Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 8 pot

Result 13.7.1 Integrals from Zero to Infinity. Let f(z) be a single-valued analytic func- tion with only isolated singularities and no singularities on the positive, real axis, [0, ∞). Let a ∈ Z. If the integrals exist then,  ∞ 0 f(x) dx = − n  k=1 Res (f(z) log z, z k ) ,  ∞ 0 x a f(x) dx = ı2π 1 − e ı2πa n  k=1 Res (z a f(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 ) ,  ∞ 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 ) ,  ∞ 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 )  , where z 1 , . . . , z n are the singularities of f(z) and there is a branch cut on the positive real axis with 0 < arg(z) < 2π. 654 13.8 Exploiting Symmetry We have already used symmetry of the integrand to evaluate certain integrals. For f(x) an even function we were able to evaluate  ∞ 0 f(x) dx by extending the range of integration from −∞ to ∞. For  ∞ 0 x α f(x) dx we put a branch cut on the positive real axis and noted that the value of the integrand below the branch cut is a constant multiple of the value of the function above the branch cut. This enabled us to evaluate the real integral with contour integration. In this section we will use other kinds of symmetry to evaluate integrals. We will discover that periodicity of the integrand will produce this symmetry. 13.8.1 Wedge Contours We note that z n = r n e ınθ is periodic in θ with period 2π/n. The real and imaginary parts of z n are odd periodic in θ with period π/n. This observation suggests that certain integrals on the positive real axis may be evaluated by closing the path of integration with a wedge contour. Example 13.8.1 Consider  ∞ 0 1 1 + x n dx 655 where n ∈ N, n ≥ 2. We can evaluate this integral using Result 13.7.1.  ∞ 0 1 1 + x n dx = − n−1  k=0 Res  log z 1 + z n , e ıπ(1+2k)/n  = − n−1  k=0 lim z→ e ıπ(1+2k)/n  (z − e ıπ(1+2k)/n ) log z 1 + z n  = − n−1  k=0 lim z→ e ıπ(1+2k)/n  log z + (z − e ıπ(1+2k)/n )/z nz n−1  = − n−1  k=0  ıπ(1 + 2k)/n n e ıπ(1+2k)(n−1)/n  = − ıπ n 2 e ıπ(n−1)/n n−1  k=0 (1 + 2k) e ı2πk/n = ı2π e ıπ/n n 2 n−1  k=1 k e ı2πk/n = ı2π e ıπ/n n 2 n e ı2π/n −1 = π n sin(π/n) This is a bit grungy. To find a spiffier way to evaluate the integral we note that if we write the integrand as a function of r and θ, it is periodic in θ with period 2π/n. 1 1 + z n = 1 1 + r n e ınθ The integrand along the rays θ = 2π/n, 4π/n, 6π/n, . . . has the same value as the integrand on the real axis. Consider the contour C that is the boundary of the wedge 0 < r < R, 0 < θ < 2π/n. There is one singularity inside the 656 contour. We evaluate the residue there. Res  1 1 + z n , e ıπ/n  = lim z→ e ıπ/n z − e ıπ/n 1 + z n = lim z→ e ıπ/n 1 nz n−1 = − e ıπ/n n We evaluate the integral along C with the residue theorem.  C 1 1 + z n dz = −ı2π e ıπ/n n Let C R be the circular arc. The integral along C R vanishes as R → ∞.      C R 1 1 + z n dz     ≤ 2πR n max z∈C R     1 1 + z n     ≤ 2πR n 1 R n − 1 → 0 as R → ∞ We parametrize the contour to evaluate the desired integral.  ∞ 0 1 1 + x n dx +  0 ∞ 1 1 + x n e ı2π/n dx = −ı2π e ıπ/n n  ∞ 0 1 1 + x n dx = −ı2π e ıπ/n n(1 − e ı2π/n )  ∞ 0 1 1 + x n dx = π n sin(π/n) 657 13.8.2 Box Contours Recall that e z = e x+ıy is periodic in y with period 2π. This implies that the hyperbolic trigonometric functions cosh z, sinh z and tanh z are periodic in y with period 2π and odd periodic in y with period π. We can exploit this property to evaluate certain integrals on the real axis by closing the path of integration with a box contour. Example 13.8.2 Consider the integral  ∞ −∞ 1 cosh x dx =  ı log  tanh  ıπ 4 + x 2  ∞ −∞ = ı log(1) − ı log(−1) = π. We will evaluate this integral using contour integration. Note that cosh(x + ıπ) = e x+ıπ + e −x−ıπ 2 = −cosh(x). Consider the box contour C that is the boundary of the region −R < x < R, 0 < y < π. The only singularity of the integrand in sid e the contour is a first order pole at z = ıπ/2. We evaluate the integral along C with the residue theorem.  C 1 cosh z dz = ı2π Res  1 cosh z , ıπ 2  = ı2π lim z→ıπ/2 z − ıπ/2 cosh z = ı2π lim z→ıπ/2 1 sinh z = 2π 658 The integrals along the sides of the box vanish as R → ∞.      ±R+ıπ ±R 1 cosh z dz     ≤ π max z∈[±R ±R+ıπ]     1 cosh z     ≤ π max y∈[0 π]     2 e ±R+ıy + e ∓R−ıy     = 2 e R − e −R ≤ π sinh R → 0 as R → ∞ The value of the integrand on the top of the box is the negative of its value on the bottom. We take the limit as R → ∞.  ∞ −∞ 1 cosh x dx +  −∞ ∞ 1 −cosh x dx = 2π  ∞ −∞ 1 cosh x dx = π 13.9 Definite Integrals Involving Sine and Cosine Example 13.9.1 For real-valued a, evaluate the integral: f(a) =  2π 0 dθ 1 + a sin θ . What is the value of the integral for complex-valued a. Real-Valued a. For −1 < a < 1, the integrand is bounded, hence the integral exists. For |a| = 1, the integrand has a second order pole on the path of integration. For |a| > 1 the integrand has two first order poles on the path of integration. The integral is divergent for these two cases. Thus we see that the integral exists for −1 < a < 1. 659 For a = 0, the value of the integral is 2π. Now consider a = 0. We make the change of variables z = e ıθ . The real integral from θ = 0 to θ = 2π becomes a contour integral along the unit circle, |z| = 1. We write the sine, cosine and the differential in terms of z. sin θ = z − z −1 ı2 , cos θ = z + z −1 2 , dz = ı e ıθ dθ, dθ = dz ız We write f(a) as an integral along C, the positively oriented unit circle |z| = 1. f(a) =  C 1/(ız) 1 + a(z − z −1 )/(2ı) dz =  C 2/a z 2 + (ı2/a)z − 1 dz We factor the denominator of the integrand. f(a) =  C 2/a (z − z 1 )(z − z 2 ) dz z 1 = ı  −1 + √ 1 − a 2 a  , z 2 = ı  −1 − √ 1 − a 2 a  Because |a| < 1, the second root is outside the unit circle. |z 2 | = 1 + √ 1 − a 2 |a| > 1. Since |z 1 z 2 | = 1, |z 1 | < 1. Thus th e pole at z 1 is inside the contou r and the pole at z 2 is outside. We evaluate the contour integral with the residue theorem. f(a) =  C 2/a z 2 + (ı2/a)z − 1 dz = ı2π 2/a z 1 − z 2 = ı2π 1 ı √ 1 − a 2 660 f(a) = 2π √ 1 − a 2 Complex-Valued a. We note that the integral converges except for real-valued a satisfying |a| ≥ 1. On any closed subset of C \ {a ∈ R | |a| ≥ 1} the integral is uniformly convergent. Thus except for the values {a ∈ R | |a| ≥ 1}, we can differentiate the integral with respect to a. f(a) is analytic in the complex plane except for the set of points on the real axis: a ∈ (−∞. . . − 1] and a ∈ [1 . . . ∞). The value of the analytic function f(a) on the real axis for the interval (−1 . . . 