set up

e(x^n)=∫(-∞->； +∞) x^nf(x)dx

Then when n= 1, it is E(X).

Because e (ax) = 1+ax+a 2x 2/2! +......+a^nx^n/n！ + .......

So e (e ax) = ∫ (-∞-> +∞) e (ax) f (x) dx = ∫ (-∞->; +∞) ( 1+ax+a^2x^2/2！ +......+a^nx^n/n！ +.......)f(x)dx

= 1+aE(X)+a^2E(X^2)/2！ +........+a^2E(X^n)/n！

= 1+aE(X)+o(a)

According to ln (1+x) = x-x 2/2+ ....

So ln (e (ax)) = ln (1+AE (x)+o (a)) = AE (x)+o (a)-[AE (x)+o (a)] 2/2. .......

=ae(x)+o(a)-a^2(e(x))^2+o(a^2)+ ...

=aE(X)+o(a)

So lim (a->; 0)ln(e^ax)/a=lim(a->； 0) [E(X)+(o(a)/a)]=E(X)