Answer
$2$
(converges)
Work Step by Step
$\displaystyle \int e^{-\sqrt{y}}dy=\left[\begin{array}{ll}
u=y^{1/2} & \\
du=\frac{1}{2}y^{-1/2}du & dy=2udu
\end{array}\right]=\displaystyle \int e^{-u}(2udu)$
Table integral 96: $\displaystyle \int ue^{au}du=\frac{1}{a^{2}}(au-1)e^{au}+C$
$2\displaystyle \int ue^{-u}du=2(u-1)e^{u}=2(\sqrt{y}-1)e^{-\sqrt{y}}$
The integral is a type I improper integral, so
$I=\displaystyle \int_{0}^{\infty}e^{-\sqrt{y}}dy=\lim_{t\rightarrow\infty}\int_{0}^{t}e^{-\sqrt{y}}dy=\lim_{t\rightarrow\infty}\left[2(\sqrt{y}-1)e^{-\sqrt{y}}\right]_{0}^{t}$
$=\displaystyle \lim_{t\rightarrow\infty}\left[\frac{2(\sqrt{y}-1)}{e^{\sqrt{y}}}\right]_{0}^{t}$
$=\displaystyle \lim_{t\rightarrow\infty}\left[\frac{2(\sqrt{t}-1)}{e^{\sqrt{t}}}\right]-\lim_{t\rightarrow\infty}\left[\frac{-2}{1}\right]$
...L'Hospital's rule
$\displaystyle \lim_{t\rightarrow\infty}\frac{\sqrt{t}-1}{e^{\sqrt{t}}} =\displaystyle \lim_{t\rightarrow\infty}\frac{\frac{1}{2\sqrt{t}}}{e^{\sqrt{t}}\cdot\frac{1}{2\sqrt{t}}}=\lim_{t\rightarrow\infty}\frac{1}{e^{\sqrt{t}}}=0$
$I=0+2$
$=2$
(converges)