Show that $$\lim\limits_{n \to \infty} \int_{\alpha}^{\infty} \sqrt n e^{-nx^2}\ dx = \int_{\alpha}^{\infty} \lim\limits_{n \to \infty} \sqrt n e^{-nx^2}\ dx$$ for all $\alpha > 0$ but not for $\alpha = 0.$
As far as I have understood the question it turns out that we need only to prove that the sequence of functions $\{f_n \}$ defined by $f_n (x) = \sqrt n e^{-nx^2},\ \ n \geq 1$ is uniformly convergent on $[\alpha, \infty)$ for any $\alpha > 0$ but not uniformly convergent on $[0, \infty).$
The first part is easier. Let $\alpha > 0.$ Choose $\varepsilon > 0$ arbitrarily. Let $k \in \Bbb N$ be any positive integer such that $k > \left \lfloor \frac {4} {{\alpha}^4 {\varepsilon}^2} \right \rfloor.$ Then for any two positive integers $m,n \geq k$ we have \begin{align*}\left |f_n(x) - f_m(x) \right | & = \left | \sqrt n e^{-nx^2} - \sqrt m e^{-mx^2} \right| \\ & \leq \sqrt n e^{-nx^2} + \sqrt m e^{-mx^2} \\ & \leq \frac {\sqrt n} {nx^2} + \frac {\sqrt m} {mx^2} \\ & \leq \frac {1} {\sqrt n {\alpha}^2} + \frac {1} {\sqrt m {\alpha^2}} \\ & \leq \frac {2} {\sqrt k {\alpha}^2} < \varepsilon \end{align*} Hence by Cauchy's criterion for uniform convergence it follows that the sequence of functions $\{f_n \}$ is uniformly convergent on $[\alpha, \infty)$ for any $\alpha > 0,$ as required.
Now how do I prove the second part i.e. the sequence of functions $\{f_n \}$ is not uniformly convergent on $[0, \infty)$?
Any help in this regard will be highly appreciated. Thanks for your time.
For the second part, $$\int_0^\infty \sqrt{n}e^{-nx^2}=\int_0^\infty e^{-u^2}\,\mathrm d u=\frac{\sqrt\pi}{2},$$ and thus $$\frac{\sqrt\pi}{2}=\lim_{n\to \infty }\int_0^\infty \sqrt{n}e^{-nx^2}\,\mathrm d x\neq 0=\int_0^\infty \lim_{n\to \infty }\sqrt ne^{-nx^2}\,\mathrm d x.$$