$\lim_{{n \to \infty}} \int_{{\frac{1}{n^2}}}^{{n}} (n^2x - 1)e^{-n^2x^2} \,dx$

Question

The limit of the integral is given

$\lim_{{n \to \infty}} \int_{{\frac{1}{n^2}}}^{{n}} (n^2x - 1)e^{-n^2x^2} \,dx$

(a) Express the integral in the form $\int_0^\infty f_n(y) \, dy$ where $f_n(y)$ represents the general term of the sequence of functions.

(b) Utilize the properties of the sequence of functions and apply either the Monotone Convergence Theorem or the Dominated Convergence Theorem to compute the given limit.

Attempt:

1. Introduce a new variable $t=nx$:

$\lim_{{n \to \infty}} \int_{{\frac{1}{n}}}^{{n^2}} (nt - 1)e^{-t^2} \,dt$

2. How do I continue? Can someone please share her/his solution :).

Answer 1

You need to fix your substitution:

$$\int_{\frac1n}^{n^2} \left(nt - 1\right)e^{-t^2}\frac{\mathrm dt}{n}$$

so $$f_n(t) = \left(t - \frac1n\right)e^{-t^2}\mathbf 1_{\left[\frac1{n}, n^2\right]} (t)\le te^{-t^2}$$

Now you can use Dominated Convergence the limit will be:

$$\int_{0}^{\infty} te^{-t^2}\mathrm d t = \frac12$$

Answer 2

The OP missed to carry a factor $\frac1n$ produced by the change of variable $t=nx$. Then the new form of the integral of interest is $$I_n=\int_{{\frac{1}{n}}}^{{n^2}} \Big(t - \frac1{n}\Big)e^{-t^2} \,dt=\int^\infty_0f_n(t)\,dt$$ where $f_n(t)=\Big(t-\frac1n\Big)_+e^{-t^2}\mathbb{1}_{(0,n^2]}(t)\leq (t+1)e^{-t^2}\in L_1$. As $f_n(t)\xrightarrow{n\rightarrow\infty}te^{-t^2}$ a.s (in fact everywhere), by dominated convergence we obtain $$\lim_nI_n=\int^\infty_0te^{-t^2}\,dt=\frac12$$

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The limit can also be estimated directly without retorting to dominated convergence: $$I_n=\int^n_{1/n^2}n^2te^{-(nt)^2}\,dt-\int^n_{1/n^2}e^{-(nt)^2}$$ The second internal can be estimated as $$0\leq \int^n_{1/n^2}e^{-(nt)^2}\,d=\frac{1}{n}\int^{n^2}_{1/n}e^{-u^2}\,du\leq \frac1n\int^\infty_0e^{-u^2}\,du\xrightarrow{n\rightarrow\infty}0$$ The first integral can be estimated by substitution $u=n^2t^2$ $$\int^n_{1/n^2}n^2te^{-(nt)^2}\,dt=\frac12\int^{n^4}_{1/n^2}e^{-u}\,du=\frac12(e^{-1/n^2}-e^{-n^4})\xrightarrow{n\rightarrow\infty}\frac12$$