因此,按傅氏积分公式,在f1)的连续点就有 f(tu(t)e r +oo 2兀 f(o)u(r)e re jo dt ejo do (B+jOt dT 2丌 edo f(r)e 2兀 F(B+joedo, t>0 等式两边同乘以e,则 (0)-1f(+jo)e+)d,>0 2丌因此, 按傅氏积分公式, 在f(t)的连续点就有 ( j ) e d , 0 2 1 e d ( ) e d 2 1 ( ) ( ) e e d e d 2 1 ( ) ( ) e j 0 j ( j ) j j = +      =     =      + - + - + - + + - + - - - - F t f f u f t u t t t t t b               b   b   b ( j )e d , 0 2 1 ( ) ( j ) = +   + - + f t F t t b    b  等式两边同乘以e bt , 则