Question

设f(x)连续,g(x) =∫(1,0)f(xt)dt,且lim x→0 f(x)/x =A,求 g'(x).

Answer 1

x>0时,g(x)=∫(0,1)f(xt)(1/x)dxt=1/x∫(0,x)f(y)dy.

所以

g'(x)=(-1/x^2)∫(0,x)f(y)dy + f(x)/x.

x=0时,lim x→0 f(x)/x =A,所以lim x→0 f(x)=0,所以f(0)=0,所以g(0)=0.所以

g'(0)=lim x→0 g(x)/x=lim x→0 ∫(0,1)f(xt)/x dt=∫(0,1)lim x→0 f(xt)/(xt)*tdt=∫(0,1)Atdt=A/2.