Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 9 ppt

Since the nearest singularity is at z = −ı, the Laurent series will converge in the annulus 0 < |z − ı| < 2. z 2 − 1 = ((z − ı) + ı) 2 − 1 = (z − ı) 2 + ı2(z − ı) −2 ı/2 z + ı = ı/2 ı2 + (z − ı) = 1/4 1 − ı(z − ı)/2 = 1 4 ∞  n=0  ı(z − ı) 2  n = 1 4 ∞  n=0 ı n 2 n (z − ı) n This geometric series converges for |ı(z −ı)/2| < 1, or |z − ı| < 2. The series expansion of f(z) is f(z) = −ı/2 z − ı − 2 + ı2(z − ı) + (z −ı) 2 + 1 4 ∞  n=0 ı n 2 n (z − ı) n . z 4 z 2 + 1 = −ı/2 z − ı − 2 + ı2(z − ı) + (z −ı) 2 + 1 4 ∞  n=0 ı n 2 n (z − ı) n for |z − ı| < 2 Laurent Series about z = ∞. Since the nearest singularities are at z = ±ı, the Laurent series will converge in 694 the annulus 1 < |z| < ∞. z 4 z 2 + 1 = z 2 1 + 1/z 2 = z 2 ∞  n=0  − 1 z 2  n = 0  n=−∞ (−1) n z 2(n+1) = 1  n=−∞ (−1) n+1 z 2n This geometric series converges for | − 1/z 2 | < 1, or |z| > 1. The series expansion of f(z) is z 4 z 2 + 1 = 1  n=−∞ (−1) n+1 z 2n for 1 < |z| < ∞ Solution 13.7 Method 1: Residue Theorem. We factor P (z). Let m be the number of roots, counting multipl ici ties, that lie inside the contour Γ. We find a simple expression for P  (z)/P (z). P (z) = c n  k=1 (z − z k ) P  (z) = c n  k=1 n  j=1 j=k (z − z j ) 695 P  (z) P (z) = c  n k=1  n j=1 j=k (z − z j ) c  n k=1 (z − z k ) = n  k=1  n j=1 j=k (z − z j )  n j=1 (z − z j ) = n  k=1 1 z − z k Now we do the integration using the residue theorem. 1 ı2π  Γ P  (z) P (z) dz = 1 ı2π  Γ n  k=1 1 z − z k dz = n  k=1 1 ı2π  Γ 1 z − z k dz =  z k inside Γ 1 ı2π  Γ 1 z − z k dz =  z k inside Γ 1 = m Method 2: Fundamental Theorem of Calculus. We factor the polynomial, P (z) = c  n k=1 (z −z k ). Let m be 696 the number of roots, counting multiplicities, that lie inside the contour Γ. 1 ı2π  Γ P  (z) P (z) dz = 1 ı2π [log P (z)] C = 1 ı2π  log n  k=1 (z − z k )  C = 1 ı2π  n  k=1 log(z − z k )  C The value of the logarithm changes by ı2π for the terms in which z k is inside the contour. Its value does not change for the terms in which z k is outside the contour. = 1 ı2π   z k inside Γ log(z − z k )  C = 1 ı2π  z k inside Γ ı2π = m Solution 13.8 1.  C e z (z − π) tan z dz =  C e z cos z (z − π) sin z dz The integrand has first order poles at z = nπ, n ∈ Z, n = 1 and a double pole at z = π. The only pole inside 697 the contour occurs at z = 0. We evaluate the integral with the residue theorem.  