Proof: Because when x tends to 0, it can be known from L'H?pital's law.

Lim g (x)/x = lim g' (x) = f (0), so it is assumed that x = 0 in the generalized integral is not deficient.

In addition, limg2 (x)/x = lim2gg' (x) = 2g (0) * g' (0) = 0.

Therefore, for any X>0, it is

Integral (from 0 to x) g (x)/x 2dx = integral (from 0 to x) g 2 (x) d (- 1/x)

=-g 2 (x)/x | upper limit x lower limit 0+ integral (from 0 to X)2g(x)g'(x)/xdx

Because -g 2 (x)/x

& lt=2 integer (from 0 to X)g(x)/x *f(x) dx.

According to Cauchy-Schwartz inequality

& lt=2 integral (from 0 to x) g 2 (x)/x 2dx (1/2) * integral (from 0 to x) f 2 (x) dx (1/2)

In order to solve this inequality

Integral (from 0 to x) g 2 (x)/x 2dx < =4 Integral (from 0 to x) f 2 (x) dx,

So the generalized integral converges and the inequality is established.