Define rv $X$ and function $f_n$ such that $I_n=E(f_n(X))=\int\limits_0^n\frac{x^2+5}{x^3+nx^2+1}\,\mathrm{d}x$ and evaluate $\lim I_n$

Question

Define random variables $X,\,Y$ and functions $f_n,\,g_n$ such that $$I_n=E(f_n(X))=\int\limits_0^n\frac{x^2+5}{x^3+nx^2+1}\,\mathrm{d}x,~~~~J_n=E(g_n(X))=\int\limits_0^nne^{-nx}\,\frac{\cos x}{\sqrt{x^2+1}}\,\mathrm{d}x.$$ Then evaluate
$$\lim_{n\to +\infty} I_n ~~~\textrm{and}~~~\lim_{n\to +\infty} J_n.$$

Attempt. I am aware of the formula: $$E_{\mathrm{P}}\big(f(X)\big)=E_{\mathrm{P}^X}(f)=\int_{\mathbb{R}}f(x)\mathrm{d}\mathrm{P}^X(x).$$ According to this, regarding the first integral, we could use: $$f_n(x)=\frac{x^2+5}{x^3+nx^2+1}\,I_{[0,n]}(x)$$ and a suitable random variable such that $\mathrm{P}^X$ is the Lebesgue measure on $\mathbb{R}.$ But 1) the Lebesgue measure on $\mathbb{R}$ is not probability measure and 2) if so, what would $(\Omega,\mathcal{F},P)$ and $X:\Omega \to \mathbb{R}$ be?

Regarding the second part, $(f_n)$ converges pointwise to $0$, but $|f_n(x)|\leq \frac{1}{x}$ for all $n$ and $x>0$, whereas $\int\limits_0^{+\infty}\frac{\mathrm{d}x}{x}=+\infty,$ so we may not apply the domination convergence theorem this way.

Thanks in advance.

Answer 1

Start with, for $n\ge2$, $$\begin{align}0&\le\int_0^n\frac{5x+5n-1}{(x^3+nx^2+1)(x+n)}dx\\ &\le\int_0^{n^{-1/2}}\frac{10n}{(1)(n)}dx+\int_{n^{-1/2}}^n\frac{10n}{(nx^2)(n)}dx\\ &=10n^{-1/2}+10n^{-1/2}-10n^{-2}\end{align}$$ So $$\begin{align}\lim_{n\rightarrow\infty}\int_0^n\frac{x^2+5}{x^3+nx^2+1}dx &=\lim_{n\rightarrow\infty}\int_0^n\frac{dx}{x+n}+\lim_{n\rightarrow\infty}\int_0^n\frac{5x+5n-1}{(x^3+nx^2+1)(x+n)}dx\\ &=\ln2+0=\ln2\end{align}$$ Then we have, for $0\le x\le1$, $$1-\frac{x^2}2\le\cos x\le1$$ $$1-\frac{x^2}2\le\frac1{\sqrt{x^2+1}}\le1$$ So $$1-x^2\le\left(1-\frac{x^2}2\right)^2\le\frac{\cos x}{\sqrt{x^2+1}}\le1$$ So for $n\ge2$ $$0\le\int_0^{n^{-1/2}}ne^{-nx}\left(\frac{\cos x}{\sqrt{x^2+1}}-(1-x^2)\right)dx\le\int_0^{n^{-1/2}}nx^2dx=\frac13n^{-1/2}$$ So $$\begin{align}\lim_{n\rightarrow\infty}\int_0^{n^{-1/2}}ne^{-nx}\frac{\cos }{\sqrt{x^2+1}}dx&=\lim_{n\rightarrow\infty}\int_0^{n^{-1/2}}ne^{-nx}(1-x^2)dx\\ &+\lim_{n\rightarrow\infty}\int_0^{n^{-1/2}}ne^{-nx}\left(\frac{\cos x}{\sqrt{x^2+1}}-(1-x^2)\right)dx\\ &=\lim_{n\rightarrow\infty}-\left.\left(1-x^2-\frac{2x}n-\frac2{n^2}\right)e^{-nx}\right|_0^{n^{-1/2}}+0\\ &=1\end{align}$$ And then since $$\left|ne^{-nx}\frac{\cos x}{\sqrt{x^2+1}}\right|\le ne^{-nx}$$ We have $$\begin{align}\left|\int_{n^{-1/2}}^nne^{-nx}\frac{\cos x}{\sqrt{x^2+1}}dx\right|&\le\int_{n^{-1/2}}^n\left|ne^{-nx}\frac{\cos x}{\sqrt{x^2+1}}\right|dx\\ &\le\int_{n^{-1/2}}^nne^{-nx}dx=e^{-n^{1/2}}-e^{-n^2}\end{align}$$ So $$\lim_{n\rightarrow\infty}\int_{n^{-1/2}}^nne^{-nx}\frac{\cos x}{\sqrt{x^2+1}}dx=0$$ Adding these last two limits, $$\lim_{n\rightarrow\infty}\int_0^nne^{-nx}\frac{\cos x}{\sqrt{x^2+1}}dx=1+0=1$$ So our strategy in both cases was to find an estimate of the integrand valid in the region where it mattered and integrate the estimate.

Answer 2

The first thing to note about $I_n$ is that the limit is going to be non-zero: for $n\ge 2$, $$ \int_0^n \frac{x^2 + 5}{x^3 + nx^2 + 1} \ge \int_1^n \frac{x^2}{3nx^2} \ge 1/6,$$ where we have used that for $x \in [1,n]$ $1 <x^3 < nx^2.$ As you correctly saw, the integrand goes to $0$ pointwise, so this suggests that DCT style approaches are not viable.

The thing to see is that if the upper limit of the integral was something that was $o(n)$, then the integral would go to $0$ - you can basically upper bound the integrand by $1/n$, except on a region of size $o(1)$, and you get the bound $I_n = o(1)$. This suggests that the most interesting part of the integral is in the linear region. The reason that the direct DCT approach doesn't work is that this region with most of the contribution 'runs off' to $\infty$ with $n$. Let us reparametrise to in instead focus our inverstigation on this region. Before I do this, I'll leave it to you to argue that $\int_0^n \frac{5}{1 + nx^2} \to 0$. This saves me writing the $+5$ terms which are irritating.

Now, using the subtitution $x = ny$, $$ \int_0^n \frac{x^2 }{nx^2 + x^3 + 1} = \int_0^1 \frac{n^3 y^2 }{n^3 y^2 + n^3y ^3 + 1} = \int_0^1 \frac{y^2}{y^2 + y^3 + n^{-3}}.$$

This new intgrand monotonically converges to $\frac{1}{1 + y}$, which is integrable on the space $[0,1]$. Now we can apply the DCT (or, I guess, MCT) to say that $$ \lim I_n = \int_0^1 \frac{1}{1 + y} = \ln 2.$$

I'd guess similar approaches for $J_n$ are viable. Haven't really thought about it. Key point is to figure out where the integral has some meat, and reparametrise to bring that section into focus.