MATHEMATICAL METHOD IN SCIENCE AND ENGINEERING Episode 11 ppt

COMPLEX TECHNIQUES IN TAKING SOME DEFINITE INTEGRALS 353 a> 1. d8 '7r I=J~ a+cose' Using Equations (13.86) and (13.87) we can write this integral as I= -i dz (13.89) (13.90) The denominator can be factorized as (2 - a) (z - P) i (13.91) where a = -a + (2 - I) 4 , (13.92) p = -a - (a2 - 1) . (13.93) 1 For a > 1 we have la1 < 1 and 101 > 1; thus only the root z = a is present inside the unit circle. We can now use the Cauchy integral theorem to find 1 I = -2i (274 - a-P (13.94) (13.95) Example 13.5. Cornplea: contour integml technique: We now consider the integral We can use Equations (13.86) and (13.87) to write I as a contour integral over the unit circle as (13.96) We can now evaluate this integral by using the residue theorem as I= (-')' (-2'2~2 [ residue of : (z - :) '' at z = 0 ] . (13.97) 2a 22' 354 COMPLEX INTEGRALS AND SERIES Using the binomial formula we can write (13.98) z k=O where the residue we need is the coefficient of the 1/z term. This can be easily found as and the result of the definite integral I becomes (21)! 221 (l!)2. I= - fig. 13.13 Contour for the type I1 integrals 11. Integrals of the type I = s_”, dxR (x) , where R (x) is a rational function of the form ao + a*x+ U2Z2 +. . . + a,xn R(x) = bo + biz + b2x2 +. . . + bmxm ’ a ) With no singular points on the real axis, b ) IR (z)l goes to zero at least as - in the limit as IzI + 00 . 1. 1z21 (13.99) (13.100) (13.101) Under these conditions I has the same value with the complex contour integral I = R (z) dz, COMPLEX TECHNIQUES IN TAKING SOME DEFINITE INTEGRALS 355 fig. 13.14 Contour for Example 13.6 where C is a semicircle in the upper half of the z-plane considered in the limit as the radius goes to infinity (Fig. 13.13). Proof is fairly straightforward if we write I as I=h R(z)dz =lmR(x)dz+i R(z)dz (13.102) and note that the integral over the semicircle vanishes in the limit as the radius goes to infinity. We can now evaluate I using the residue theorem. 03 Example 13.6. Complex contour integral technique: Let us evaluate the integral dx n = 1,2, (1 fX2)" (13.103) Since the conditions of the above technique are satisfied, we write I=f dz c (z+2)n(z-2)n- (13.104) Only the singular point z = 2 is inside the contour C (Fig. 13.14); thus we can write I as ) at z = 21. (13.105) (2 + 2)" (z - 2)" To find the residue we write m (13.106) __ (13.107) k = XAk (z - 2) k=O 356 COMPLEX INTEGRALS AND SERIES fig. 13.15 Contour C in the limit R -+ 00 for type I11 integrals and extract the An-l coefficient as (13.108) n(n+ 1) (n + 2). (2n - 2) (- l)n- 1 - (n- I)! (z + i)2n- I=&. This gives the value of the integral I as (13.109) 27ri (-1)- ' (2n - 2)!i I= (n - I)! 22n-1 (n - I)! 111. Integrals of the type I = s-ma dxR (z) einz, where K. is a real parameter and R (x) is a rational function with a ) No singular points on the real axis, b ) In the limit as IzI + 00, IR (.)I -+ 0 independent of 8. Under these conditions we can write the integral I as the contour integral I = R(z) einzdz, (13.110) where the contour C is shown in Figure 13.15. To show that this is true, we have to show the limit R(z) einzdz + 0. (13.111) We start by taking the moduli of the quantities in the integrand to put an upper limit to this integral as I IpieiOI a. (13.112) COMPLEX TECHNIQUES IN JAKtNG SOME DEFINITE INTEGRALS 357 fig. 13.16 Upper limit calculation We now call the maximum value that R(z) takes in the interval [0,2w] M (P) = m= IR (211 (13.113) and improve this bound as (13.115) Since the straight line segment shown in Figure 13.16, in the interval [0,7r/2] , is always less than the sin 6 function, we can also write Equation (13.115) as IA 5 2pM (p) 1% e-2np$d3. (13.116) 0 This integral can easily be taken to yield (13.117) (13.118) From here we see that in the limit as p t (~f the value of the integral IA goes to zero, that is, 7F IA 5 2pM (p) - (1 - ePnp) , ZA 5 M (p) lc (1 - epnp). 2KP ?r This result is also called Jordan's lemma. 