Finding $lim_{n to infty} int_{2}^{infty} frac{nsinleft(frac{x-2}{n}right)}{(x-2)+(1+(x-2)^2)} dx$

Question

I have to calculate the following limits, using a theorem but I don't really know what theorem to use (it is for the subject of measurement and integration, for the unit "Measurable functions, integration and its properties"). $space$
$$lim_{n to infty} int_{2}^{infty} frac{nsinleft(frac{x-2}{n}right)}{(x-2)+(1+(x-2)^2)} dx$$

Do I have to use the dominated convergence theorem of Lebesgue?
I have first of all calculated $limlimits_{n to infty}frac{nsinleft(frac{x-2}{n}right)}{(x-2)+(1+(x-2)^2)}$ and I've obtained $frac{1}{2x-3}$
Now, I want to calculate $int_{2}^{infty} frac{1}{2x-3},dx$ but $ln(infty)$ doesn't exist... so what am I doing wrong?

metamorphy · Answer

I would "drop the tail" and use the monotone convergence theorem (which doesn't require the finiteness).
Our integral is $$I_n=int_0^inftyfrac{nsin(x/n)}{1+x+x^2},dx=J_n+R_n$$ with $J_n=int_0^{npi}$ and $R_n=int_{npi}^infty$. Now $|R_n|leqslant nint_{npi}^inftyfrac{dx}{x^2}=frac1pi$, i.e. $R_n$ is bounded, and $limlimits_{ntoinfty}J_n=+infty$ by MCT (indeed, $J_n=int_0^infty f_n(x),dx$ with $f_n(x)=0$ for $xgeqslant npi$, and $nmapsto f_n(x)$ is nondecreasing for each fixed $x$, because $tmapsto(sin t)/t$ is decreasing for $tin[0,pi]$). Thus, $limlimits_{ntoinfty}I_n=+infty$.

Finding $lim_{n to infty} int_{2}^{infty} frac{nsinleft(frac{x-2}{n}right)}{(x-2)+(1+(x-2)^2)} dx$

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