Suppose:
(i) converges forall . That is the following funtion is well defined:
(ii) is continuous on
(iii) uniformly converges on
Then:
(a) There exits for all .
(b) .
Regards.
Fernando Revilla
Hello
I know when the limits of integration are finite its possible to bring the derivative sign inside the integral (i.e. the generalized Leibniz integral rule). In its basic form the rule says
.
But what if and ?
I've seen people bring the derivative sign inside the integral in this case, but since the limits are not finite the Leibniz rule does not apply directly. So how can this operation be performed. What theorem applies??
I have a suspicion it may have something to do with Lebesgues Dominated Convergence Theorem.
Suppose:
(i) converges forall . That is the following funtion is well defined:
(ii) is continuous on
(iii) uniformly converges on
Then:
(a) There exits for all .
(b) .
Regards.
Fernando Revilla
See for example Tom Apostol Mathematical Analysis, chapter 14.
Regards.
Fernando Revilla