cauchy s integral theorem: examples

Physics 4 Spring 17 cauchy s integral theorem: examples lecture notes, spring semester 17 http://www.phys.uconn.edu/ rozman/courses/p4_17s/ Last modified: April 6, 17 Cauchy s theorem states that if f (z) is analytic at all points on and inside a closed complex contour C, then the integral of the function around that contour vanishes: f (z)dz =. (1) C 1 A trigonometric integral Problem: Show that cos(αφ)[cosφ] α 1 dφ = α B(α,α) = α Γ (α) Γ (α). () Solution: Recall the definition of Beta function, B(α,β) = 1 x α 1 (1 x) β 1 dx, (3) Page 1 of 13

Physics 4 Cauchy s integral theorem: examples Spring 17 and consider the integral: J = C [z (1 z)] α 1 dz =, α > 1, (4) where the integration is over closed contour shown in Fig. 1. Since the integrand in Eq. (4) is analytic inside C, On the other hand, J =. (5) J = J I + J II, (6) where J I is the integral along the segment of the positive real axis, x 1; J II is the integral along the circular arc or radius R = 1 centered at z = 1. Along the real axis z = x, dz = dx, thus Along the semi-circular arc where < θ <, and J I = 1 z = 1 + 1 eiθ = 1 ei θ x α 1 (1 x) α 1 dx = B(α,α). (7) ( ) e i θ + e i θ = e i θ θ cos, (8) 1 z = 1 1 eiθ = i e i θ sin θ. (9) y z = 1 ( 1 + e iθ ) Figure 1: Integration contour for Problem 1 II C z = x I 1 x Page of 13

Physics 4 Cauchy s integral theorem: examples Spring 17 Hence, z(1 z) = i e iθ cos θ sin θ = i eiθ sinθ. (1) Therefore, J II = ( i ) α dz = i eiθ dθ. (11) e iαθ (sinθ) α 1 dθ = α e i α e iαθ (sinθ) α 1 dθ, (1) where we used that i = e i. (13) Combining Eqs. (5), (6), (7), and (1) we obtain: α e i α Transforming the expression on the left as following: e iαθ (sinθ) α 1 dθ = B(α,α). (14) e iα(θ ) (sinθ) α 1 dθ = = e iα(θ ) [ ( cos θ )] α 1 dθ (15) e iαφ [cosφ] α 1 dφ (16) = cos(αφ)[cosφ] α 1 dφ (17) finally obtain the relation: cos(αφ)[cosφ] α 1 dφ = α B(α,α) = α Γ (α) Γ (α). (18) Page 3 of 13

Physics 4 Cauchy s integral theorem: examples Spring 17 Euler s log-sine integral. Problem: Show that log(sinx) dx = log. (19) Integral Eq. (19), is called Euler s log-sine integral. It was first evaluated (by Euler) in 1769. Solution: We start by integrating the function f (z), f (z) = log ( 1 e iz), () along the rectangular contour C with the corners at,, +ir, ir, indented at the corners when necessary (see Fig. ), and letting R. J = log ( 1 e iz) dz. (1) C On the one hand, the integrand in Eq. (1) is an analytic function inside C, therefore On the other hand, J =. () J = J I + J II + J III + J IV + J V + J VI, (3) where the subscripts corresponds to integration contours labeled in Fig.. Consider first the integrals J II and J IV. The integrand f (z) is a periodic function with the period, f (z) = f (z + ). (4) Indeed, f (z + ) = log ( 1 e i(z+)) = log ( 1 e iz e i) = log ( 1 e iz) = f (z). (5) Therefore in J II and J IV we have equal integrands but we are integrating in the opposite directions. Therefore, J II = J IV, (6) Page 4 of 13

Physics 4 Cauchy s integral theorem: examples Spring 17 y ir III + ir Figure : Integration contour for Problem IV II ir V VI ir r I r x Page 5 of 13

Physics 4 Cauchy s integral theorem: examples Spring 17 or J II + J IV =. (7) Next, observe that f (z) as y = Im(z) + : f (z) = log ( 1 e i(x+iy)) = log ( 1 e ix e y) e ix e y. (8) Therefore, J III =. (9) Next, let s show that J V lim r C V Indeed, z = r e iθ, dz = i r e iθ dθ, θ : J V = i lim r r Similarly we can show that log ( 1 e ireiθ) e iθ dθ i lim r r Combining Eqs (), (3), (7), (9), (3), and (3), we get that J I = Rewriting the integrand in Eq. (33) as following, log ( 1 e ix) = log [ e ( ix e ix e ix)] [ ( e = log ( i)e ix ix e ix )] i ( log ) 1 e iz dz =. (3) log ( ire iθ) e iθ dθ =. (31) J VI =. (3) log ( 1 e ix) dx =. (33) ) = log (e i(x ) sinx ( = log + i x ) + log(sinx), (34) we obtain [ ( log + i x ) ] + log(sinx) dx =, (35) Page 6 of 13

Physics 4 Cauchy s integral theorem: examples Spring 17 or log(sin x) dx = log. (36) 3 Another Euler integral Problem: evaluate the following integral: I(α) = sin(x) x α dx, < α < 1. (37) Solution: Let s consider the following integral: J(α) = C e iz dz, (38) zα where the integration contour C is sketch in Fig. 3. The integrand in Eq. (38) is an analytic function inside C, therefore J(α) =. (39) On the other hand, J(α) = J I + J II + J III + J IV. (4) where the subscripts corresponds to integration contours labeled in Fig. 3. Let s consider J I, J II, J III, and J IV separately: J I : the integration is along the real axis, so z = x, dz = dx, r x R: so R e ix J I = lim lim r R x α dx = r e ix dx, (41) xα I(α) = ImJ I. (4) Page 7 of 13

