For (a,b))∈(R∗+)2. Let (In)n∈N be the sequence of improper integrals defined by
(∫bae−nt2dt)1/n
I'm asked to calculate the limit of In when n→+∞.
I've shown that
∫+∞xe−t2dt∼(+∞)e−x22x
However, how can I use it ? I wrote that
∫bae−nt2dt=1√n∫√nb√nae−t2dt
Hence I wanted to split it in two integrals to use two times the equivalent but i cannot sum them so ... Any idea ?
Answer
First answer. This has some problems but now it is fixed.
So you have the result:
∫∞xe−t2dt=e−x22x+o(e−x2x) as x→∞
In your last step, you had a mistake. It would be:
∫bae−nt2dt=1√n∫√nb√nae−t2dt=1√n(∫∞√nae−t2dt−∫∞√nbe−t2dt)
Assume $0∫bae−nt2dt=e−na22na+o(e−na2n)
For n large enough we can take n-th root on both sides of (2) to get:
(∫bae−nt2dt)1/n=[e−na22na+o(e−na2n)]1/n=e−a21n1/n(2a)1/n[1+o(1)]1/n→e−a2
Where we have used c1/nn→1 for cn strictly positive and bounded away from 0 and the fact that n√n→1.
(⋆): If you allow a=0, then something similar can be done which is even easier.
Edit One can also come up with the asymptotics of the integral:
Inn=∫bae−nt2dt
Assume $0The Laplace Method, we get:
Inn∼e−na22an
Taking n-th root we obtain the result:
lim
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