Answer
$$2$$
Work Step by Step
Let us consider that $I=\int_{0}^{\infty} \dfrac{e^{-\sqrt x}}{\sqrt x} dx$
Plug $a=\sqrt x \implies dx=2a \ da$
Integrate the integral by using limits.
$I=\lim\limits_{z \to 0^{+}}\int_{z}^1 \dfrac{e^{-\sqrt x}}{\sqrt x}\ dx+\lim\limits_{z \to \infty}\int_{1}^z \dfrac{e^{-\sqrt x}}{\sqrt x} dx \\=\lim\limits_{z \to 0^{+}}[-2e^{-\sqrt x}]_z^1+\lim\limits_{z \to \infty}[-2 e^{-\sqrt x}]_1^z\\=-2 e^{-1}+2+0+2e^{-1}\\=2$
