Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 10 ppt

Solution 13.28 Convergence. If x a f(x)  x α as x → 0 for some α > −1 then the integral  1 0 x a f(x) dx will converge absolutely. If x a f(x)  x β as x → ∞ for some β < −1 then the integral  ∞ 1 x a f(x) will converge absolutely. These are sufficient conditions for the absolute convergence of  ∞ 0 x a f(x) dx. Contour Integration. We put a branch cut on the positive real axis and choose 0 < arg(z) < 2π. We consider the integral of z a f(z) on the contour in Figure ??. Let the singularities of f(z) occur at z 1 , . . . , z n . By the residue theorem,  C z a f(z) dz = ı2π n  k=1 Res (z a f(z), z k ) . On the circle of radius , the integrand is o( −1 ). Since the length of C  is 2π, the integral on C  vanishes as  → 0. On the circle of radius R, the integrand is o(R −1 ). Since the length of C R is 2πR, the integral on C R vanishes as R → ∞. The value of the integrand below the branch cut, z = x e ı2π , is f(x e ı2π ) = x a e ı2πa f(x) In the limit as  → 0 and R → ∞ we have  ∞ 0 x a f(x) dx +  0 −∞ x a e ı2πa f(x) dx = ı2π n  k=1 Res (z a f(z), z k ) . 734  ∞ 0 x a f(x) dx = ı2π 1 − e ı2πa n  k=1 Res (z a f(z), z k ) . Solution 13.29 In the interval of uniform convergence of th integral, we can differentiate the formula  ∞ 0 x a f(x) dx = ı2π 1 − e ı2πa n  k=1 Res (z a f(z), z k ) , with respect to a to obtain,  ∞ 0 x a f(x) log x dx = ı2π 1 − e ı2πa n  k=1 Res (z a f(z) log z, z k ) , − 4π 2 a e ı2πa (1 − e ı2πa ) 2 n  k=1 Res (z a f(z), z k ) .  ∞ 0 x a f(x) log x dx = ı2π 1 − e ı2πa n  k=1 Res (z a f(z) log z, z k ) , + π 2 a sin 2 (πa) n  k=1 Res (z a f(z), z k ) , Differentiating the solution of Exercise 13.26 m times with resp ect to a yields  ∞ 0 x a f(x) log m x dx = ∂ m ∂a m  ı2π 1 − e ı2πa n  k=1 Res (z a f(z), z k )  , Solution 13.30 Taking the limit as a → 0 ∈ Z in the solution of Exercise 13.26 yields  ∞ 0 f(x) dx = ı2π lim a→0   n k=1 Res (z a f(z), z k ) 1 − e ı2πa  The numerator vanishes because the sum of all residues of z n f(z) is zero. Thus we can use L’Hospital’s rule.  ∞ 0 f(x) dx = ı2π lim a→0   n k=1 Res (z a f(z) log z, z k ) −ı2π e ı2πa  735  ∞ 0 f(x) dx = − n  k=1 Res (f(z) log z, z k ) This suggests that we could have derived the result directly by considering the integral of f(z) log z on the contour in Figure ??. We put a branch cut on the positive real axis and choose the branch arg z = 0. Recall that we have assumed that f(z) has only isolated singularities and no singularities on the positive real axis, [0, ∞). By the residue theorem,  C f(z) log z dz = ı2π n  k=1 Res (f(z) log z, z = z k ) . By assuming that f(z)  z α as z → 0 where α > −1 the integral on C  will vanish as  → 0. By assuming that f(z)  z β as z → ∞ where β < −1 the integral on C R will vanish as R → ∞. The value of the integrand below the branch cut, z = x e ı2π is f (x)(log x + ı2π). Taking the limit as  → 0 and R → ∞, we have  ∞ 0 f(x) log x dx +  0 ∞ f(x)(log x + ı2π) dx = ı2π n  k=1 Res (f(z) log z, z k ) . Thus we corroborate the result.  ∞ 0 f(x) dx = − n  k=1 Res (f(z) log z, z k ) Solution 13.31 Consider the integral of f(z) log 2 z on the contour in Figure ??. We put a branch cut on the positive real axis and choose the branch 0 < arg z < 2π. Let z 1 , . . . z n be the singularities of f(z). By the residue theorem,  C f(z) log 2 z dz = ı2π n  k=1 Res  f(z) log 2 z, z k  . If f(z)  z α as z → 0 for some α > −1 then the integral on C  will vanish as  → 0. f(z)  z β as z → ∞ for some β < −1 then the integral on C R will vanish as R → ∞. Below the branch cut the integrand is f(x)(log x + ı2π) 2 . 736 Thus we have  ∞ 0 f(x) log 2 x dx +  0 ∞ f(x)(log 2 x + ı4π log x −4π 2 ) dx = ı2π n  k=1 Res  f(z) log 2 z, z k  . −ı4π  ∞ 0 f(x) log x dx + 4π 2  ∞ 0 f(x) dx = ı2π n  k=1 Res  f(z) log 2 z, z k  .  ∞ 0 f(x) log x dx = − 1 2 n  k=1 Res  f(z) log 2 z, z k  + ıπ n  k=1 Res (f(z) log z, z k ) Solution 13.32 Convergence. We conside r  ∞ 0 x a 1 + x 4 dx. Since the integrand behaves like x a near x = 0 we must have (a) > −1. Since the integrand behaves like x a−4 at infinity we must have  (a − 4) < −1. The integral converges for −1 < (a) < 3. Contour Integration. The function f(z) = z a 1 + z 4 has first order poles at z = (±1 ±ı)/ √ 2 and a branch point at z = 0. We could evaluate the real integral by putting a branch cut on the positive real axis with 0 < arg(z) < 2π and integrating f(z) on the contour in Figure 13.12. Integrating on this contour would work because the value of the integrand below the branch cut is a constant times the value of the integrand above the branch cut. After demonstrating that the integrals along C  and C R vanish in the limits as  → 0 and R → ∞ we would see that the value of the integral is a constant times the sum of the residues at the four poles. However, this is not the only, (and not the best), contour that can be used to evaluate the real integral. Consider the value of the integral on the line arg(z) = θ. f(r e ıθ ) = r a e ıaθ 1 + r 4 e ı4θ 737 C R C ε Figure 13.12: Possible path of integration for f(z) = z a 1+z 4 If θ is a integer multiple of π/2 then the integrand is a constant multiple of f(x) = r a 1 + r 4 . Thus any of the contours in Figure 13.13 can be used to evaluate the real integral. The only difference is how many residues we have to calculate. Thus we choose the first contour in Figure 13.13. We put a branch cut on the negative real axis and choose the branch −π < arg(z) < π to satisfy f(1) = 1. We evaluate the integral along C with the Residue Theorem.  