358 COMPLEX INTEGRALS AND SERIES Example 13.7. Complex contouy integral technique: In calculating dis- persion relations we frequently encounter integrals like la f (z) = - 1 dkg (k)eikx. 6 -a Let us consider a case where g (k) is given as i ) For x > 0 we can write (13.119) (13.120) (13.121) In this integral k is now a point in the complex k-plane. Because we have a pole (k = ip) inside our contour, we use the Cauchy integral theorem [Eq. (13.6)] to find ii ) For z < 0, we complete our contour C from below to find (13.122) (13.123) (13.124) (13.125) IV. Integrals of the type I = s," dzzX-l R (z), where a) X # integer, b) R(z) is a rational function with no poles on the positive real axis and the origin, c) In the limit as 121 -+ 0, lzXR (z)I -+ 0 and d) In the limit as 121 -+ 00 , IzXR(z)l + 0. COMPLEX TECHNIQUES IN TAKING SOME DEFINITE INTEGRALS 359 Fig. 13.17 Contour for the integrals of type IV Under these conditions we can evaluate the integral I as (13.126) 7r (-1y residues of [zA- ' R (z)] , - - sin 7rA inside C where C is the closed contour shown in Figure 13.17 Proof: Let us write the integral I as a complex contour integral: i zX-'R (z) dz. In the limit as the radius of the small circle (13.127) goes to zero the integral over the contour Ci goes to zero because of c. Similarly, in the limit as the radius of the large circle goes to infinity the integral over Co goes to zero because of d. This leaves us with We can now evaluate the integral on the left-hand side by using the residue theorem. On the other hand, the right-hand side can be written 360 COMPLEX INTEGRALS AND SERIES as f zX-'R(z)dz *L1++L2 = xX-l R (x) d~ (13.128) Thus we obtain 27ri residues of [z'-'R(z)] inside C dxxx-' R (x) . (13.130) Rearranging this, we write the final result as 7r (-1y residues of [zX-'R(z)] . sin 7rX inside C (13.131) 13.8 GAMMA AND BETA FUNCTIONS 13.8.1 Gamma Function For an important application of the type IV integrals we now introduce the gamma and beta functions, which are frequently encountered in applications. The gamma function is defined for all x values as I- N! N" X[Z + 1][~ + 21 . . . [X + N] r(x) = lim (13.132) Integral definition of the gamma function, even though restricted to x > 0, is also very useful: r (x) = Lmyz-' exp(-y)dy. (13.133) Using integration by parts we can write this as GAMMA AND BETA FUNCTIONS 361 PW (13.134) (13.135) This gives us the formula r = (. - i)r (x- I), ( 1 3.136) which is one of the most important properties of the gamma function. For the positive integer values of x, this formula gives us (13.137) (13.138) (13.139) Besides, if we write we can also define the gamma function for the negative integer arguments. Even though this expression gives infinity for the values of r (0) , r (-1) and for all the other negative integer arguments, their ratios are finite: r(-n) - [-N] [-N+ 11 . [-N - 21 [-N - 11 r (-N) For some n values, the gamma function takes the values: I r(-$)= QJiF I r(i)=i I I r(-1) = *OO I r($) = $& I The inverse of the finite with the limit (13.140) gamma function, l/r (x) , is single valued and always (13.141) 362 COMPLEX INTEGRALS AND SERIES 13.8.2 Beta Function Let us write the multiplication of two gamma functions as r(n + 1) r (m + 1) = e-"u"d~]~ e-"vmdv. lo Using the transformation u = x2 and v = y2, we can write 13.142) 13.143) In plane polar coordinates this becomes r(n+i)r(m+i) (13.145) The first term on the right-hand side is r (m + n + 2) and the second term is called the beta function B (m + 1, n + 1). The beta function is related to the gamma function through the relation (13.146) Another definition of the beta function is obtained by the substitutions sin2 8 = t (13.147) and as (13.148) X t=- 1-2 xmdx 00 .I (1 + x)m+n+2. B(m+ l,n+ 1) = (13.149) [...]