Methods of Mathematical Physics(2016.10) napter 5 Calculations on definite integrals YLMaaPhys FDU 在4上ag==0,·上=“Q(=x(xx 在l2上arg eI"o(x)dx xo(x dx 当二→0和z→>∞时,z“Q(z)一致地趋于零,即 lm-(]=0,lmn=[(-)=0,由引理1,2, mL:“(=0,如m[==0 因此xxdx-/"axo"o(x)dx=2m∑Rsv x"-IOGx)dr=- 2ni ∑Res[=o( Alat aciar 2 Res=-oe) al sarg=s2r ∑Res(-)"oe) sin a/<arg(-:)cr (2)正实轴上存在一阶极点的情况 设Q(二)在正实轴上有单极点 z=c,所取积分路径如图所示,则 取积分闭曲线如左图 G 当取极限R→∞,δ>0,E→0时 在1上aEg=0,[==[x“(x)Methods of Mathematical Physics (2016.10) Chapter 5 Calculations on definite integrals YLMa@Phys.FDU 21 在 1 l 上 arg z = 0, = 0 -1 -1 ( )d ( )d 1 z Q z z x Q x x l , 在 2 l 上 arg z = 2 , ( ) 2 0 0 -1 -1 2 2( 1) -1 0 2 -1 2 -1 0 ( )d ( )d ( )d ( )d ( )d . i i l i i z Q z z xe Q x x e x Q x x e x Q x x e x Q x x − = = = = − , 当 z →0 和 z → 时, z Q(z) 一致地趋于零,即 lim ( ) 0 1 0 = − → z z Q z z ,lim ( ) 0 1 = − → z z Q z z ,由引理 1,2, lim ( )d 0 -1 0 = → C z Q z z , lim ( )d 0 -1 = R→ CR z Q z z . 因此 − = 0 arg 2 -1 0 2 -1 0 -1 ( )d ( )d 2 Res ( ) z i x Q x x e x Q x x i z Q z ( ) ( ) -1 -1 2 0 0 arg 2 ( 1) -1 0 arg 2 -1 0 arg 2 -1 arg( ) 2 ( )d Res ( ) 1 2 Res ( ) Res ( ) sin Res ( ) . sin i i z i i i z z z i x Q x x z Q z e ie z Q z e e ze Q z z Q z − − − − − − = − = − = = − (2)正实轴上存在一阶极点的情况。 设 Q(z) 在正实轴上有单极点 z = c ,所取积分路径如图所示,则 取积分闭曲线如左图 1 2 3 4 -1 -1 ' -1 0 arg 2 ( )d ( )d 2 Res ( ) R l C l C C l C l C z z Q z z z Q z z i z Q z + + + = + + + = 当取极限 R → → → , 0, 0 时, 在 1 l 上 arg z = 0, = c l z Q z z x Q x x 0 -1 -1 ( )d ( )d 1