limit of $\int_{-\frac{n}{2}}^{\frac{n}{2}}\exp\left(-\frac{(x-n)^2}{2n^2}\right)dx$ and its summation analog.

Question

I'm getting confused about taking the following limits I would be appreciated if somebody could guide me through

\begin{equation} \lim_{n\to\infty}\int_{-\frac{n}{2}}^{\frac{n}{2}}\exp\left(-\frac{(x-n)^2}{2n^2}\right)dx \end{equation}

and also the discrete version:

\begin{equation} \lim_{n\to\infty}\sum_{x=-\frac{n}{2}}^{\frac{n}{2}}\exp\left(-\frac{(x-n)^2}{2n^2}\right) \end{equation}

Can anybody help me with this?

Edited: I am more looking for an asymptotic behavior so if the answer is $\infty$ an asymptotic behavior would also suffice

Answer 1

For the integral, the change of variables $u=\frac{x-n}{n}$ gives $$ n\int^{-\frac{1}{2}}_{-\frac{3}{2}}e^{-\frac{u^2}{2}}\,du=n\int^{\frac{1}{2}}_{-\frac{1}{2}}e^{-\frac{(u-1)^2}{2}}\,du $$ So the first limit is $\infty$.

The limit in the discrete version is also $\infty$ since $$ \sum^{\tfrac{n}{2}}_{k=-\tfrac{n}{2}}\frac{1}{n} e^{-\frac{1}{2}\big(\tfrac{k}{n}-1\big)^2}\xrightarrow{n\rightarrow\infty}\int^{\tfrac{1}{2}}_{-\frac{1}{2}}e^{-\tfrac{1}{2}(u-1)^2}\,du $$

If $I_n$ denotes the first sequence in your problem and $D_n$ its discrete version, what escapes me is how $I_n$ and $D_n$ differ as $n\rightarrow\infty$ (asymptotics), that is

• how does $\Delta_n:=I_n-D_n$ behave? for instance, is $I_n-D_n=O(n^{-\alpha})$ for some $\alpha>0$?
• Clearly $\lim_n \frac{I_n}{D_n}=1$, but how fast? for instance, is $\frac{I_n}{D_n}-1 =o(n^{-\beta})$ for some $\beta>0$

$\Delta_n$ is the error in approximating an integral by Riemman sums, i.e. \begin{aligned} \Delta_n&=\sum^{\tfrac{n}{2}}_{k=-\tfrac{n}{2}}\frac{1}{n} e^{-\frac{1}{2}\big(\tfrac{k}{n}-1\big)^2}-\int^{\tfrac{1}{2}}_{-\frac{1}{2}}e^{-\tfrac{1}{2}(u-1)^2}\,du\\ &=\sum^{\tfrac{n}{2}}_{k=-\tfrac{n}{2}}\int^{\tfrac{k+1}{n}}_{\tfrac{k}{n}}\Big(e^{-\frac{1}{2}\big(\tfrac{k}{n}-1\big)^2}-e^{-\frac{1}{2}(u-1)^2}\Big)\,du \end{aligned}

To answer the firs question about asymptotics, one may try to use the mean value theorem to see if we can get bounds from below and above. But still some work to do.