Consider $$ \lim_{n\rightarrow \infty}\int_{\mathbb R}e^{-|x|n}e^{\frac{x^2}{2}}dx $$ The goal of an exercise I am working on is to compute the limit of the the integral above. By intuition it should not converge i.e $\lim_{n\rightarrow \infty}\int_{\mathbb R}e^{-|x|n}e^{\frac{x^2}{2}}dx=\infty$.
I think I found an argument proving it, namely, ( I am not sure if this is valid)
Since the integrand is even we have, $$ \int_{\mathbb R}e^{-|x|n}e^{\frac{x^2}{2}}dx=2\int_{0}^{\infty}e^{-xn}e^{\frac{x^2}{2}}dx $$ Now for some fixed $n\in\mathbb N$ and for all $x\geq 2n$ the following holds, $$ nx\leq \frac{x^2}{2} $$ Consequently, $$ e^{-\frac{x^2}{2}}\leq e^{-xn} $$ for all $x\geq 2n$.
Hence, $$ 2\int_{2n}^{\infty}e^{-xn}e^{\frac{x^2}{2}}dx\geq2\int_{2n}^{\infty}e^{-\frac{x^2}{2}}e^{\frac{x^2}{2}}dx=\int_{0}^{\infty}\chi_{(2n,\infty)}(x)dx=\infty $$ holds for all $n\in\mathbb N$.
Finally this yields $$ \int_{\mathbb R}e^{-|x|n}e^{\frac{x^2}{2}}dx=\infty $$ for all $n\in\mathbb N$ and consequently $$ \lim_{n\rightarrow \infty}\int_{\mathbb R}e^{-|x|n}e^{\frac{x^2}{2}}dx=\infty $$
I am not sure if this argumentation is valid. Could someone look over it?
I am pretty sure there are much better ways to show that the upper limit does not converge and I really would appreciate to see one of those if someone wants to post one.
Thanks in advance!
First you should note that $$e^{\frac12 (-2xn + x^2)} = e^{\frac12 (x-n)^2 - \frac12 n^2} = e^{-n^2/2} e^{\frac{(x-n)^2}{2}}$$
Which means that you can rewrite your integral as: $$2 e^{-n^2/2} \int_0^\infty e^{\frac{(x-n)^2}{2}} dx$$
Now the integrand diverges to infinity for each $n$. Which means each term in your sequence is infinite.