The original problem is to compute $\lim_{n \to \infty} \int_{\mathbb{R}}\frac{\cos^n x}{1+x^2} dm(x),$ where $m$ is the Lebesgue measure. I think this can be done by observing that $f_n(x):= \frac{\cos^n x}{1+x^2}$ is bounded above by $g(x) := \frac{1}{1+x^2}$, which is integrable on $\mathbb{R}$, and also that $f_n(x)$ converges to $0$ a.e. Hence, by the dominated convergence theorem, the answer is $0$.
My question is whether the following limit of Riemann integral (instead of Lebesgue integral) also makes sense and is equal to the above : $$\lim_{n \to \infty} \int_{-\infty}^{\infty}\frac{\cos^n x}{1+x^2} dx.$$
Here is my attempt: Using $f_n(x) \leq g(x)$ for all $x$ and $\int_{-\infty}^{\infty}g(x) dx = \pi$, we can say that the improper (Riemann) integral exists (and is less than or equal to $\pi$), which depends on $n$. But it is not clear to me why the limit exists.
Another thing I know is the following result:
Let $f$ be a bounded function on a compact interval $[a, b]$. If $f$ is Riemann integrable on $[a, b]$, then $f$ is Lebesgue integrable on $[a, b]$, and $$\int_a^b f(x) dx = \int_{[a, b]} f(x) dm(x).$$
But this result requires a compact interval, which is not the case here.
To summarize, could anyone give a rigorous explanation of how to show $$\lim_{n \to \infty} \int_{\mathbb{R}}\frac{\cos^n x}{1+x^2} dm(x) = \lim_{n \to \infty} \int_{-\infty}^{\infty}\frac{\cos^n x}{1+x^2} dx,$$ if it is indeed true? Thank you.