Physics 4 Cauchy s integral theorem: examples Spring 17 J II : the integration is counterclockwise along the quarter-circle of radius R, z = Re iθ, dz = i Re iθ dθ, θ : J II = lim i R (1 α) R e i R cosθ e R sinθ e i(1 α)θ dθ. (43) y ir II Figure 3: Integration contour for Problem 3 III ir IV r I R x Page 8 of 13

Physics 4 Cauchy s integral theorem: examples Spring 17 For the absolute value of J II we have the following estimates: J II = lim R R (1 α) e i R cosθ e R sinθ e i(1 α)θ dθ (44) lim R (1 α) e i R cosθ e R sinθ e i(1 α)θ dθ (45) R = lim R (1 α) R = lim R R (1 α) R where we used the inequalities e Rsin(θ) dθ lim R (1 α) R R e R θ dθ (46) e u du = lim R R α ( 1 e R) =, (47) sin(φ) θ e sin(θ) e θ e Rsin(θ) e R θ, (48) that are valid within the integration range θ, and introduce a new integration variable u = R θ. Thus, J II =. (49) J III : the integration is along the imaginary axis, so z = iy, dz = i dy, r y R: r J I = lim lim i (1 α) r R R e y y α dy = ei (1 α) e y y α dy = e i (1 α) Γ (1 α). (5) J IV : the integration is clockwise along the quarter-circle of radius r, z = r e iθ, dz = i r e iθ dθ, θ : J IV = lim i r (1 α) r e i r eiθ e i(1 α)θ dθ lim i r (1 α) r Page 9 of 13 e i(1 α)θ dθ =. (51)

Physics 4 Cauchy s integral theorem: examples Spring 17 Combining Eqs. (39), (4), and (51), we get Taking the imaginary part, and using Eq. (4), we obtain For the case α = 1, sin(x) x sin(x) x α J I = e i (1 α) Γ (1 α). (5) ( ) dx = sin (1 α) Γ (1 α). (53) ( ) dx = lim sin α 1 (1 α) Γ (1 α) lim α 1 (1 α)γ (1 α) = lim Γ ( α) = α 1 Γ (1) =. 4 Fresnel integrals Problem: Assuming that the value of the Gaussian integral is known, evaluate the Fresnel integrals, I = C = e x dx =, (54) cos ( x ) dx (55) and S = sin ( x ) dx. (56) The integrals C and S are named after the Fresnel (French physicist, 1788-187). They were first evaluated by Euler in 1781. Solution: Let s pack C and S together: F C + i S = [ cos ( x ) + i cos ( x )] dx = e ix dx, (57) Page 1 of 13

Physics 4 Cauchy s integral theorem: examples Spring 17 such that and C = ReF (58) S = ImF. (59) Consider the integral J = e i z dz, (6) where C is the contour in the complex plane shown in Fig. 4. Since the integrand in Eq. (6) is analytic inside C, On the other hand, C J =. (61) J = J I + J II + J III, (6) where J I is the integral along the positive real axis, J II is the integral along the circular arc or radius R, θ 4, and J III is the integral from infinity to the origin along the ray that makes the angle θ = 4 with the real axis. Let s consider J I, J II, and J III separately: J I : the integration is along the real axis, so z = x, dz = dx, x : y Figure 4: Integration contour for Problem 4 z = re i 4 III II z = Re iθ z = x I R x Page 11 of 13

Physics 4 Cauchy s integral theorem: examples Spring 17 J I = C I i z e dz = e i x dx = F. (63) J II : the integration is along the circular arc of radius R so z = Re iθ, dz = i Re iθ dθ, z = R e iθ = R ( cos(θ) + i sin(θ) ), θ 4 : J II = C II e 4 i z dz = i R For the absolute value of J II we have the following estimates: J II = 4 R 4 = R = R R e i R cos(θ) e R sin(θ) dθ e R sin(θ) dθ = R R e u du = 4R e i R cos(θ) e R sin(θ) dθ. (64) 4 R e R sin(φ) dφ < R e i R cos(θ) e R sin(θ) dθ (65) e R φ dφ (66) ( 1 e R ) < 4R, (67) where we introduced a new integration variable φ = θ, used the inequalities sin(φ) φ e sin(φ) e φ e R sin(φ) e R φ, (68) that are valid within the integration range φ, and introduce a new integration variable u = R φ. Thus we obtained that Therefore, as R. JII < 4R. (69) J II = (7) Page 1 of 13

Physics 4 Cauchy s integral theorem: examples Spring 17 J III : the integration is along the ray making the angle 4 with the real axis so z = rei 4, z = r e i = i r, dz = e i 4 dr, r <. i z J III = dz = e i 4 C III e e r dr = e i 4 e r dr = e i 4. (71) Combining Eqs. (61), (63), (7), and (71) we obtain: Finally, the Fresnel s integrals are: F = e 4. (7) C = ReF = ( ) cos = 4 8 (73) and S = ImF = ( ) sin = 4 8. (74) Page 13 of 13