Question

ab设(0,+∞)上的连续函数f使得积分值∫ f(x)dx 与a无关,其中a,b>0,求证：f(x)=c/xa其中c为常

Answer 1

g'(x) = f(x)

从a到ab f（x）的积分 = g(ab) - g(a)

g(ab) - g(a) 与a无关,

d(g(bx) - g(x))/dx = 0

g'(bx)*b - g'(x) = 0

g'(t)*t -g'(1) = 0

g'(t) = g'(1)/t

f(x) = g'(x) = C/x.(C=g'(1))