1) is f(a) = 2π √ 1 − a 2 . By analytic continuation we see that the value of f(a) in the complex plane is the branch of the function f(a) = 2π (1 − a 2 ) 1/2 where f(a) is positive, real-valued for a ∈ (−1 . . . 1) and there are branch cuts on the real axis on the intervals: (−∞. . . − 1] and [1 . . . ∞). Result 13.9.1 For evaluating integrals of the form  a+2π a F (sin θ, cos θ) dθ it may be useful to make the change of variables z = e ıθ . This gives us a contour integral along the unit circle about the origin. We can write the sine, cosine and differential in terms of z. sin θ = z −z −1 ı2 , cos θ = z + z −1 2 , dθ = dz ız 661 13.10 Infinite Sums The function g(z) = π cot(πz) has simple poles at z = n ∈ Z. The residues at these points are all unity. Res(π cot(πz), n) = lim z→n π(z −n) cos(πz) sin(πz) = lim z→n π cos(πz) − π(z − n) sin(πz) π cos(πz) = 1 Let C n be the square contour with corners at z = (n + 1/2)(±1 ± ı). Recall that cos z = cos x cosh y −ı sin x sinh y and sin z = sin x cosh y + ı cos x sinh y. First we bound the modulus of cot(z). |cot(z)| =     cos x cosh y −ı sin x sinh y sin x cosh y + ı cos x sinh y     =  cos 2 x cosh 2 y + sin 2 x sinh 2 y sin 2 x cosh 2 y + cos 2 x sinh 2 y ≤  cosh 2 y sinh 2 y = |coth(y)| The hype rbolic cotangent, coth(y), has a simple pole at y = 0 and tends to ±1 as y → ±∞. Along the top and bottom of C n , (z = x ± ı(n + 1/2)), we bound the modulus of g(z) = π cot(πz). |π cot(πz)| ≤ π   coth(π(n + 1/2))   662 [...]... at z = 0 and z = 2 lie inside the contour We evaluate the integral with Cauchy’s residue theorem eız eız dz = 2 Res ,z = 0 2 z 2 (z − 2) (z + ı5) C z (z − 2) (z + ı5) eız + Res ,z = 2 z 2 (z − 2) (z + ı5) eız eız d = 2 + 2 dz (z − 2) (z + ı5) z=0 z (z + ı5) z =2 eız eız d = 2 + 2 dz (z − 2) (z + ı5) z=0 z (z + ı5) z =2 ı (z 2 + (ı7 − 2) z − 5 − ı 12) eız 1 5 e 2 + −ı = 2 2 (z + ı5 )2 (z − 2) 58 116 z=0... 116 z=0 3 ı 1 5 e 2 = 2 − + + −ı 25 20 58 116 π 5 1 6π 1 5 = − + π cos 2 − π sin 2 + ı − + π cos 2 + π sin 2 10 58 29 25 29 58 687 3 We consider the integral e1/z sin(1/z) dz C where C is the positive circle |z| = 1 There is an essential singularity at z = 0 We determine the residue there by expanding the integrand in a Laurent series 1 +O z 1 1 = +O z z2 e1/z sin(1/z) = 1+ 1 z2 1 +O z 1 z3 The residue... Fourier Cosine and Sine Integrals Hint 13 .20 Consider the integral of eıx ıx Hint 13 .21 Show that ∞ −∞ ∞ 1 − cos x 1 − eıx dx = − dx x2 x2 −∞ Hint 13 .22 Show that ∞ 0 eıx sin(πx) ı ∞ dx = − − dx x(1 − x2 ) 2 −∞ x(1 − x2 ) Contour Integration and Branch Cuts Hint 13 .23 Integrate a branch of log2 z/(1 + z 2 ) along the boundary of the domain < r < R, 0 < θ < π Hint 13 .24 6 82 Hint 13 .25 Note that 1 xa... contour Γ and branch for log z ] Exercise 13 .24 Show that ∞ 0 From this derive that xa πa dx = 2 (x + 1) sin(πa) ∞ 0 for − 1 < (a) < 1 ∞ log x dx = 0, (x + 1 )2 0 log2 x 2 dx = (x + 1 )2 3 Hint, Solution Exercise 13 .25 Consider the integral ∞ I(a) = 0 xa dx 1 + x2 1 For what values of a does the integral exist? 