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

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,
$$\lim_{n\to+\infty}I_n=\int_0^{+\infty}\exp\bigl(-2t^2\bigr)\,\mathrm{d}t=\frac{\sqrt{2\pi}}4.$$
A: 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).  
