I am working on my research about approximation a function. I come up with the following integral. I run some simulations and saw that the integral would converge to zero as n goes to infinty. Here is my prediction. Hopefully someone can give me the idea to deal with this kind of upper bound. Thanks a lot.
Problem. Given $a \in [0,1]$. Prove that $\int_{0}^{1}e^{-n(t-a)^2} - e^{\frac{-n(t-a)^2}{1-(t-a)^2}}dt \leq \frac{c}{n},$ where $c >0$ is a constant.
The integral in question can be rewritten as
$$
\begin{aligned}
I&:=\frac1{\sqrt n}\,\int_{-a\sqrt n}^{(1-a)\sqrt n} e^{-u^2}\Big(1-\exp\Big\{-\frac{u^4/n}{1-u^2/n}\Big\}\Big)\,du \\
&\le\frac1{\sqrt n}\,\int_{-a\sqrt n}^{(1-a)\sqrt n} e^{-u^2}\min\Big(1,\frac{u^4/n}{1-u^2/n}\Big)\,du \\
&\le \frac1{\sqrt n}\,(I_1+I_2),
\end{aligned}$$
where
\begin{equation}
I_1:=\int_{|u|\le\sqrt n/2} e^{-u^2}2u^4/n\,du\le\int_{|u|<\infty} e^{-u^2}2u^4/n\,du=O(1/n)
\end{equation}
and
\begin{equation}
I_2:=\int_{|u|>\sqrt n/2} e^{-u^2}\,du
\le\int_{|u|>\sqrt n/2} \frac{u^2}{(\sqrt n/2)^2}e^{-u^2}\,du
\le\int_{|u|>0} \frac{u^2}{(\sqrt n/2)^2}e^{-u^2}\,du
=2\sqrt\pi/n
=O(1/n)
\end{equation}
(as $n\to\infty$). So,
\begin{equation}
I=O(1/(n\sqrt n)),
\end{equation}
which is better than desired.
