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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"). $\space$

$$\lim_{n \to \infty} \int_{2}^{\infty} \frac{n\sin\left(\frac{x-2}{n}\right)}{(x-2)+(1+(x-2)^2)} dx$$

Do I have to use the dominated convergence theorem of Lebesgue?

I have first of all calculated $\lim\limits_{n \to \infty}\frac{n\sin\left(\frac{x-2}{n}\right)}{(x-2)+(1+(x-2)^2)}$ 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?

StubbornAtom
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User160
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1 Answers1

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I would "drop the tail" and use the monotone convergence theorem (which doesn't require the finiteness).

Our integral is $$I_n=\int_0^\infty\frac{n\sin(x/n)}{1+x+x^2}\,dx=J_n+R_n$$ with $J_n=\int_0^{n\pi}$ and $R_n=\int_{n\pi}^\infty$. Now $|R_n|\leqslant n\int_{n\pi}^\infty\frac{dx}{x^2}=\frac1\pi$, i.e. $R_n$ is bounded, and $\lim\limits_{n\to\infty}J_n=+\infty$ by MCT (indeed, $J_n=\int_0^\infty f_n(x)\,dx$ with $f_n(x)=0$ for $x\geqslant n\pi$, and $n\mapsto f_n(x)$ is nondecreasing for each fixed $x$, because $t\mapsto(\sin t)/t$ is decreasing for $t\in[0,\pi]$). Thus, $\lim\limits_{n\to\infty}I_n=+\infty$.

metamorphy
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