不妨设a < b.
对任意0 < α < A,有∫{α,A} (f(ax)-f(bx))/x dx
= ∫{α,A} f(ax)/x dx -∫{α,A} f(bx)/x dx
= ∫{αa,Aa} f(x)/x dx -∫{αb,Ab} f(x)/x dx
= ∫{αa,αb} f(x)/x dx -∫{Aa,Ab} f(x)/x dx
= ∫{αa,αb} f(0)/x dx +∫{αa,αb} (f(x)-f(0))/x dx -∫{Aa,Ab} f(x)/x dx
= f(0)ln(b/a)+∫{αa,αb} (f(x)-f(0))/x dx -∫{Aa,Ab} f(x)/x dx.
当A → +∞,有Aa,Ab → +∞,于是由f(x)/x在[c,+∞)广义可积,可得∫{Aa,Ab} f(x)/x dx → 0.
而当α → 0,有αa,αb → 0,由f(x)在[0,+∞)连续,可证得∫{αa,αb} (f(x)-f(0))/x dx → 0.
因此∫{0,+∞} (f(ax)-f(bx))/x dx = f(0)ln(b/a).
过程没有写得很详细,
