Let us derive the integration formula ( I ) 1 -x. x" ( I -

Chapter 0, Problem 7.27

(choose chapter or problem)

Let us derive the integration formula ( I ) 1 -x. x" ( I - a );r , , dx = ----- 0 (x- + I)- 4 cos(a;r /2) (-l 0. To do this. \\IC shall use the function exp( a In x) when . :'j cxp(a log:.) f(-) = = --- .. (:.2+J)2 (:2+J)2 ( ;r 3;r) I: I > 0. - 2 < a rg : < T whose branch cul is the origin and the negative imaginary axis. The path of integration is shown in Fig. I 09. where p < I < R aml the branch cul is indicated v.:ith a hollow dot and dashes. -R \' -P 01 I I I x Starting with Cauchy"s residue theorem. we write ( 2) l f (:) d: + / f ( : ) d: = 2;r i ~.sf ( : ) - l f (:.) d:. - l_ .f ( :. ) d:.. J,_, },_~ . Jr. }, Ii If we use the parametric equations 11 . :. =re' = r (p s r s R) and :. =re'' = -r (p s r s R) for L 1 and -L2. respectively. we can write the left-hand side of equation (2) as 1 _ --/ _ --f,Rcxpla(lnr+iO)]. /Rcxp(a(lnr+i;r)]. /( . ) d.. /(..) d . - ' ") d1 + ' ' d1 ., L~ ,, (r-+1)- . 1 (r-+l)- 1 R ,-o . JR ru - ---..,-,--~, d,. + ('"' :r ' ' d,.. - 1 , (r- + I)- " (r"" + I)- Thus ( 3) i . /(:.) d::. + ; f(:.)d:. =(I +e':r);R ,r 0 , dr. (r"" + I)- ,_, I.~ '' SEC. 90 Also. (4) Al' ll'DENTATIOl'\ AROl'l\D .\ 13RAl'\Cll POINT Re sf(:.) = ' (i) ;~1 \vhcre ' (o lllo!!.[(ll-2):.+llil (:.) = {' ' . . (:.+t)' and. therefore. (5) Res/(:.)= -ie"':r/_ -- . . .,(l-ll) ~~ 4 Upon substituting expressions (3) m1d (5) into equation (2). we have ; l'T ;R r" I ;r( l - l/) ia:r:2 1 1 (6) (l+e") , , 0 asp --> 0 since a + I > 0. As for the second of limits (7). lj. ',, -" ) d::: I Ci: ( ;::-'- + I)- R" 7r ~ 11 R~ < ( R2 _ 1)2 n R = ( R2 _ I )2 . -1- = R~ 7r-- , R-' " . (1 - ~,)" and \\IC sec that I/ R-' ti --> 0 as R --> oo since 3 - a > 0.