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I want to show that $\lim_{n \rightarrow \infty} \int_{0}^{\infty} ne^{-nx} \frac{x^2+1}{x^2+x+1} dx$ exists and compute its value.

First I need to show that $ne^{-nx} \frac{x^2+1}{x^2+x+1}$ is integrable.

I think that the problem comes from when $x$ is large. Thus I wanted to say something like, if $x \rightarrow \infty$, then $\frac{x^2+1}{x^2+x+1} \sim 1$ and we have that $ne^{-nx} \frac{x^2+1}{x^2+x+1} \sim ne^{-nx}$ which is integrable for all $n$.

Does that make sense?

If I then apply the dominated convergence theorem to $f_n = ne^{-nx} \frac{x^2+1}{x^2+x+1}$ I'd get $\lim_{n \rightarrow \infty} \int_{0}^{\infty} ne^{-nx} \frac{x^2+1}{x^2+x+1}dx = 0$ right?

I'm really confused and would be thankful if someone could comment on this.

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    $\begingroup$ First, your justification for integrability is right, but it's pretty superficial. Do you know how to fill in the details? So what if $ne^{-nx} \frac{x^2+1}{x^2+x+1} \sim ne^{-nx}$? If $f(x) \sim g(x)$ and $g$ is integrable, what lets you conclude that $f$ is integrable? Second, just saying "apply the dominated convergence theorem" isn't enough. You actually have to apply it. If you did, you would find that the limit you guessed is incorrect. $\endgroup$ – Antonio Vargas Oct 22 '18 at 0:17
  • $\begingroup$ A hint: Substitute $y = nx$. $\endgroup$ – Antonio Vargas Oct 22 '18 at 0:19
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    $\begingroup$ Pointwise, sure. But what's the key part of the dominated convergence theorem? What's your domination function? The dominated convergence theorem is not just "$f_n \to 0$ pointwise implies $\int f_n \to 0$". There's a reason it has "dominated" in the name. $\endgroup$ – Antonio Vargas Oct 22 '18 at 0:30
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    $\begingroup$ First, $g$ must not have an $n$ in it. Second, the point I'm trying to make is that there is no such $g$. It seems like it would be a good exercise for you to convince yourself of this. You can't prove that $\int f_n \to 0$ in this case because it's not true! $\endgroup$ – Antonio Vargas Oct 22 '18 at 0:47
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    $\begingroup$ Correct. But you have been given two different approaches where you can eventually use the DCT. The first was in my second comment, and the second was in Umberto's answer. $\endgroup$ – Antonio Vargas Oct 22 '18 at 0:50
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By the dominated convergence theorem, if $f(x)$ is a bounded function on $\mathbb{R}^+$ we have (provided that the RHS makes sense) $$ \lim_{n\to +\infty}\int_{0}^{+\infty} n e^{-nx} f(x)\,dx = \lim_{x\to 0^+} f(x) $$ since $\int_{0}^{+\infty} n e^{-nx}\,dx = 1$ but $n e^{-nx}$ gets more and more concentrated around the origin as $n$ increases. Approximate identities are pretty useful for showing $$ \lim_{x\to 0^+}\sum_{n\geq 1}\frac{\sin(nx)}{n} = \frac{\pi}{2}$$ and similar identities, for instance.

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It is evident that $\int_0^\infty ne^{-nx} \, dx = 1$ for all $n$ and that $f_n(x) = ne^{-nx} \dfrac{x^2+1}{x^2 + x + 1}$ is integrable, since $0 \le f_n(x) \le ne^{-nx}$ for all $n$. The difference between these integrals satisfies $$\left| \int_0^\infty n e^{-nx} \, dx - \int_0^\infty ne^{-nx} \frac{x^2+1}{x^2 + x + 1} \, dx \right| = \int_0^\infty \frac{nxe^{-nx}}{x^2+1} \, dx.$$ The maximum value of $te^{-t}$ for $t > 0$ occurs when $t=1$, so that $$0 \le \frac{nx e^{-nx}}{x^2 + 1} \le \frac{e^{-1}}{x^2 + 1}$$for all $n$ and all $x > 0$. Since $g(x) = \frac{1}{x^2+1}$ is integrable, the dominated convergence theorem gives you $$\lim_{n \to \infty} \int_0^\infty \frac{nx e^{-nx}}{x^2 + 1} \, dx = 0.$$

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  • $\begingroup$ thank you Umberto! $\endgroup$ – riri92 Oct 22 '18 at 1:07
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Too long for a commant.

Starting from Umberto P.'s good answer, we can approximate the last integral.

$$I_n=\int\frac{nx e^{-nx}}{x^2 + 1} \, dx=n \int\frac{x e^{-nx}}{(x+i)(x-i)} \, dx=\frac n 2\int \left(\frac{e^{-nx}}{x+i}+\frac{e^{-nx}}{x-i} \right)\,dx$$ $$I_n=\frac n 2\left(e^{i n} \text{Ei}(-n (x+i))+e^{-i n} \text{Ei}(-n (x-i))\right)$$ Then $$J_n=\int_0^\infty \frac{nx e^{-nx}}{x^2 + 1} \, dx=n \left(\frac{1}{2} (\pi -2 \text{Si}(n)) \sin (n)-\text{Ci}(n) \cos (n)\right)$$ Using thr series expansions of the sine and cosine integrals for large values of $n$, we ned with $$J_n=\frac{1}{n}-\frac{6}{n^3}+\frac{120}{n^5}+O\left(\frac{1}{n^7}\right)$$

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