x α dx = 2πi e απi (1 e 2απi ) = π sin απ. 41

42 Example Evaluate the real integral Solution e αx dx, 0 < α < ex The integrand function in its complex extension has infinitely many poles in the complex plane, namely, at z = (2k + 1)πi, k is any integer. We choose the rectangular contour as shown l 1 : y = 0, R x R, l 2 : x = R, 0 y 2π, l 3 : y = 2π, R x R, l 4 : x = R, 0 y 2π. 42

43 The chosen closed rectangular contour encloses only one simple pole at z = πi. 43

44 The only simple pole that is enclosed inside the closed contour C is z = πi. By the Residue Theorem, we have C e αz 1 + e z dz = R R R + R e αx = 2πi Res = 2πi eαz e z e α(r+iy) 2π 1 + e x dx e e α(x+2πi) e x+2πi dx + ( e αz 1 + ez, πi ) z=πi = 2πie απi. 2π R+iy idy e α( R+iy) idy R+iy 1 + e 44

45 Consider the bounds on the moduli of the integrals as follows: 2π 0 e α(r+iy) 2π idy R+iy e e αr e R 1 dy O(e (1 α)r ), 0 2π e α( R+iy) idy R+iy 1 + e 2π 0 e αr 1 e R dy O(e αr ). As 0 < α < 1, both e (1 α)r and e αr tend to zero as R tends to infinity. Therefore, the second and the fourth integrals tend to zero as R. On taking the limit R, the sum of the first and third integrals becomes so (1 e 2απi ) e αx 1 + e x dx = 2πieαπi ; e αx 1 + e x dx = 2πi e απi e απi = π sin απ. 45

46 Example Evaluate x 3 dx. Solution Since the integrand is not an even function, it serves no purpose to extend the interval of integration to (, ). Instead, we consider the branch cut integral Log z C 1 + z 3 dz, where the branch cut is chosen to be along the positive real axis whereby 0 Arg z < 2π. Now C Log z R 1 + z 3 dz = ǫ + ln x ǫ 1 + x 3 dx + R Log z C R 1 + z 3 dz + Log (xe 2πi ) 1 + (xe 2πi ) 3 dx C ǫ Log z 1 + z 3 dz 46

47 = 2πi 3 j=1 Res ( ) Log z 1 + z 3, z j, where z j, j = 1,2,3 are the zeros of 1/(1 + z 3 ). Note that Hence lim R ǫ 0 thus giving Log z C ǫ 1 + z 3 dz Log z C R 1 + z 3 dz C Log z 1 + z 3 dz = 0 0 ( ) ǫln ǫ = O 0 as ǫ 0; 1 ( ) Rln R = O 0 as R. = 2πi R 3 ln x 1 + x 3 dx x 3 dx = 0 j= x 3 dx, Res ( ) Log z 1 + z 3, z j. Log (xe 2iπ ) 1 + (xe 2iπ ) 3 dx 47

48 The zeros of 1 + z 3 are α = e iπ/3, β = e iπ and γ = e 5πi/3. Sum of residues is given by ( ) ( ) ( ) Log z Log z Log z Res 1 + z 3, α + Res 1 + z 3, β + Res 1 + z 3, γ Log α = (α β)(α γ) + Log β (β α)(β γ) + Log γ (γ α)(γ β) [ π3 (β γ) + π(γ α) + 5π 3 (α β)] = i = 2π (α β)(β γ)(γ α) 3 3. Hence, 0 1 2π dx = 1 + x

49 Evaluation of Fourier integrals A Fourier integral is of the form eimx f(x) dx, m > 0, 1. lim z f(z) = 0, 2. f(z) has no singularity along the real axis. Remarks 1. The assumption m > 0 is not strictly essential. The evaluation method works even when m is negative or pure imaginary. 2. When f(z) has singularities on the real axis, the Cauchy principal value of the integral is considered. 49

50 Jordan Lemma We consider the modulus of the integral for λ > 0 C R f(z)e iλz dz π 0 f(reiθ ) e iλreiθ R dθ max z C R f(z) R = 2R max z C R f(z) 2R max z C R f(z) = 2R max z C R f(z) π 0 e λrsin θ dθ π 2 0 e λrsin θ dθ π 2 0 e λr2θ π dθ π 2Rλ (1 e λr ), which tends to 0 as R, given that f(z) 0 as R. 50

