st 0 0 0
t
t
t时,f1 (t ) 1(t ) 0 f1 (t ) f 2 ( )d f1 (t ) 1(t ) f 2 ( )d
0 0
10
t
L[ f 1 (t ) f 2 ( ) d ]
19
, bl i
1 d M (s) l { [ ( s p1 ) ]}s p1 i! ds D( s )
l 1
i
1 d M (s) l b1 { [ ( s p1 ) ]}s p1 (l 1)! ds D( s ) 系数cl 1 , , cn , 仍按以前的方法计算
0
t
[ f 1 (t )1(t ) f 2 ( ) d ]e st dt
0 0
f 2 ( ) d f 1 (t )1(t )e st dt
0 0
令t , 则 L[ f 1 (t ) f 2 ( ) d ]
0
0 st 0
等式两边对s趋向于0取极限
st 左边 lim f ( t ) e dt lim f ( t ) e dt s 0 s 0
f (t ) dt f (t ) 0 lim f (t ) f (0) t
0
右边 lim [ sF ( s ) f (0)] lim sF ( s ) f (0) s 0 s 0 lim f (t ) lim sF ( s ) t s 0
sn
即原函数 f(t)的n重积分的拉氏变换等于其象 n 函数除以 s 。