I found a proof here. For possible future reference, I copy it here along with several key words: improper integral, uniform convergence, compact, Rudin, P167, 12.
Note that the domain here is , so is not a compact subspace, so we have to split the domain into two parts: (0,1] and . We first focus on the latter.
Since which is a fixed number independent of , is improper Riemann integrable on for all . Since for all , so for all , so exists too.
Now we want to prove converges uniformly to on . From the hypothesis We know that converges in the topology of compact convergence, which is unfortunately coarser than uniform topolgy, so we cannot deduce from general topology the desired conclusion and have to prove it from scratch: Given any , since , there is a such that by Cauchy criterion. Because is now compact, we can choose an such that for all and . We show this will do the job. For any and we can write , where . Since , the first term . The second term exists only if , in which case, in view of for all , . Then we get for all whenever which is independent of . Therefore for any this , , as desired.
Now we have proved , applying the same argument above shows . Summing up these two equalities and noting that all improper integrals involved are valid, we have , i.e., , which completes the whole proof.
The same idea applies if we want to prove convergence of an uncountable sequence of improper integrals, which is used at the end of the proof of Stirling's formula in page 195 of this book.
Thank Pere Callahan for this brilliant proof, also thank myself for the hard search.