$\lim_n (\int\frac{e^{-nx}}{\sqrt{x}}d\mu)_{n\in\mathbb{N}}$

Question

Let $(0,\infty)$ measure space with borelians in $(0,\infty)$ and $\mu$ Lebesgue's measure. Show that $(\int\frac{e^{-nx}}{\sqrt{x}}d\mu)_{n\in\mathbb{N}}$ converges and find the limit.

Why converges?

Answer 1

For $x \in (0, \infty)$, we have

$e^{-nx} \le \frac{1}{e^{x}} < \frac{1}{1+x+\frac{1}{2}x^2}<\frac{1}{1+x}$

Let $f_n(x)=\frac{e^{-nx}}{\sqrt{x}}$, let $g(x)=\frac{1}{\sqrt{x}(1+x)}$.

$f_n$ are measurable, $f_n(x) \rightarrow f(x) = 0$ for each $x \in (0, \infty)$. And we could check that $g$ is integrable ($2\arctan(\sqrt{x})\bigg |_0^{\infty}=\pi$). Plus we have $|f_n(x)| \le g(x)$. By Dominated Convergence Theorem, we know that:

$$\lim_{n \rightarrow \infty} \int f_n d\mu = \int f d\mu = 0$$

Answer 2

This can be done using the dominated convergence theorem, for example using $e^{-nx} \leq e^{-x} \leq 1 + e^{-x}$. So for example ${1 + e^{-x} \over \sqrt{x}}$ is a dominating function.

Another approach is to change variables to $y = nx$. Then you are looking at $\lim_{n \rightarrow \infty} {1 \over \sqrt{n}} \int_0^{\infty} {e^{-y} \over \sqrt{y}}\,dy = 0$.