51 51

52 To evaluate the Fourier integral, we integrate e imz f(z) along the closed contour C that consists of the upper half-circle C R and the diameter from R to R along the real axis. We then have C eimz f(z) dz = R R eimx f(x) dx + C R e imz f(z) dz. Taking the limit R, the integral over C R vanishes by virtue of the Jordan Lemma. Lastly, we apply the Residue Theorem to obtain eimx f(x) dx = 2πi [sum of residues at all the isolated singularities of f in the upper half-plane] since C encloses all the singularities of f in the upper half-plane as R. 52

53 Example Evaluate the Fourier integral sin2x x 2 + x + 1 dx. Solution It is easy to check that f(z) = 1 z 2 +z+1 has no singularity along the real axis and lim 1 z z 2 +z+1 poles, namely, z = e 2πi 3 in the upper half-plane and e 2πi = 0. The integrand has two simple half-plane. By virtue of the Jordan Lemma, we have 3 in the lower 53

54 sin2x x 2 + x + 1 dx = Im e 2ix x 2 + x + 1 dx = Im C e 2iz z 2 + z + 1 dz, where C is the union of the infinitely large upper semi-circle and its diameter along the real axis. Note that Hence, C e 2iz z 2 + z + 1 sin2x x 2 + x + 1 dz = 2πi Res = 2πi dx = Im ( e 2iz 2z + 1 2πi e 2iz z 2 + z + 1, e2πi 3 2πi z=e e 2ie 2πi 3 2e 2πi 3 ) = 2πi e 2ie2πi 3 2e 2πi = 2 πe 3 sin

55 Example Show that Solution 0 sin x2 dx = 0 cos x2 dx = 1 2 π 2. 55

56 0 = C eiz2 dz = R 0 eix2 dx π 4 R eir2 e iπ/2 e iπ/4 dr. 0 eir2 e 2iθ ire iθ dθ Rearranging R R π/4 (cos 0 x2 +isin x 2 ) dx = e iπ/4 0 e r2 dr e ir2 cos2θ R 2 sin2θ ire iθ dθ. 0 Next, we take the limit R. We recall the well-known result π e iπ 4 0 e r2 dr = 2 eiπ/4 = 1 π i π 2 2. Also, we use the transformation 2θ = φ and observe sin φ 2φ π,0 φ π 2, to obtain 56

57 π/4 0 e ir2 cos2θ R 2 sin 2 θ ire iθ dθ We then obtain = R 2 π/4 0 π/2 0 π/2 e R2 sin2θ R dθ e R2 sin φ dφ R e 2R2 φ/π dφ 2 0 = π 4R (1 e R2 ) 0 as R. π π so that 0 (cos x2 + isin x 2 ) dx = cos x2 dx = 2 + i1 2 0 sin x2 dx = 1 2 π

58 Example Evaluate 0 ln(x 2 + 1) x dx. Hint: Use Log(i x) + Log(i + x) = Log(i 2 x 2 ) = ln(x 2 + 1) + πi. Solution Consider C Log(z + i) z dz around C as shown. 58

59 Log(z + i) The only pole of z 2 in the upper half plane is the simple + 1 pole z = i. Consider ( ) Log(z + i) 2πiRes z 2 + 1, i = 2πi lim z i (z i)log(z + i) (z i)(z + i) Log(z + i) C R z R 0 dz = O R ( (ln R)R R 2 ) = πlog 2i = π ln2 + π2 2 i. 0 as R Log(i x) Log(x + i) Log(z + i) x 2 dx x 2 dx C R z and Log(i x) + Log(i + x) = ln(x 2 + 1) + πi. dz = π ln2 + π2 2 i From so that 0 ln(x 2 + 1) x dx πi x ln(x 2 + 1) x π2 dx = π ln2 + 2 i and 0 dx = π ln2. dx 1 + x 2 = π 2 59

60 Cauchy principal value of an improper integral Suppose a real function f(x) is continuous everywhere in the interval [a, b] except at a point x 0 inside the interval. The integral of f(x) over the interval [a, b] is an improper integral, which may be defined as [ b x0 ǫ ] 1 b f(x) dx = lim f(x) dx + f(x) dx, ǫ 1, ǫ 2 > 0. a ǫ 1,ǫ 2 0 x 0 +ǫ 2 a In many cases, the above limit exists only when ǫ 1 = ǫ 2, and does not exist otherwise. 60