C e z cos z (z − π) sin z dz = ı2π Res  e z cos z (z − π) sin z , z = 0  = ı2π lim z=0 z e z cos z (z − π) sin z = −ı2 lim z=0 z sin z = −ı2 lim z=0 1 cos z = −ı2  C e z (z − π) tan z dz = −ı2 2. The integrand has a first order poles at z = 0, −π and a second order pole at z = π inside the contour. The value of the integral is ı2π times the sum of the residues at these points. From the previous part we know that residue at z = 0. Res  e z cos z (z − π) sin z , z = 0  = − 1 π We find the residue at z = −π with the residue formula. Res  e z cos z (z − π) sin z , z = −π  = lim z→−π (z + π) e z cos z (z − π) sin z = e −π (−1) −2π lim z→−π z + π sin z = e −π 2π lim z→−π 1 cos z = − e −π 2π 698 We find the residue at z = π by finding the first few terms in the Laurent series of the integrand. e z cos z (z − π) sin z = ( e π + e π (z − π) + O((z − π) 2 )) (1 + O((z − π) 2 )) (z − π) (−(z − π) + O((z − π) 3 )) = − e π − e π (z − π) + O((z − π) 2 ) −(z − π) 2 + O((z − π) 4 ) = e π (z−π) 2 + e π z−π + O(1) 1 + O((z − π) 2 ) =  e π (z − π) 2 + e π z − π + O(1)   1 + O  (z − π) 2  = e π (z − π) 2 + e π z − π + O(1) With this we see that Res  e z cos z (z − π) sin z , z = π  = e π . The integral is  C e z cos z (z − π) sin z dz = ı2π  Res  e z cos z (z − π) sin z , z = −π  + Res  e z cos z (z − π) sin z , z = 0  + Res  e z cos z (z − π) sin z , z = π   = ı2π  − 1 π − e −π 2π + e π   C e z (z − π) tan z dz = ı  2π e π −2 − e −π  Cauchy Principal Value for Real Integrals 699 Solution 13.9 Consider the integral  1 −1 1 x dx. By the definition of improper integrals we have  1 −1 1 x dx = lim →0 +  − −1 1 x dx + lim δ→0 +  1 δ 1 x dx = lim →0 + [log |x|] − −1 + lim δ→0 + [log |x|] 1 δ = lim →0 + log  − lim δ→0 + log δ This limit diverges. Thus the integral diverges. Now consider the integral  1 −1 1 x − ıα dx where α ∈ R, α = 0. Since the integrand is bounded, the integral exists.  1 −1 1 x − ıα dx =  1 −1 x + ıα x 2 + α 2 dx =  1 −1 ıα x 2 + α 2 dx = ı2  1 0 α x 2 + α 2 dx = ı2  1/α 0 1 ξ 2 + 1 dξ = ı2 [arctan ξ] 1/α 0 = ı2 arctan  1 α  700 Figure 13.8: The real and imaginary part of the integrand for several values of α. Note that the integral exists for all nonzero real α and that lim α→0 +  1 −1 1 x − ıα dx = ıπ and lim α→0 −  1 −1 1 x − ıα dx = −ıπ. The integral exists for α arbitrarily close to zero, but diverges when α = 0. The real part of the integrand is an odd function with two humps that get thinner and taller with decreasing α. The imaginary part of the integrand is an even function with a hump that gets thinner and taller with decreasing α. (See Figure 13.8.)   1 x − ıα  = x x 2 + α 2 ,   1 x − ıα  = α x 2 + α 2 Note that   1 0 1 x − ıα dx → +∞ as α → 0 + and   0 −1 1 x − ıα dx → −∞ as α → 0 − . 