2 Evaluate the integral Show that I(a) = π 2 cos(πa /2) 3 Deduce from your answer in part (b)... Exercise 13 .20 Evaluate ∞ −∞ 1 − cos x dx x2 Hint, Solution Exercise 13 .21 Evaluate ∞ 0 sin(πx) dx x(1 − x2 ) Hint, Solution Contour Integration and Branch Cuts Exercise 13 .22 Evaluate the following integrals ∞ ln2 x π3 dx = 1 + x2 8 ∞ ln x dx = 0 1 + x2 1 0 2 0 6 72 Hint, Solution Exercise 13 .23 By methods of contour integration find ∞ 0 [ Recall the trick of considering Hint, Solution Γ x2 dx + 5x +... = 2 cos (2 ), 0 ≤ θ ≤ 2 (See Figure 13.7.) There are first order 686 poles at z = ±1 We evaluate the integral with Cauchy’s residue theorem C dz 1 = 2 Res , z = 1 + Res 2 1 −1 z 1 1 = 2 + z + 1 z=1 z − 1 z=−1 z2 z2 1 , z = −1 −1 =0 2 We consider the integral eız dz, 2 C z (z − 2) (z + ı5) where C is the positive circle |z| = 3 There is a second order pole at z = 0, and first order poles at z = 2 and. .. Axis Exercise 13. 12 Evaluate the following improper integrals ∞ π x2 dx = 2 + 1)(x2 + 4) (x 6 ∞ dx , (x + b )2 + a2 1 0 2 −∞ a>0 Hint, Solution Exercise 13.13 Prove Result 13.4.1 Hint, Solution Exercise 13.14 Evaluate ∞ − −∞ x2 2x +x+1 Hint, Solution Exercise 13.15 Use contour integration to evaluate the integrals 670 ∞ 1 −∞ ∞ 2 −∞ ∞ 3 −∞ dx , 1 + x4 x2 dx , (1 + x2 )2 cos(x) dx 1 + x2 Hint, Solution... (n + a )2 n=−∞ a ∈ Z By Result 13.10.1 with f (z) = 1/(z + a )2 we have ∞ 1 1 = − Res π cot(πz) , −a (n + a )2 (z + a )2 n=−∞ d cot(πz) z→−a dz −π sin2 (πz) − π cos2 (πz) = −π sin2 (πz) = −π lim ∞ 1 2 = (n + a )2 sin2 (πa) n=−∞ Example 13.10 .2 Derive π/4 = 1 − 1/3 + 1/5 − 1/7 + 1/9 − · · · 664 Consider the integral In = 1 2 Cn dw w(w − z) sin w where Cn is the square with corners at w = (n + 1 /2) (±1... Hint, Solution 677 Definite Integrals Involving Sine and Cosine Exercise 13.40 Evaluate the following real integrals π 1 −π √ dθ = 2 1 + sin2 θ π /2 sin4 θ dθ 2 0 Hint, Solution Exercise 13.41 Use contour integration to evaluate the integrals 2 1 0 π 2 −π dθ , 2 + sin(θ) cos(nθ) dθ 1 − 2a cos(θ) + a2 for |a| < 1, n ∈ Z0+ Hint, Solution Exercise 13. 42 By integration around the unit circle, suitably indented,... proper rational function, (numerator has a lower degree than the denominator) and expand in partial fractions Hint 13.6 Hint 13.7 Cauchy Principal Value for Real Integrals Hint 13 .8 680 Hint 13.9 For the third part, does the integrand have a term that behaves like 1/x2 ? Cauchy Principal Value for Contour Integrals Hint 13.10 Expand f (z) in a Laurent series Only the first term will make a contribution . theorem. f(a) =  C 2/ a z 2 + ( 2/ a)z − 1 dz = 2 2/ a z 1 − z 2 = 2 1 ı √ 1 − a 2 660 f(a) = 2 √ 1 − a 2 Complex-Valued a. We note that the integral converges except for real-valued a satisfying. y     =  cos 2 x cosh 2 y + sin 2 x sinh 2 y sin 2 x cosh 2 y + cos 2 x sinh 2 y ≤  cosh 2 y sinh 2 y = |coth(y)| The hype rbolic cotangent, coth(y), has a simple pole at y = 0 and tends to. at z = ıπ /2. We evaluate the integral along C with the residue theorem.  C 1 cosh z dz = 2 Res  1 cosh z , ıπ 2  = 2 lim z→ıπ /2 z − ıπ /2 cosh z = 2 lim z→ıπ /2 1 sinh z = 2 6 58 The integrals