C z a 1 + z 4 dz = ı2π Res  z a 1 + z 4 , z = 1 + ı √ 2  Let a = α + ıβ and z = r e ıθ . Note that |z a | = |(r e ıθ ) α+ıβ | = r α e −βθ . 738 C C C C C C R ε ε R ε R Figure 13.13: Possible Paths of Integration for f(z) = z a 1+z 4 The integral on C  vanishes as  → 0. We demonstrate this with the maximum modulus integral bound.      C  z a 1 + z 4 dz     ≤ π 2 max z∈C      z a 1 + z 4     ≤ π 2  α e π|β|/2 1 −  4 → 0 as  → 0 The integral on C R vanishes as R → ∞.      C R z a 1 + z 4 dz     ≤ πR 2 max z∈C R     z a 1 + z 4     ≤ πR 2 R α e π|β|/2 R 4 − 1 → 0 as R → ∞ 739 The value of the integrand on the positive imaginary axis, z = x e ıπ/2 , is (x e ıπ/2 ) a 1 + (x e ıπ/2 ) 4 = x a e ıπa/2 1 + x 4 . We take the limit as  → 0 and R → ∞.  ∞ 0 x a 1 + x 4 dx +  0 ∞ x a e ıπa/2 1 + x 4 e ıπ/2 dx = ı2π Res  z a 1 + z 4 , e ıπ/4   1 − e ıπ(a+1)/2   ∞ 0 x a 1 + x 4 dx = ı2π lim z→ e ıπ/4  z a (z − e ıπ/2 ) 1 + z 4   ∞ 0 x a 1 + x 4 dx = ı2π 1 − e ıπ(a+1)/2 lim z→ e ıπ/4  az a (z − e ıπ/2 ) + z a 4z 3   ∞ 0 x a 1 + x 4 dx = ı2π 1 − e ıπ(a+1)/2 e ıπa/4 4 e ı3π/4  ∞ 0 x a 1 + x 4 dx = −ıπ 2( e −ıπ(a+1)/4 − e ıπ(a+1)/4 )  ∞ 0 x a 1 + x 4 dx = π 4 csc  π(a + 1) 4  Solution 13.33 Consider the branch of f(z) = z 1/2 log z/(z + 1) 2 with a branch cut on the positive real axis and 0 < arg z < 2π. We integrate this function on the contour i n Figure ??. We use the maximum modulus integral bound to show that the integral on C ρ vanishes as ρ → 0.       C ρ z 1/2 log z (z + 1) 2 dz      ≤ 2πρ max C ρ     z 1/2 log z (z + 1) 2     = 2πρ ρ 1/2 (2π −log ρ) (1 − ρ) 2 → 0 as ρ → 0 740 The integral on C R vanishes as R → ∞.      C R z 1/2 log z (z + 1) 2 dz     ≤ 2πR max C R     z 1/2 log z (z + 1) 2     = 2πR R 1/2 (log R + 2π) (R − 1) 2 → 0 as R → ∞ Above the branch cut, (z = x e ı0 ), the integrand is, f(x e ı0 ) = x 1/2 log x (x + 1) 2 . Below the branch cut, (z = x e ı2π ), we have, f(x e ı2π ) = −x 1/2 (log x + ıπ) (x + 1) 2 . Taking the limit as ρ → 0 and R → ∞, the residue theorem gives us  ∞ 0 x 1/2 log x (x + 1) 2 dx +  0 ∞ −x 1/2 (log x + ı2π) (x + 1) 2 dx = ı2π Res  z 1/2 log z (z + 1) 2 , −1  . 2  ∞ 0 x 1/2 log x (x + 1) 2 dx + ı2π  ∞ 0 x 1/2 (x + 1) 2 dx = ı2π lim z→−1 d dz (z 1/2 log z) 2  ∞ 0 x 1/2 log x (x + 1) 2 dx + ı2π  ∞ 0 x 1/2 (x + 1) 2 dx = ı2π lim z→−1  1 2 z −1/2 log z + z 1/2 1 z  2  ∞ 0 x 1/2 log x (x + 1) 2 dx + ı2π  ∞ 0 x 1/2 (x + 1) 2 dx = ı2π  1 2 (−ı)(ıπ) − ı  741 2  ∞ 0 x 1/2 log x (x + 1) 2 dx + ı2π  ∞ 0 x 1/2 (x + 1) 2 dx = 2π + ıπ 2 Equating real and imaginary parts,  ∞ 0 x 1/2 log x (x + 1) 2 dx = π,  ∞ 0 x 1/2 (x + 1) 2 dx = π 2 . Exploiting Symmetry Solution 13.34 Convergence. The integrand, e az e z − e −z = e az 2 sinh(z) , has first order poles at z = ınπ, n ∈ Z. To study convergence, we split the domain of integration.  ∞ −∞ =  −1 −∞ +  1 −1 +  ∞ 1 The principal value integral −  1 −1 e ax e x − e −x dx exists for any a because the integrand has only a first order pole on the path of integration. Now consider the integral on (1 . . . ∞).      ∞ 1 e ax e x − e −x dx     =  ∞ 1 e (a−1)x 1 − e −2x dx ≤ 1 1 − e −2  ∞ 1 e (a−1)x dx This integral converges for a −1 < 0; a < 1. 742 [...]... 4π √ (2 c2 − 1)3 πc = 2 − 1)3 (c G(c) = − 2 2 Res √ 758 √ c2 − 1 is inside the unit circle and −c + √ c2 − 1 is outside the unit circle √ z √ √ G(c) = − 2 2 Res , z = −c − c2 − 1 (z + c + c2 − 1 )2 (z + c − c2 − 1 )2 d z √ = 4π lim √ z→−c− c2 −1 dz (z + c − c2 − 1 )2 2z 1 √ √ − = 4π lim √ z→−c− c2 −1 (z + c − c2 − 1 )2 (z + c − c2 − 1)3 √ c − c2 − 1 − z √ = 4π lim √ z→−c− c2 −1 (z + c − c2 − 1)3 2c √... ı eıθ dθ, 764 θ ∈ (0 2 ) We write sin θ and the differential dθ in terms of z Then we evaluate the integral with the Residue theorem 2 0 1 dθ = 2 + sin θ 1 dz C 2 + (z − 1/z)/( 2) ız 2 = dz 2 + ı4z − 1 C z 2 √ √ = dz 3 z+ı 2 3 C z+ı 2+ √ √ = 2 Res z + ı 2 + 3 z+ı 2 3 , z = ı 2 + 2 = 2 √ 2 3 2 =√ 3 2 First consider the case a = 0 π cos(nθ) dθ = −π 0 2 for n ∈ Z+ for n = 0 Now we consider... at z = ±1 ± 2 The poles at z = −1 + 2 and z = 1 − 2 are inside the path of integration We evaluate the integral with Cauchy’s Residue Formula √ = C z4 ı4z dz = 2 Res − 6z 2 + 1 + Res = −8π + √ √ ı4z , z = −1 + 2 2+1 − 6z √ ı4z ,z = 1 − 2 z 4 − 6z 2 + 1 z4 z−1− z−1− √ √ 2 z √ z−1+ 2 2 z √ z+1− 2 1 1 = −8π − √ − √ 8 2 8 2 √ = 2 2 First we use symmetry to expand the domain of integration π /2 sin4 θ dθ... (0 2 ) and make the change of variables, z = eıθ to integrate along the unit circle in the complex plane G(c) = 1 2 2π 0 dθ (c + cos θ )2 For this change of variables, we have, cos θ = z + z −1 , 2 dz ız dz/(ız) −1 2 C (c + (z + z ) /2) z = − 2 dz 2 2 C (2cz + z + 1) z √ √ = − 2 dz 2 − 1 )2 (z + c − c c2 − 1 )2 C (z + c + G(c) = 1 2 dθ = 757 √ √ If c > 1, then −c − c2 − 1 is outside the unit circle and. .. x2 √ dx 2 2 0 (1 + x ) 1 − x We make the change of variables x = sin ξ to obtain, π /2 sin2 ξ (1 + sin2 ξ) 1 − sin2 ξ 0 π /2 0 π /2 0 1 4 Now we make the change of variables z = eıξ 1 4 sin2 ξ dξ 1 + sin2 ξ 1 − cos (2 ) dξ 3 − cos (2 ) 2 0 cos ξ dξ 1 − cos ξ dξ 3 − cos ξ to obtain a contour integral on the unit circle C 1 − (z + 1/z) /2 3 − (z + 1/z) /2 −ı z dz (z − 1 )2 √ √ dz C z(z − 3 + 2 2)(z − 3 − 2 2)... R 1 (R2 − 1 )2 =0 Then we evaluate the integral with the residue theorem ∞ −∞ 1 ,z = ı (1 + z 2 )2 1 = 2 Res ,z = ı 2 (z + ı )2 (z − ı) d 1 = 2 lim z→ı dz (z + ı )2 2 = 2 lim z→ı (z + ı)3 π = 2 dx = 2 Res (1 + x2 )2 ∞ 0 dx π = 2 )2 4 (1 + x 2 We wish to evaluate ∞ 0 dx x3 + 1 Let the contour C be the boundary of the region 0 < r < R, 0 < θ < 2 /3 We factor the denominator of the integrand to... −c + c2 − 1 is inside the unit circle The integrand has a second order pole inside the path of integration We evaluate the integral with the residue theorem √ z √ , z = −c + c2 − 1 (z + c + c2 − 1 )2 (z + c − c2 − 1 )2 d z √ = 4π lim √ z→−c+ c2 −1 dz (z + c + c2 − 1 )2 1 2z √ √ = 4π lim − √ z→−c+ c2 −1 (z + c + c2 − 1 )2 (z + c + c2 − 1)3 √ c + c2 − 1 − z √ = 4π lim √ z→−c+ c2 −1 (z + c + c2 − 1)3 2c =... e−x 2 2 −∞ e ∞ ıaπ ax e ıπ(1 − e ) dx = x − e−x 2( 1 + eıaπ ) −∞ e ∞ eax π ı(e−ıaπ /2 − eıaπ /2 ) dx = x −x 2 eıaπ /2 + eıaπ /2 −∞ e − e ∞ −∞ eax aπ π dx = tan x − e−x e 2 2 Solution 13.35 1 ∞ dx 1 ∞ dx = 2 2 −∞ (1 + x2 )2 (1 + x2 ) 0 We apply Result 13.4.1 to the integral on the real axis First we verify that the integrand vanishes fast enough in the upper half plane lim R→∞ R max z∈CR 1 (1 + z 2 )2 744... integrand is analytic inside C the integral along C is zero Taking the limit as R → ∞, the integral from r = 0 to ∞ along θ = 0 is equal to the integral from r = 0 to ∞ along θ = π/4 ∞ e −x2 ∞ 0 0 0 ∞ 0 1 2 e−x dx = √ 2 “ 1+ı √ x 2 0 ∞ ∞ e dx = − ∞ ∞ cos(x2 ) dx + 0 1+ı √ dx 2 ∞ 1+ı 2 e−x dx = √ 2 1+ı 2 e−x dx = √ 2 2 2 e−ıx dx 0 ∞ cos(x2 ) − ı sin(x2 ) dx 0 ı sin(x2 ) dx + √ 0 ∞ ∞ cos(x2 ) dx − 2 0... theorem The integrand has first order poles at z = eıπ(1+2k)/6 , 747 k = 0, 1, 2, 3, 4, 5 Three of these poles are in the upper half plane For R > 1, we have 2 Γ 1 dz = 2 Res z6 + 1 k=0 1 , eıπ(1+2k)/6 z6 + 1 2 = 2 lim k=0 z→eıπ(1+2k)/6 z − eıπ(1+2k)/6 z6 + 1 Since the numerator and denominator vanish, we apply L’Hospital’s rule 2 = 2 lim k=0 2 z→eıπ(1+2k)/6 1 6z 5 ıπ e−ıπ5(1+2k)/6 3 k=0 ıπ −ıπ5/6 . z) 2  ∞ 0 x 1 /2 log x (x + 1) 2 dx + 2  ∞ 0 x 1 /2 (x + 1) 2 dx = 2 lim z→−1  1 2 z −1 /2 log z + z 1 /2 1 z  2  ∞ 0 x 1 /2 log x (x + 1) 2 dx + 2  ∞ 0 x 1 /2 (x + 1) 2 dx = 2  1 2 (−ı)(ıπ). 1) 2 dx +  0 ∞ −x 1 /2 (log x + 2 ) (x + 1) 2 dx = 2 Res  z 1 /2 log z (z + 1) 2 , −1  . 2  ∞ 0 x 1 /2 log x (x + 1) 2 dx + 2  ∞ 0 x 1 /2 (x + 1) 2 dx = 2 lim z→−1 d dz (z 1 /2 log z) 2  ∞ 0 x 1 /2 log. ı  741 2  ∞ 0 x 1 /2 log x (x + 1) 2 dx + 2  ∞ 0 x 1 /2 (x + 1) 2 dx = 2 + ıπ 2 Equating real and imaginary parts,  ∞ 0 x 1 /2 log x (x + 1) 2 dx = π,  ∞ 0 x 1 /2 (x + 1) 2 dx = π 2 . Exploiting