701 However, lim α→0   1 −1 1 x − ıα dx = 0 because the two integrals above cancel each other. Now note that when α = 0, the integrand is real. Of course the integral doesn’t converge for this case, but if we could assign some value to  1 −1 1 x dx it would be a real number. Since lim α→0  1 −1   1 x − ıα  dx = 0, This number should be zero. Solution 13.10 1. −  1 −1 1 x 2 dx = lim →0 +   − −1 1 x 2 dx +  1  1 x 2 dx  = lim →0 +   − 1 x  − −1 +  − 1 x  1   = lim →0 +  1  − 1 − 1 + 1   The principal value of the integral does not exist. 702 [...]... dx (x2 + 1 )2 The integrand is analytic on the real axis and has second order poles at z = ±ı Since the integrand decays sufficiently fast at infinity, lim R→∞ R max z∈CR z2 (z 2 + 1 )2 = lim R→∞ R R2 (R2 − 1 )2 =0 we can apply Result 13.4.1 ∞ −∞ Res x2 dx = 2 Res (x2 + 1 )2 z2 ,z = ı (z 2 + 1 )2 ∞ −∞ z2 ,z = ı (z 2 + 1 )2 d z2 (z − ı )2 2 z→ı dz (z + 1 )2 d z2 = lim z→ı dz (z + ı )2 (z + ı )2 2z − z 2 2(z +... integrand is an even function and extend the domain of integration ∞ 0 x2 1 dx = 2 + 1)(x2 + 4) (x 2 ∞ −∞ x2 dx (x2 + 1)(x2 + 4) Next we close the path of integration in the upper half plane Consider the integral along the boundary of the 706 domain 0 < r < R, 0 < θ < π 1 2 C 1 z2 dz = 2 + 1)(z 2 + 4) (z 2 z2 dz C (z − ı)(z + ı)(z − 2) (z + 2) 1 z2 = 2 ,z = ı Res 2 (z 2 + 1)(z 2 + 4) z2 + Res , z = 2. .. Residue Theorem ∞ π 2 0 1 2 ∞ 1 dx = dx 1 + x2 2 −∞ 1 + x2 1 2 ,z = ı = 2 Res 2 1 + z2 1 = ıπ 3 lim z→ı z + ı π3 = 2 Now we return to Equation 13.1 ∞ 2 0 ∞ ln2 x dx + 2 1 + x2 0 π3 ln x dx = 1 + x2 4 We equate the real and imaginary parts to solve for the desired integrals ∞ 0 ln2 x π3 dx = 1 + x2 8 ∞ 0 ln x dx = 0 1 + x2 Solution 13 .25 We consider the branch of the function f (z) = z2 log z + 5z +... 6 2 2 2 2 2 −ı = 2 −√ − − √ (2) (ı4)( 2) 1 + ı 2 2 + 2 ı 2 √ 3 = 2 (1 − ı) 2 32 3π = √ (1 + ı) 8 2 The integral along the circular part of the contour, CR , vanishes as R → ∞ We demonstrate this with the maximum 714 modulus integral bound CR z6 πR z6 dz ≤ max (z 4 + 1 )2 4 z∈CR (z 4 + 1 )2 πR R6 = 4 (R4 − 1 )2 → 0 as R → ∞ Taking the limit R → ∞, we have: ∞ 0 x6 (ıy)6 3π dx + ı dy = √ (1 + ı) 4 2. .. (r + 1 )2 f (r e 2 ) = e 2 a ra (r + 1 )2 Below the branch cut, (z = r e 2 ), we have, Now we use the residue theorem ∞ 0 ra dr + (r + 1 )2 ∞ 0 0 e 2 a ra za dr = 2 Res , −1 2 (z + 1 )2 ∞ (r + 1) ∞ ra d a 1 − e 2 a dr = 2 lim (z ) 2 z→−1 dz (r + 1) 0 ∞ a eıπ(a−1) ra dr = 2 (r + 1 )2 1 − e 2 a 0 ∞ a r − 2 a dr = −ıπa 2 e (r + 1) − eıπa 0 xa πa dx = for − 1 < (a) < 