# Calculate $\lim\limits_{n \rightarrow +\infty} \int_{0}^{1}f_n(x) e^{-x^2}dx$.

$$f_n(x)=nxe^{-\sqrt{n}x}$$ 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 $$\lim\limits_{n \rightarrow +\infty} \int_{0}^{1}f_n(x)e^{-x^2} dx$$=$$\int_{0}^{1}\lim\limits_{n \rightarrow +\infty}f_n(x)e^{-x^2} dx$$?

• i have correct the text – Giulia B. Feb 11 at 18:06

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