integral $\begin{equation} \int \limits_{0}^{\infty} \frac{e^{-\sqrt{x}}}{\sqrt{x}} dx \end{equation}$

Question

I've been tasked with calculating the following improper integral:

$$\begin{equation} \int \limits_{0}^{\infty} \frac{e^{-\sqrt{x}}}{\sqrt{x}} dx \end{equation}$$

There are several things here which arise questions:

1) I need first need to determine whether this integral converges or diverges, for that I need to calculate the limit $\lim_{x \to \infty}\frac{f(x)}{g(x)}$ but I'm not sure which part should be the g and which the f. Also if either f or g converge then they should both behave similarly, however on which of them should I check the convergence?

2)In order to determine the convergence/divergence of the integral, and in order to calculate it I need to split the range $(0,\infty)$ at some point, however I do not understand where and why?

Answer 1

$$\sqrt{x}=t$$ $$\frac{1}{\sqrt{x}}dx=2dt$$ $$\implies \begin{equation} \int \limits_{0}^{\infty} {2e^{-t}}dt \end{equation}$$

You can integrate it easily now and find relevant properties.

Answer 2

Let $x=t^2\implies dx=2t\ dt$ $$=\int_0^{\infty} \frac{e^{-\sqrt x}}{\sqrt x}\ dx$$ $$=\int_0^{\infty} \frac{e^{-t}}{t}(2t\ dt)$$ $$=2\int_0^{\infty} e^{-t}\ dt$$ $$=2[-e^{-t}]_0^{\infty} $$ $$=2[-\lim_{t\to \infty}e^{-t}+\lim_{t\to 0}e^{-t}]$$ $$=2(0+1)$$ $$=2$$