# Convergence of $\lim_{n \rightarrow \infty} \int_{0}^{\infty} ne^{-nx} \frac{x^2+1}{x^2+x+1}dx$

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.

• 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. – Antonio Vargas Oct 22 '18 at 0:17
• A hint: Substitute $y = nx$. – Antonio Vargas Oct 22 '18 at 0:19
• 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. – Antonio Vargas Oct 22 '18 at 0:30
• 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! – Antonio Vargas Oct 22 '18 at 0:47
• 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. – Antonio Vargas Oct 22 '18 at 0:50

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.
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 $$\lim_{n \to \infty} \int_0^\infty \frac{nx e^{-nx}}{x^2 + 1} \, dx = 0.$$
$$I_n=\int\frac{nx e^{-nx}}{x^2 + 1} \, dx=n \int\frac{x e^{-nx}}{(x+i)(x-i)} \, dx=\frac n 2\int \left(\frac{e^{-nx}}{x+i}+\frac{e^{-nx}}{x-i} \right)\,dx$$ $$I_n=\frac n 2\left(e^{i n} \text{Ei}(-n (x+i))+e^{-i n} \text{Ei}(-n (x-i))\right)$$ 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 $$J_n=\frac{1}{n}-\frac{6}{n^3}+\frac{120}{n^5}+O\left(\frac{1}{n^7}\right)$$