I am attempting to compute the limit of
$$\displaystyle\lim_{n \to \infty} \int_0^1 \frac{e^{-nt}}{\sqrt{t}} \, dt$$
Proof Attempt
Recall that $\int_a^b f(x) \, dx = \displaystyle\lim_{n \to \infty} \int_a^b f_n(x) \, dx$ provided that each $f_n$ is continuous and $f_n \to f$ converges uniformly on $[a,b]$ (Thomson & Bruckner, p. 388).
Denote $\langle f_n \rangle$ be a sequence of functions defined by $f_n(x)= \frac{e^{-nt}}{\sqrt{t}}$ where $n \in \mathbb{N}$ and $t \in [0,1]$. Note that each $f_n$ is continuous on $[0,1]$.
Consider, the fact that $f_n \to f$ pointwise, but does not converge to the function $f$ uniformly on $[0,1]$, however. This is because $t=0$ is undefined for the function $f_n \hspace{0.3cm} \forall n \in \mathbb{N}$. But notice that $f_n \to f$ uniformly on $[a,1]$ for $a>0$. So Theorem 9.26 is applicable for $t \in [a,1]$. Thus, consider the following:
$$(A.) \hspace{0.5cm} \displaystyle\lim_{n \to \infty} \int_0^1 f_n(x) = \displaystyle\lim_{n \to \infty} \int_{\frac{1}{n}}^1 f_n(x)$$
since, for $n$ large enough, $\frac{1}{n} < \varepsilon$ for any $\varepsilon > 0$ (Archimedean Property). Since $\displaystyle\lim_{n \to \infty} \langle \frac{1}{n}: n \in \mathbb{N} \rangle = 0$, the above statement $(A.)$ holds.
NOTE: This proof is not complete.
The basic idea of my proof is to observe that $t=0$ breaks the functions $f_n$. So the intuition is to split up $\int f_n \, dt$ into some other integrals that satisfy uniform convergence so that we may use the result from Thomson & Bruckner.
Does this motivation for a proof seem in the right ballpark? Or is there a better means of computing this limit?
Best Answer
Υοu can find the limit without using uniform convergence and so many estimations.
Just make the change of variables $y=\sqrt{t}$ and another change of variables and using the fact that $e^{-y^2}$ is positive, the integral $I_n$ becomes: $$0 \leq I_n=\frac{2}{\sqrt{n}}\int_0^{\sqrt{n}}e^{-y^2}dy \leq \frac{2}{\sqrt{n}}\int_0^{\infty}e^{-y^2}dy \to 0$$
You just have to use that the integral $\int_0^{\infty}e^{-y^2}dy$ converges (and we also know its value)