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$f_n(x)=nxe^{-\sqrt{n}x}$ for $x \in [0,+\infty)$. The sequence of functions pointwise converges in $x \in [0,+\infty)$ to the null function but not uniformly. There is uniformly convergence in sub-interval $[a,+\infty)$ with a>0. To calculate the limit I can't said $\lim\limits_{n \rightarrow +\infty} \int_{0}^{1}f_n(x)e^{-x^2} dx$=$ \int_{0}^{1}\lim\limits_{n \rightarrow +\infty}f_n(x)e^{-x^2} dx$?

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  • $\begingroup$ i have correct the text $\endgroup$ – Giulia B. Feb 11 at 18:06
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Let $$ I_n = n\int_0^1 xe^{-\sqrt{n}x}e^{-x^2}\mathrm dx.$$ Make change of variable $u=\sqrt{n}x$ to obtain $$\begin{align*} I_n =\int_0^\sqrt{n} ue^{-u}e^{-\frac{u^2}n} \mathrm du=\int_0^\infty ue^{-u}e^{-\frac{u^2}n}1_{\{u\le \sqrt{n}\}} \mathrm du. \end{align*}$$ Then we find that $$ 0\le e^{-\frac{u^2}n}1_{\{u\le \sqrt{n}\}} \le e^{-\frac{u^2}{n+1}}1_{\{u\le \sqrt{n+1}\}} \xrightarrow{n\to\infty} 1. $$ Thus by monotone convergence theorem, we have $$\begin{align*} \lim_{n\to\infty}I_n &=\lim_{n\to\infty}\int_0^\infty ue^{-u}e^{-\frac{u^2}n}1_{\{u\le \sqrt{n}\}} \mathrm du\\ &=\int_0^\infty ue^{-u}\mathrm du=\left[-ue^{-u}\right]^\infty_0+\int_0^\infty e^{-u}\mathrm du\\ &=1. \end{align*}$$

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Hint: you can explicitly evaluate the integral:

$$\int_c^d xe^{-ax}dx = -\left.\frac{1}{a}xe^{-ax}\right|_c^d+\int_c^d\frac{1}{a} e^{-ax}dx=...$$

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