I have $f_n(x)=nxe^{-\sqrt{n}x}$ with $x\in [0,+\infty)$.
$\lim\limits_{n \rightarrow +\infty} \int_{0}^{1} f_n(x)e^{-x^2}dx$=?
The succession doesn't uniformly converge in [0,1] so i can't change limit with integral?
Can I say $\int_{0}^{1} f_n(x)e^{-x^2}dx<\int_{0}^{1} f_n(x) dx=1-e^{-\sqrt{n}}(\sqrt{n}+1)$?
Let $u=\sqrt n x$ then the integral becomes
$$\int_0^{\sqrt n}u e^{-u}e^{-\frac{u^2}n}du\xrightarrow{n\to\infty}\int_0^\infty ue^{-u}du=\Gamma(2)=1$$ using the DCT.