... is undefined (divergent) at z = a However, because the problem is only a t x = a, we can modify this integral by first integrating up to an infinitesimally close point, ( a - a), to a and then continue integration on the other side from an arbitrarily close point, ( a + S), to infinity, that is, define the integral I as (13.175) This is called taking the Cauchy principal value of the integral, and it... equilibrium fractal curves and surfaces are encountered, where ordinary mathematical techniques are not sufficient In this regard the relation between fractional calculus and fractals is also being actively investigated Fractional calculus also offers us some useful mathematical techniques in evaluating definite integrals and finding sums of infinite series In this chapter, we introduce some of the basic... analytic function in and on the closed contour C in a simply connected domain 13.4 Find the Laurent expansions of the function about the origin for the regions IzI < 1, 11 2 > 2, and 1 < I < 2 z I Use two different methods and show that the results agree with each other 13.5 Using the path in Figure 13.23 evaluate the integral 13.6 Evaluate the following integrals: 374 COMPLEX INTEGRALS AND SERIES Fig... DIFFERINTEGRALS Compared t o Equation (14.29), the binomial coefficients in this equation are going as and all the terms are positive 14.1.4 Unification of Derivative and Integral Operations for Integer Orders Using Equations (14.29) and (14.36) and also making use of the relation (14.37) we can write a single expression for both the derivative and integral of order nas In this equation n takes integer... 14.2.1 DIFFERINTEGRALS Griinwald's Definition of Differintegrals Considering that the gamma function in the above formula is valid for all n, we obtain the most general and basic definition of differintegral given by Griinwald as In this expression q can take all values A major advantage of this definition (also called the Griinwald-Letnikov definition) is that the differintegral is found by using only... research areas in science and engineering that make use of fractional calculus Chemical analysis of fluids, heat transfer, diffusion, the Schrodinger equation, and material science are some areas where fractional calculus is used Interesting applications t o economy, finance, and earthquake science should also be expected It is well known that in the study of nonlinear situations and in the study of processes... derivatives or integrals On the other hand, evaluation of the infinite series could pose practical problems in applications In this formula even though the gamma function r(-q) is infinite for the pcxsitive values of q, their ratio ' ( j - 9) is finite r(-4 386 FRACTIONAL DERIVATIVES AND INTEGRALS: "DIFFERINTEGRALS" We now show that for a positive integer n and for all 4 values the following relation... be true for (n- 1) and then showing it for n DIFFERINTEGRALS 14.2.2 387 Riemann-Liouville Definition of Differintegrals Another commonly used definition of the differintegral is given by Riemann and Liouville Assume that the following integral is given: In( %) = L z ( x - ~n-lf(t)@, (14.48) where n is a n integer greater than zero and a is a constant Using the formula (14.49) , we find the derivative... ) is analytic in the upper half z-plane, that is as IzI + 00, f(z) -+ 0 for y > 0, 366 COMPLEX INTEGRALS AND SERIES t' fig 13.18 Contour G for the Cauchy principal value integral we can evaluate the Cauchy principal value of the integral (13.174) by using the contour in Figure 13.18 In this case we write (13.177) z] and evaluate this integral by using the residue theorem as dz = 27~2 inside C residues... n = integer > 0 Show that x x2 23 n=O where the double factorial means 13.16 Use the factorization method (Chapter 9) to show that the spherical Hankel functions of the first kind, (1) h l can be expressed as Hint First define in - -Ji fin1, 378 COMPLEX INTEGRALS AND SERIES Using this result, define hil)(z) a contour integral in the complex j’-plane as (j’= t’ z ’ ,where s) + d I d dt xdx‘ Indicate . this integral by first integrating up to an infinitesimally close point, (a - a), to a and then continue integration on the other side from an arbitrarily close point, (a + S), to infinity,. R(z) einzdz + 0. (13 .111 ) We start by taking the moduli of the quantities in the integrand to put an upper limit to this integral as I IpieiOI a. (13 .112 ) COMPLEX TECHNIQUES IN JAKtNG. bound as (13 .115 ) Since the straight line segment shown in Figure 13.16, in the interval [0,7r/2] , is always less than the sin 6 function, we can also write Equation (13 .115 ) as IA