Evaluation of the Integral ??1+1?Plm,n(x)Plm?2,n(x)?dx

(1)

Texturesand Microstructures,1987,Vol.7, pp.207-210

Photocopyingpermittedbylicenseonly

SHORT

COMMUNICATION

Eva

lu

atio

n of

th

e nteg

ral

i+,

_P;’(x)P;

x)

dx

C.

M. BRAKMAN

DelftUniversity ofTechnology,Laboratory ofMetallurgy,Rotterdarnseweg 137, 2628ALDelft, TheNetherlands

(Received21December1986;infinalform30 April1987)

Integralsof thetypeindicatedcanbeevaluated using the Fourier-seriesdevelopment

of theJacobi-polynomials. Analternative procedureisgiven yielding an analytical

expressionin termsofl, m andn.

KEY WORDS: Jacobi-polynomials,definiteintegral,texturedevelopment.

In

texture analysis, for instance in the case of the

series-development

of crystallite rotations used for the prediction of

texture-development,

one

frequently

encounters integrals of the

type:

lP’’"(x)PT’-2’"(x)dx

The expression can be evaluated using the Fourier-series

develop-ment of the

_PT’"

given by

Bunge (1982,

p.

359).

However,

an

analytical expression can be obtainedrelatively easy.

Assume

>

0 oddoreven,m even and

>-2,

n even and _->0. Using

the definition of the

_P7

given

by

Bunge (1982,

p.

351)

the factor 207

(2)

208 C. M. BRAKMAN

independentof

x

in the integralis written:

-(l

+

{(1-

m)!

(l-

m

+

2)! (1

+

m)!

(l

+

m

2)!}

-1/2

(1)

C

22/(

1

n)!

Writingf

P(x)

(1

X)-"+m-(1

+ X)-"-m+Dq-"){(1

x)l--m(1

V

_X)

_l+m)

1,

_(l-n)

)l-n-k

(1+ m)!

(l-k=0 -k

(-1

(1

+

rn

_k)!

(n

rn

+

x

(1-

x)k-(1

+

x)

l-n+l-k

(2)

Q(x)

(1

X)/-m+2(1

-t-

X)

,+m-2

(3)

Theintegral reads:

c

P(x)

_x

D(I-"-)Q(x)

dx

(4)

Integrating by parts yields:

d

cP(x)Dt--X)Q(x)l+

c

D)P(x)

x

D’--2)Q(x)

dx

₍₅₎

Evaluation of the stock term requires analyzing whether negative exponents of

(1-

x)

or

(1

+

x)

occur.

It

can be shown that negative exponents do not arise.

Conse-quently,

the stock term

equals

zero.

It

can beshown for the

general

case

(1

-<_

_q

<-

_n)

that:

c(-1)q-D(q-1)e(x)D(l-n-q)Q(x)

1_+

0

(6)

Eventually, after

repeatedly

integrating

by

parts

Eq. (5)

yields:

c(-1)l-n

+lD(l-n)p(x)D()a(x)

dx

₍₇₎

For

_D(-’P(x)

itcan be written"

/)(-,o

_A[(-x

_+""

₊

₁₁

k--1

+ Dq-’){A0(1

X)-1(1

h-

_X)

’-n+l}

_(7a)

fLeibniz’s rule is applied, thesymbolD(’)_{stands for}_d"/dx’*_{and in the sum}

onlyk values leadingtonon-negative factorials are allowed: rn n-<_k<-_ ₊rnforn<m.

(3)

EVALUATION OF THE INTEGRAL 2O9 Using l(1-x)P(l+x)qdx=

P!q!

2p+q+l

_(p>O’q>O)

_(8c)

(p+q+

1)!

it follows for the n

<

rn case for

Eq. (7)

2

((l+m-2)!(l-m+2)!}

1/2

21

+

1

(1

+

m)! (l- m)!

(9)

For

the n

->_

rn case evaluationof

Eq. (7)

requires calculation of the

-

Usehas been made of the well-known formula

_t=0

r- r

r, s, and u are integers. For u-r<0 the sum starts at r-u but the result

expressionremainsthe same.

where

(l--

n)

l)l_n_k

(l

+

m)! (l-

m)!

Ak

k

(-

(l

+

m

k)!

(n

m

+

k)!

accordingto

Eq.

(2).

Two

cases can be distinguished:

(i)

n

<

m"

Ao

does not existand"

D(l_n)p(x)

(1)11{(i

n)!}2X

(l+m)(l-m)

(8a)

k k l-n -k

(ii)

n->

m: the k-sum occurring in

Eq. (8a)

must start at k 0 now.

Consequentlyf

Dl-")p(x)

Dl-")(Ao(1

x)-l(l +

x)

’-n+l)

+

the result of

Eq.

(8a)

(_1)(,_._1){(/_n)!)

2 l+rn 1-0 n

(-1)’-"

(l- m)!

(n

m)

D(’-){(1

-x)-l(1 +

+(-1)’-n-l(l-n)!(t-n)’{(

21

)_(l-m)}

l-n l n

(8b)

(4)

210 C.M. BRAKMAN

integral

f_Q(x)Df’-n)((1-

x)-(1

+

x)

1-n+l}

dx

(10)

Repeatedly

integrating

by

parts it can be shown that the

general

stockterm

(_l)q-lo(q-1)a(x)O(l-n-q)((

1

x)-1(1

+

x)l-n+l}[+__.

(l

<_-

_q

_<-

_n)

does not exhibit negative exponents of

(l-x)

or

(1

+

x).

Consequently,

it

equals

zero.

As

aresult,

Eq.

(10)

reads:

(-1)/-n

D(1-n)a(x)O()((1-

x)-l(1 +

x)

’-=+1)

dx

_(10a)

Using Leibniz’s rule and

Eq.

(8c)

it isfound for

Eq.

(10a):

(/+m-2)(

l-m+2

)

k n-m+2+k

22/+1

1-n

(l-

n)!

(-1)

k

"21

+

1k=O -re+l+

(11)

It

can be shown

("trial

and

error"

methods,

no proof found

yet)

that the sum in

Eq. (11)

equals:

(l+m-2)!(l-n+l)!(n-m+l)!{(

21+1

)_(l-m+12)}

(20!

l-n

+

1 l-n

+

(lla)

For

then

>-rn

case itfollows

(via Eqs.

(8b)

and

_(lla))

for

Eq.

(7):

2

{(l+m-2)!(l-m+2)!}1/2{

(2l+l)(n-m+l)

}

21

+

1

(1

+

m)! (l-

m)!

1

(l-

m

+

1)(1-

rn

+

2)

(12)

Equations

(9)

and

(12)

have been checked using the Fourier-series

development

ofthe

_P7

for

_4(1)23,

rn

2(2)1,

n

0(2)1.

References