Determine if $\lim_{n\rightarrow \infty }\int_{-\infty}^{\infty}\frac{(\sin(x))^{n}}{x^2+1}d\lambda(x)$ exists and calculate its value if it exists.

Determine if the next limit exists and calculate its value if it exists. $$\lim_{n\rightarrow \infty }\int_{-\infty}^{\infty}\frac{(\sin(x))^{n}}{x^2+1}d\lambda(x)$$ where $\lambda$ is the Lebesgue measure in $\mathbb{R}$

My attepmt: We consider $f_{n}(x)=\frac{(\sin(x))^{n}}{x^2+1}$, and $g(x)=\frac{1}{x^2+1}$. Note that $g$ is absoltely improper Riemann integragle, that is, $\int_{-\infty}^{\infty}|g(x)|dx<\infty$ because $$\int_{-\infty}^{\infty}|g(x)|dx=\int_{-\infty}^{\infty}g(x)dx=\int_{-\infty}^{\infty}\frac{1}{x^2+1}dx=2\int_{0}^{\infty}\frac{1}{x^2+1}dx=\pi<\infty.$$ Therefore, $g$ is Lebegsue integrable in $[-\infty,\infty]$ and $$\int_{-\infty}^{\infty}g(x)d\lambda(x)=\int_{-\infty}^{\infty}g(x)dx=\pi <\infty.$$

Note that $|f_{n}(x)|\leq g(x)$ for all $x\in [-\infty,\infty]$. Also, $\{f_{n}(x)\}_{n}$ converge for each $x\in [-\infty,\infty]$, then we define $f$ as $$f(x)=\lim_{n\rightarrow \infty }f_{n}(x)=\left\{\begin{array}{ll}1 & \mathrm{if}\:x=\frac{\pi}{2}+2m\pi \mbox{for some }m\in\mathbb{N} \\ -1 & \mathrm{if}\:x=\frac{3\pi}{2}+2m\pi \mbox{ for some }m\in\mathbb{N}\\ 0 & \mbox{otherwise} \end{array}\right.$$

We have that $\int_{-\infty}^{\infty}f d\lambda(x)=0$ because $f$ is not zero in a discrete set. Is it correct?

Then, by Dominated Convergence Theorem we have that $f$ is Lebesgue integrable and

$$\begin{array}{rcl}\lim_{n\rightarrow \infty }\int_{-\infty}^{\infty}\frac{(\sin(x))^{n}}{x^2+1}d\lambda(x)&=&\lim_{n\rightarrow \infty }\int_{-\infty}^{\infty}f_{n}(x)d\lambda(x)\\ &=&\int_{-\infty}^{\infty}\lim_{n\rightarrow \infty }f_{n}(x)d\lambda(x) \\ &=&\int_{-\infty}^{\infty}f(x) d\lambda(x)\\&=&0.\end{array}$$

The question: Is my attempt correct? Do I have an error? and if I have it, what is it?

• You may just say that $f_n\to 0$ almost everywhere on $\Bbb R$, that's enough for DCT.
– A.Γ.
Nov 19, 2017 at 21:06

The only small mistake is that for $x$ of the form $3\pi/2+2\pi m$, $f_n(x)$ is of the form $(-1)^n c$ where $c$ do not depend on $n$ hence the sequence $\left(f_n(x)\right)_{n\geqslant 1}$ is not convergent.
However, the almost everywhere convergence is sufficient to apply the dominated convergence theorem. Actually, we do not even need to see what happens on the set $N:=\left\{\pi/2+2\pi m,m\in\mathbb Z\right\}\cup \left\{3\pi/2+2\pi m,m\in\mathbb Z\right\}$ since its measure is zero.