MATHEMATICAL METHOD IN SCIENCE AND ENGINEERING Episode 11 ppt

13.17 Contour for the integrals of type IV Under these conditions we can evaluate the integral I as 13.126 7r - 1 y residues of [zA- ' R z] , - sin 7rA inside C where C is the clos

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) 1 (13.93) 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

Using the binomial formula we can write

fig 13.13 Contour for the type I1 integrals

11 Integrals of the type I = s_”, d x R ( 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 a t least as - 1 in the limit as IzI + 00

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 ) d z = l m R ( x ) d z + i R ( z ) d z (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

fig 13.15 Contour C in the limit R -+ 00 for type I11 integrals

and extract the An-l coefficient as

( 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

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

From here we see that in the limit as p t ( ~ f the value of the integral

I A goes to zero, that is,

Example 13.7 Complex contouy integral technique: In calculating dis-

persion relations we frequently encounter integrals like

f (z) = - 1 dkg (k)eikx

Let us consider a case where g ( k ) is given as

i ) For x > 0 we can write

theorem [Eq (13.6)] to find

ii ) For z < 0, we complete our contour C from below to find

(13.122) (13.123)

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 ( - 1 y

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:

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

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

GAMMA AND BETA FUNCTIONS 361

P W

(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)

The inverse of the

finite with the limit

(13.140)

gamma function, l/r (x) , is single valued and always

(13.141)

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

GAMMA AND BETA FUNCJIONS 363

Using the substitution

(13.157) The function 9 ( z ) satisfies the recursion relation

@(x + 1) = 9 ( z ) + 2 - 1 , (13.158) from which we obtain

13.8.3

Among the useful relations of the gamma function we can write

Useful Relations of the Gamma Functions

-7r csc(7rx) r(x + 1) ’

CAUCHY PRINCIPAL VALUE INTEGRAL 365

On the other hand, the incomplete gamma function is defined by

y* (c, X ) = - y("-') exp( -y)dy

Sometimes we encounter integrals with poles on the real axis, such as the integral

(13.174)

which 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

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

inside C

(13.179)

f -

From the condition f ( z ) + 0 as IzI -+ 00 for y > 0, the second integral over

CR on the right-hand side is zero To evaluate the integral over the small arc

c6 we write

(13.180) (13.181)

CAUCHY PRINCIPAL VALUE INTEGRAL 367

tz

Fig 13.19 Another path for the Cauchy principal value calculation

and find the Cauchy principal value as

(13.182) (13.183)

Another contour that we can use to find the Cauchy principal value is given

in Figure 13.19 In this case the pole at x = u is inside our contour Using the residue theorem we obtain

(13.185)

In this case we again have two choices for the detour around the singular point on the real axis Again the Cauchy principal value is - i x f ( u ) for both

choices

Z

-kr kr

fig 13.20 Contour for 11

Example 13.8 Cauchy principal value integral: Let us now evaluate the integral

m

2

CONTOUR INTEGRAL REPRESENTATIONS OF SOME SPECIAL FUNCTIONS 369

Fig 13.21 Contour for 12

x sin x d x Hence the divergent integral

its Cauchy principal value as

can now be replaced with

13.10 CONTOUR INTEGRAL REPRESENTATIONS OF SOME

Fig 13.22 Contour for the Schlijfli formula

This gives us the complex contour representation of the Legendre polynomials

as

1 1 ( 2 ' - 1)ldz'

fi (z) = 27rz2' c

@ I +l+l (13.194) This is also called the Schl8fli integral formula, where the contour is given

in Figure 13.22 Using the Schlofli formula [Eq (13.194)] and the residue theorem, we can obtain the Legendre polynomials as

For the residue we need the coefficient of ( z - z) ; hence we need the j = 1

term in the above series, which is

(13.197)

coefficient of ( z - z)' = C k! (1 - k)! (1 - 2k)!l!

k=O

CONTOUR INTEGRAL REPRESENTATIONS OF SOME SPECIAL FUNCTIONS 371

Using this in Equation (13.195) we finally obtain P' (x) as

about the origin for a contour with unit radius is given as

O 0 1

n!

f ( t ) = c -f'"'(O)tn,

n=O where

Note that this is valid for a region enclosed by a circle centered at the origin with unit radius To obtain L,(z) valid for the whole complex plane one might expand f ( t ) about t = 1 in Laurent series

Another contour integral representation of the Laguerre polynomials can

be obtained by using the Rodriguez formula

ex d"

L,(x) = (x"e-.) n! dxn Using the formula

d(xne-") - n! f f ( z ) d z dxn 2Ti c ( z - %)"+'

(13.206)

(13.207) and taking as a point on the real axis and

f(z) = zne-' (13.208)

we can write

2 K i z"ec"dz

n! ( z - X)"+l' where C is a circle centered at some point z = x , thus obtaining

(13.209)

(13.210)

polynomials: Establish the following contour integral representation for the Hermite

where C encloses the point x, and use it to derive the series expansion

13.3 Using Taylor series prove the Cauchy-Goursat theorem

where f ( z ) is an 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, 121 > 2, and 1 < IzI < 2

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:

Fig 13.23 Contour for problem 13.5

2 x (cos 36) dB

1 5-4cos6

ii)

sin xdx x2 + 42 + 5

J -, x(a2 + x2)

