Determine whether the integral $ \int^{+\infty}_0\frac{e^{-t}} {\sqrt t} \, dt$ converges or diverges?

If you simply want to decide convergence and you don't notice the trick in Henry Lee's answer, you can look at the following. Near zero, $$ \int_0^1 \frac{e^{-t}}{\sqrt t}\,dt\leq \int_0^1\frac1{\sqrt t}\,dt, $$ which is convergent.

At infinity, $$ \int_1^\infty \frac{e^{-t}}{\sqrt t}\,dt\leq\int_1^\infty e^{-t}\,dt<\infty. $$ So the integral converges.

$$I=\int_0^\infty\frac{e^{-t}}{\sqrt{t}}\,dt$$ $u=\sqrt{t}\,,dt=2\sqrt{t}du$ $$I=2\int_0^\infty e^{-u^2}\,du=\sqrt{\pi}$$ As it is a standard integral

EDIT: $$I=2\int_0^\infty e^{-x^2}\,dx$$ then $$I^2=4\left(\int_0^\infty e^{-x^2}\,dx\right)^2=4\left(\int_0^\infty e^{-x^2}\right)\left(\int_0^\infty e^{-y^2} \, dy\right) = 4\int_0^\infty \int_0^\infty e^{-(x^2+y^2)} \, dx \, dy$$ now we can use polar coordinates to simplify this. $x^2+y^2=r^2\,$ and $dA=dx\,dy=r\,dr\,d\theta$ so our integral becomes: $$I^2=4\int_0^{\frac{\pi}{2}} \int_0^\infty e^{-r^2}r\,dr\,d\theta$$ now $u=-r^2$ so $\frac{du}{dr}=-2r\, \therefore\,dr=\frac{du}{-2r}$ and the integral becomes: $$I^2=-2\int_0^{\frac{\pi}{2}} \int_0^{-\infty}e^u \, du \, d\theta = 2\int_0^{\frac{\pi}{2}} \int_{-\infty}^0 e^u\,du\,d\theta = 2\int_0^{\frac{\pi}{2}} \left[e^u\right]_{-\infty}^0 \, d\theta = 2\int_0^{\frac{\pi}{2}} \, d\theta =2\cdot\frac{\pi}{2}=\pi$$ so if $I^2=\pi$ then $I=\sqrt{\pi}$

$$\int_0^{+\infty}\frac{e^{-t}}{\sqrt{t}}dt=\int_{0}^{1}\frac{e^{-t}}{\sqrt{t}} \, dt + \int_1^{+\infty}\frac{e^{-t}}{\sqrt{t}} \, dt=A+B$$ for $A$ when $e^{-t}\sim1$ then $$A=\int_0^1 \frac{e^{-t}}{\sqrt{t}}\,dt\sim\int_0^1 \frac{1}{\sqrt{t}} \, dt < \infty$$ for $B$ when $t\geq1$ then $$B=\int_{1}^{+\infty}\frac{e^{-t}}{\sqrt{t}}\,dt\leq\int_1^{+\infty}e^{-t} \, dt < \infty$$