61 Example Consider the following improper integral x 1 dx, show that the Cauchy principal value of the integral exists, then find the principal value. Solution Principal value of 2 1 lim ǫ 0 + = lim ǫ 0 + = lim ǫ 0 1 x 1 [ 1 ǫ 1 ln x 1 Hence, the principal value of ln2. dx exists if the following limit exists. 1 2 x 1 dx + 1 ǫ 1 1+ǫ +ln x 1 1 x 1 dx 2 1+ǫ +[(ln ǫ ln2) + (ln1 ln ǫ)] = ln x 1 ] dx exists and its value is 61

62 Lemma If f has a simple pole at z = c and T r is the circular arc defined by then T r : z = c + re iθ (θ 1 θ θ 2 ), lim f(z) dz = i(θ 2 θ 1 )Res(f, c). r 0 + T r In particular, for the semi-circular arc S r lim r 0 + S r f(z) dz = iπres(f, c). 62

63 Proof Since f has a simple pole at c, f(z) = a 1 z c + a k (z c) k, 0 < z c < R for some R. k=0 } {{ } g(z) For 0 < r < R, T r f(z) dz = a 1 T r 1 z c dz + T r g(z) dz. 63

64 Since g(z) is analytic at c, it is bounded in some neighborhood of z = c. That is, g(z) M for z c < r. For 0 < r < R, and so Finally, so that g(z) dz T r M arc length of T r = Mr(θ 2 θ 1 ) lim g(z) dz = 0. r 0 + T r T r 1 z c dz = θ2 θ 1 1 re iθireiθ dθ = i lim r 0 + θ2 θ 1 dθ = i(θ 2 θ 1 ) T r f(z) dz = a 1 i(θ 2 θ 1 ) = Res(f, c)i(θ 2 θ 1 ). 64

65 Example Compute the principal value of Solution lim R r 1,r xe 2ix x 2 1 dx. The improper integral has singularities at x = ±1. The principal value of the integral is defined to be ( ) 1 r1 xe 2ix + R x 2 1 dx. 1 r2 R + 1+r 1 1+r 2 65

66 Let I 1 = I 2 = I R = Sr1 ze 2iz z 2 1 dz Sr2 ze 2iz z 2 1 dz ze CR 2iz z 2 1 dz. Now, f(z) = ze2iz z 2 1 is analytic inside the above closed contour. By the Cauchy Integral Theorem ( 1 r1 R + 1 r2 ) R xe 2ix + 1+r 1 1+r 2 x 2 1 dx + I 1 + I 2 + I R = 0. 66

67 By the Jordan Lemma, and since z z as z, so lim I R = 0. R Since z = ±1 are simple poles of f, lim I r = iπres(f, 1) = iπ = ( iπ)e 2i /2. lim (z + 1)f(z) z 1 Similarly, lim r 2 0 +I 2 = iπres(f,1) = iπe2i. 2 PV xe 2ix x 2 1 dx = iπe 2i 2 + iπe2i 2 = iπ cos2. 67

68 Poisson integral formula f(z) = 1 f(s) 2πi C s z ds. Here, C is the circle with radius r 0 centered at the origin. Write s = r 0 e iφ and z = re iθ, r > r 0. We choose z 1 such that z 1 z = r0 2 and both z 1 and z lie on the same ray so that z 1 = r2 0 r eiθ = r2 0 z = ss z. 68

69 Since z 1 lies outside C, we have f(z) = 1 2πi = 1 2π ( 1 ) f(s) C s z 1 ds s z ( 1 s s z s ) f(s) dφ. s z 1 2π The integrand can be expressed as 0 s s z 1 1 s/z = s s z + z s z = r2 0 r2 s z 2 and so f(re iθ ) = r2 0 r2 2π 2π 0 f(r 0 e iθ ) s z 2 dφ. 69

70 Now s z 2 = r 2 0 2r 0r cos(φ θ) + r 2 > 0 (from the cosine rule). Taking the real part of f, where f = u + iv, we obtain u(r, θ) = 1 2π 2π 0 r 2 0 r2 r 2 0 2r 0rcos(φ θ) + r 2 }{{} P(r 0,r,φ θ) u(r 0, φ) dφ, r < r 0. Knowing u(r 0, φ) on the boundary, u(r, θ) is uniquely determined. The kernel function P(r 0, r, φ θ) is called the Poisson kernel. ( s P(r 0, r, φ θ) = r2 0 r2 s z 2 = Re s z + z s z ( s = Re s z + z ) s z ( ) s + z = Re which is harmonic for z < r 0. s z ) 70