13.7 Evaluate the following Cauchy principal value integral:

13.8 Using the generating function for the polynomials Pnm (z)

-xt

e ( 1 - t )

= Cpnm ( Z I P , < 1, m = positive, (1 - t)m+' n=O

establish a contour integral representation in the complex t-plane Use this representation to find A(n, m, k ) in

a, b are arbitrary real numbers are defined by the Rodriguez formula

The Jacobi polynomials P27b)(~~s8), where n = positive integer and

dn

- [( 1 - x)n+"( 1 + Z)rn+b] , 1x1 < 1

p p q x ) = (-W

2%!( 1 - z)"( 1 + x ) ~ dzn

Find a contour integral representation for this polynomial valid for 1x1 < 1 and use this to show that the polynomial can be expanded as

on the real axis with the property

For a function F ( z ) analytic everywhere in the upper half plane and

F ( z ) -+ b as IzI -+ 00 , b is a real constant, show the following Cauchy principal value integrals:

1 Fl(x')dz'

FR(z) = b + -P lr s_, x'-x

and

13.12

function of the first kind

Given the following contour integral definition of spherical Hankel

i) Show that this series breaks of at the k = Ith term

ii) By using the contour integral definition given above, find explicitly the constants A(k, I ) , and p(l, k)

13.13 Another definition for the gamma function is given as

Using this result, define hil)(z) as a contour integral in the complex j’-plane (j’ = t’ + zs’), where

FRA CTIONAL DERIVATIVES and

INTEGRALS

WIFFERINTEGRALS~~

The diffusion equation in integral form is given as

(14.1)

where c ( 7 , t ) is the concentration of particles and f ( 7 , t ) is the current

density The left-hand side gives the rate of change of the number of particles

in volume V, and the right-hand side gives the number of particles flowing

past the boundary S of this volume per unit time In the absence of sources

or sinks, these terms are naturally equal Using the Gauss theorem we can

write this equation as

(14.3)

This gives us a partial differential equation to be solved for concentration as

d -c(+,t) + 3~7(7,t) = 0 (14.4)

at

In order to solve this equation, we also need a relation between c( T", t ) and

f ( 7 , t ) Because particles have a tendency to flow from regions of high to

low concentration, as a first approximation we can aSsume a linear relation

between the current density and the gradient of concentration as

379

The proportionality constant k is called the diffusion constant We can now write the diffusion equation as

a

at

which is also called Fick's equation

Einstein noticed that in a diffusion process concentration is also propor- tional to the probability, P( ?, t), of finding a diffusing particle a t position

7' and time t Thus the probability distribution satisfies the same differential equation as the concentration For a particle starting its motion from the origin, probability distribution can be found as

1

(47rkt)q

This means that even though the average displacement of a particle is zero

(< 7' >= 0 ) , mean square displacement is nonzero and is given as

relation For the particle to cover twice the distance, time must be increased

by a factor of four This scaling property results from the diffusion equation

where the time derivative is of first and the space derivative is of second order However, it has been experimentally determined that for some systems this relation goes as

UNIFIED EXPRESSION OF DERIVATIVES AND INTEGRALS 381

situation on the applied side of this branch of mathematics is now changing rapidly, and there are now a growing number of research areas in science and engineering that make use of fractional calculus Chemical analysis of flu- ids, heat transfer, diffusion, the Schrodinger equation, and material science

are some areas where fractional calculus is used Interesting applications to 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 away from equilibrium fractal curves and surfaces are encountered, where or- dinary 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 properties of fractional calculus along with some mathematical techniques and their applications

14.1 UNIFIED EXPRESSION OF DERIVATIVES AND INTEGRALS 14.1.1 Notation and Definitions

In our notation we follow Oldham and Spanier, where a detailed treatment

of the subject along with a survey of the history and various applications can

be found Unless otherwise specified we use n and N for positive integers, q

and Q for any number The nth derivative of a function is shown as

flf

-

dxn '

Since an integral is the inverse of a derivative, we write

Successive integrations will be shown as

is true for derivatives, it is not true for integrals, that is,

Hence for n successive integrals we will also use

f'-"' = L- &,-I L;-;l dzn- 2 Ly dz1 LI1 f (z0)dxO (14.21)

When there is no room for confusion, we write

f ' " ( X >

The value of a differintegral at x = b is shown as

Other commonly used expressions for differintegrals are:

UNIFIED EXPRESSION OF DERIVATIVES AND INTEGRALS 383

Similarly, the second- and third-order derivatives can be written as

and

- d3f = lim {/6z]-3[f(z) - 3f(z -6z) + 3f(z - 2 6 ~ ) - f ( ~ - ~SZ)]}

dX3 6 x 1 0

(14.24)

Since the coefficients in these equations are the binomial coefficients, for the

n t h derivative we can write

In these equations we have assumed that all the derivatives exist In addition,

we have assumed that [6z] goes to zero continuously, that is, by taking all

values on its way to zero For a unified representation with the integral, we

are going to need a restricted limit For this we divide the interval [z - a] into

N equal segments;

S N X = [ z - a ] / N , N = 1,2,3 , (14.26)

In this expression a is a number smaller than z

Since the binomial coefficients

also write

are zero for the j > n values, we can

Now, assuming that this limit is also valid in the continuum limit, we write

the nth derivative as

- n N - l

- - [dz]" d"f - ~ + m lim { [y] [-l]'( '3" ) f ( z - j [y])} (14.29)

3 =O