Show $\lim_{n_\rightarrow \infty}\int_0^\infty ne^{-\frac{2n^2x^2}{x + 1}}dx = \infty$

Question

As part of an analysis qual problem, I am having a hard time showing that $\lim_{n_\rightarrow \infty}\int_0^\infty ne^{-\frac{2n^2x^2}{x + 1}}dx = \infty$. Any suggestions? Thanks in advance. I tried using Fatou. That didn't work

Answer 1

I'm not sure about the result you want to show: let $n\in\mathbb{N}^*$ and perform the substitution $t=nx$: $$I_n=\int_0^{+\infty}n\mathrm{e}^{-\frac{2n^2x^2}{1+x}}\,\mathrm{d}x=\int_0^{+\infty}\mathrm{e}^{-\frac{2t^2}{1+t/n}}\,\mathrm{d}t.$$ Clearly, $I_n\in\mathbb{R}_+$ and we easily check that the sequence $(I_n)_n$ is decreasing.

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In fact your limit is quite easy to compute: define the sequence of functions $(f_n)_n$ by: $$\forall n\in\mathbb{N}^*,\ \forall t\in\mathbb{R}_+,\ f_n(t)=\exp\left(-\frac{2t^2}{1+t/n}\right).$$ Clearly, for all $n\in\mathbb{N}^*$, $0\leq f_{n+1}\leq f_1$, all the $f_n$'s are continuous (hence measurable), $f_1\in L^1(\mathbb{R}_+)$ and $$\forall t\in\mathbb{R}_+,\ \lim_{n\to+\infty}f_n(t)=\exp\bigl(-2t^2\bigr).$$ Hence by the dominated convergence theorem, $$\lim_{n\to+\infty}I_n=\int_0^{+\infty}\exp\bigl(-2t^2\bigr)\,\mathrm{d}t=\frac{\sqrt{2\pi}}4.$$

Answer 2

The integral from $1$ to $\infty$ is $\lt \int_1^\infty ne^{-2n^2x^2/2x}\,dx$, which is very nicely behaved as $n\to\infty$.

Similarly, the integral from $0$ to $1$ is nicely behaved. After majorizing by replacing the $x+1$ by $2$, we get a pleasant integral (error function).