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Theorem: (Dominated Convergence Theorem). Let $(f_n)_{n\geq 1}$ be a sequence of integrable functions such that $\lim\limits_{n\to\infty}f_n = f$ pointwise. If there exists an integrable function $g:S\to[0,\infty]$ such that $\left|f_n\right|\leq g$ for all $n\geq 1$, then $f$ is integrable and $\lim\limits_{n\to\infty}\int_Sf_n d\mu = \int_S f d\mu$.

I have to use this theorem to solve the following exercise:

Exercise: Find $\lim\limits_{n\to\infty}\displaystyle\int_{\mathbb{R}}\mathbb{1}_{[0,1]}(x)\dfrac{nx^n\sin(nx)-n}{\sqrt{x + 2n^2}}d\lambda$, using DCT (answer is $-\frac{1}{2}\sqrt{2}$).

What I've tried: I have to find $\lim\limits_{n\to\infty}f_n = f$ and a function $g$ such that $\left|f_n\right|\leq g$ for all $n\geq 1$. This would imply that $f$ is integrable so that I can proceed to calculate the integral according to the DCT. Unfortunately I'm not able to determine $\lim\limits_{n\to\infty}f_n$. I have that $$\lim\limits_{n\to\infty}\mathbb{1}_{[0,1]}(x)\dfrac{nx^n\sin(nx)-n}{\sqrt{x + 2n^2}} = \lim\limits_{n\to\infty}\mathbb{1}_{[0,1]}(x)\dfrac{x^n\sin(nx)-1}{\sqrt{x + 2}}.$$ $-2\leq x^n\sin(nx)-1 \leq 0$, which means that the limit is oscillating. I'm not sure how to proceed from here.

Question: How do I solve this exercise using the given theorem?

Thanks in advance!

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I give you some ideas. The characteristic function you are carrying around is just there "for confusing you"; id est $$\int_{\mathbb{R}} \mathbb{1}_{[0,1]} (x)\frac{n x^n \sin(nx) - n}{\sqrt{x +2n^2}} \, dx = \int_{[0,1]} \frac{n x^n \sin(nx) - n}{\sqrt{x +2n^2}} \, dx.$$ Moreover your pointwise limit is $$\lim_n \frac{n x^n \sin(nx) - n}{\sqrt{x +2n^2}} = \begin{cases}-1/\sqrt{2} & \text{if } x \in [0,1) \\ \text{it does not exist} & \text{if } x =1 \end{cases}$$and we have also that for $x \in [0,1]$ $$ \frac{|n x^n \sin(nx) - n|}{\sqrt{x +2n^2}} \le \frac{2n}{\sqrt{x + 2n^2}}=\frac{2n}{n \sqrt{x/n^2 + 2}} \le \sqrt{2}$$so your sequence of functions is bounded (uniformly in $x$ and $n$) by a constant, which is integrable over $[0,1]$.

Remember also that even though the pointwise limit does not exist in $x=1$ we have $$ \int_{[0,1]} \frac{n x^n \sin(nx) - n}{\sqrt{x +2n^2}} \, dx = \int_{[0,1)} \frac{n x^n \sin(nx) - n}{\sqrt{x +2n^2}} \, dx$$because a point is of zero-Lebesgue measure.

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