1, a = 0 2 (x + 1) sin(πa) 727 The... + z2 ln2 R + 2 ln R + π 2 ≤ πR R2 − 1 → 0 as R → ∞ Let C be the semi-circle from − to in the upper half plane We show that the integral along C vanishes as → 0 with the maximum modulus integral bound C log2 z log2 z dz ≤ π max z∈C 1 + z 2 1 + z2 ln2 − 2 ln + π 2 ≤π 1− 2 → 0 as → 0 Now we take the limit as → 0 and R → ∞ for the integral along C ∞ 2 0 log2 z π3 dz = − 2 4 C 1+z 2 ∞ 0 (ln r + ıπ )2 π3... (±1 ± ı)/ 2 Of these only (1 + ı)/ 2 lies inside the path of integration We evaluate the contour integral with the Residue Theorem For R > 1: C z6 , z = eıπ/4 (z 4 + 1 )2 d z6 = 2 lim (z − eıπ/4 )2 4 (z + 1 )2 z→eıπ/4 dz d z6 = 2 lim (z − eı3π/4 )2 (z − eı5π/4 )2 (z − eı7π/4 )2 z→eıπ/4 dz z6 dz = 2 Res (z 4 + 1 )2 = 2 lim z→eıπ/4 z6 (z − eı3π/4 )2 (z − eı5π/4 )2 (z − eı7π/4 )2 6 2 2 2 − − − ı3π/4... and the angle range −π /2 < θ < 3π /2 We consider the integral of log2 z/(1 + z 2 ) on this contour C log2 z dz = 2 Res 1 + z2 log2 z ,z = ı 1 + z2 log2 z = 2 lim z→ı z + ı (ıπ /2) 2 = 2 2 π3 =− 4 720 Let CR be the semi-circle from R to −R in the upper half plane We show that the integral along CR vanishes as R → ∞ with the maximum modulus integral bound CR log2 z log2 z dz ≤ πR max z∈CR 1 + z 2. .. the integrand along the semi-circle vanishes as the radius tends to infinity The value of the integral is thus √ 2z −1 1 3 2z −1 , z = 0 + 2 Res ,z = − + ı ıπ Res z2 + z + 1 z2 + z + 1 2 2 − ıπ lim z→0 2 2+z+1 z ∞ − −∞ + 2 lim√ z→(−1+ı 3) /2 2z −1 √ z + (1 + ı 3) /2 2x 2 dx = − √ x2 + x + 1 3 Solution 13.17 1 Consider ∞ −∞ The integrand 1 z 4 +1 x4 1 dx +1 is analytic on the real axis and has isolated.. .2 1 − −1 1 dx = lim →0+ x3 = lim + →0 − −1 − = lim − + →0 1 dx + x3 1 2x2 1 − + − −1 1 dx x3 1 2x2 1 1 1 1 1 + − + 2 2(− )2 2(−1 )2 2(1 )2 2 =0 3 Since f (x) is real analytic, ∞ fn xn f (x) = for x ∈ (−1, 1) n=1 We can rewrite the integrand as f (x) f0 f1 f2 f (x) − f0 − f1 x − f2 x2 = 3+ 2+ + x3 x x x x3 Note that the final term is real analytic . π. 1 2  C z 2 (z 2 + 1)(z 2 + 4) dz = 1 2  C z 2 (z − ı)(z + ı)(z − 2) (z + 2) dz = 2 1 2  Res  z 2 (z 2 + 1)(z 2 + 4) , z = ı  + Res  z 2 (z 2 + 1)(z 2 + 4) , z = 2  = ıπ  z 2 (z. max z∈C R     z 2 (z 2 + 1) 2      = lim R→∞  R R 2 (R 2 − 1) 2  = 0 we can apply Result 13.4.1.  ∞ −∞ x 2 (x 2 + 1) 2 dx = 2 Res  z 2 (z 2 + 1) 2 , z = ı  Res  z 2 (z 2 + 1) 2 , z = ı  =. lim z→ı d dz  (z − ı) 2 z 2 (z 2 + 1) 2  = lim z→ı d dz  z 2 (z + ı) 2  = lim z→ı  (z + ı) 2 2z − z 2 2(z + ı) (z + ı) 4  = − ı 4  ∞ −∞ x 2 (x 2 + 1) 2 dx = π 2 3. Since sin(x) 1 + x 2 7 12