设函数f0(x)在(-∞,+∞)内连续,fn(x)=∫0xfn-1(t)dt(n=1,2,…). (1)证明:fn(x)=1/[(n-1)!]∫0xf0(t)(x-t)n-1dt(n=1,2,…); (2)证明:fn(x)绝对收敛.

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问题 设函数f0(x)在(-∞,+∞)内连续,fn(x)=∫0xfn-1(t)dt(n=1,2,…).
(1)证明:fn(x)=1/[(n-1)!]∫0xf0(t)(x-t)n-1dt(n=1,2,…);
(2)证明:fn(x)绝对收敛.

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答案(1)n=1时,f1(x)=∫0xf0(t)dt,等式成立; 设n=k时,fk(x)=1/[(k-1)!]∫0xf0(t)(x-t)k-1dt, 103 则n=k+1时,fk+1(x)=∫0xfk(t)dt=∫0xdt∫0t[1/(k-1)!]f0(u)(t-u)k-1du =[1/(k-1)!]∫0xdu∫uxf0(u)(t-u)k-1dt =(1/k!)∫0xf0(u)(t-u)kdu, 由归纳法得fn(x)=[1/(n-1)!]∫0xf0(t)(x-t)n-1dt(n=1,2,…). (2)对任意的x∈(-∞,+∞),f0(t)在[0,x]或[x,0]上连续,于是存在M>0(M与 x有关),使得|f0(t)|≤M(t∈[0,x]或t∈[x,0]),于是 [*]

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