2°当0<Rez<1 f1(z)=I(=)r(1-z)解析 f2(=)= 解析 sin2丌 而在0<x<1: f1(z)=T(x)(1-x) 丌 f2(=)= Sinx丌 f1(z)=f2( 有唯一性定理 r()r(1-2)=x -,0<Rez<1 slnx丌 在(6)中取z=1/2 则r(1/2)r(1/1)=→I(1/1)=√π
p p p p p p p p G × G = Þ G = = G G - = < < \ = = = G G - < < = = G G - < < (1 / 2 ) (1 / 1) (1 / 1) ( 6 ) 1 / 2 ,0 Re 1 sin ( ) (1 ) ( ) ( ) sin ( ) ( ) ( ) (1 ) 0 1 : sin ( ) ( ) ( ) (1 ) 2 . 0 Re 1 : 1 2 2 1 2 1 则 在 中取 有唯一性定理 而在 解析 解析 当 z z x z z f z f z x f z f z x x x z f z f z z z z o
r((+12)el」es (t+s(ts 6)2.t 2dt ds, Re z>0 令t=52,=n2,0<=0<m< dt=2ds, ds=2dn 则rr(+12157)54 同样以和n交换 r(r(2+12)14(2tmy
( ) ( ) 0 0 ( ) 2 1 0 0 ( ) 2 1 (0 ,0 ) 2 2 0 0 ( ) 0 0 1 ( ) ( 1 / 2) 4 ( ) ( ) ( 1 / 2) 4 ( ) 2 , 2 , , ( ) , Re 0 ( ) ( 1 / 2) 2 2 2 2 2 1 2 1 2 1 Ñ D ò ò ò ò ò ò ò ò ¥ ¥ - + - ¥ ¥ - + - < < ¥ < < ¥ ¥ ¥ - + - - ¥ - - ¥ - - G G + = × G G + = × = = = = = × > G G + = xh h x h x h xh x x h x h x h x h x h x h z z e d d z z e d d dt d ds d t s e ts t dtds z z z e t dt e s ds z z z t s z t z s 同样以 和 交换 则 令