Q:

Limit of an integral: Possible issue with uniform integrability?

Let $f_n(x) = \frac{ne^x\cos x}{1 + n^2x^2}$. I have an integral $\int\limits_0^1f_n$ whose limit I would like to find as $n \rightarrow \infty$. I've tried using Lebesgue Dominated Convergence, Vitali Convergence, and even Bounded Convergence, but unfortunately in all three cases, I run into the problem that the $f_n$ becomes unbounded at 0, so I don't think $\{f_n\}$ is uniformly integrable on [0, 1].  
Could anyone provide a hint? (Or perhaps rectify my understanding of uniform integrability...)
Thank you in advance!

A:

Let $y = nx$. Then
$$\int_0^1 f_n(x) \, dx = \int_0^n \frac{e^{y/n} \cos(y/n)}{1 + y^2}\, dy = \int_0^\infty g_n(y)\,dy$$ with $$g_n(y)=\frac{e^{y/n} \cos(y/n)}{1 + y^2}  1_{[0,n]}(y).$$
We have for $y \geqslant 0$
$$\left| g_n(y)\right| \leqslant eg(y) = \frac{e}{1+y^2}.$$
 and $$\lim_{n \to \infty}g_n(y)=g(y).$$
Since $g$ is integrable you can now apply Lebesgue dominated convergence theorem:
$$\lim_{n \to \infty}\int_0^\infty g_n(y)\, dy = \int_0^\infty \frac{dy}{1+y^2} = \frac{\pi}{2}$$

A:

Consider $g_n(x) = \frac{n}{1+n^2x^2}$. Then $c g_n \leq f_n \leq Cg_n$ on $[0,1]$ and $\int_0^1 g_n(x)\,dx = \arctan(n)$. Also notice that $g_n \to 0$ but $arctan(n) \to \frac{\pi}{2}$,
Since the $f_n$ are very similar, we should not expect $\int_0^1 f_n(x)\,dx$ to converge to $\int_0^1 0\,dx$. In fact, we know $\limsup_n\int_0^1 f_n(x)\,dx \geq c\lim_n \int_0^1 g_n(x)\,dx = c \frac{\pi}{2} \neq 0$, so we definitely can't use any of the standard convergence theorems.
Instead, consider $g_n-f_n$. I'll leave it to you to show that $g_n-f_n$ is uniformly integrable (in fact uniformly bounded, compute the derivative to find its maximum and show the max goes to zero with $n$), and hence $\lim_n\int_0^1f_n(x)\,dx = \lim_n\int_0^1g_n(x)\,dx = \frac{\pi}{2}$.

