Answer
Convergent $\;,\;$ 2
Work Step by Step
Let \[I=\int_{0}^{\infty}e^{-\sqrt{y}}dy\]
\[I=\lim_{t\rightarrow \infty}\int_{0}^{t}e^{-\sqrt{y}}dy\;\;\;\ldots (1)\]
Let \[I_1=\int e^{-\sqrt{y}}dy\]
Substitute $\; r=-\sqrt{y}\;\;\;\ldots (2)$
\[\Rightarrow dr=-\frac{dy}{2\sqrt y}\]
\[I_1=2\int re^{r}dr\]
Using integration by parts
\[I_1=2\left[r\int e^{r}dr-\int\left((r)'\int e^{r}dr\right)dr\right]\]
\[I_1=2[re^{r}-\int e^{r}dr]\]
\[I_1=2(r-1)e^{r}\]
From (2)
\[I_1=-2(\sqrt y+1)e^{-\sqrt y}\;\;\;\ldots (3)\]
Using (3) in (1)
\[I=\lim_{t\rightarrow \infty}\left[-2(\sqrt y+1)e^{-\sqrt y}\right]_{0}^{t}\]
\[I=\lim_{t\rightarrow \infty}\left[-2(\sqrt t+1)e^{-\sqrt t}+2\right]\]
\[I=2+\lim_{t\rightarrow \infty}\left[\frac{-2(\sqrt t+1)}{e^{\sqrt t}}\right] \;\;\;\;\left(\frac{\infty}{\infty}form\right)\]
Using L' Hopital's rule
\[I=2+\lim_{t\rightarrow \infty}\left[\frac{\frac{-2}{2\sqrt t}}{\frac{e^{\sqrt t}}{2\sqrt t}}\right]\]
\[I=2+\lim_{t\rightarrow \infty}\left[\frac{-2}{e^\sqrt t}\right]=2\]
Since limit on R.H.S. of (1) exists so $I$ is convergent and $I=2$.
