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

I have computed it getting $$ \frac{2e}{3}    \therefore \text{ it converges.}$$ using integration by parts letting $$u = \frac 1 {\sqrt t} $$ and $$ dV= e^{-t} \, dt$$
Is it the right way to do it ?

$$I=\int_0^\infty\frac{e^{-t}}{\sqrt{t}}\,dt$$
$u=\sqrt{t}\,,dt=2\sqrt{t}du$
$$I=2\int_0^\infty e^{-u^2}\,du=\sqrt{\pi}$$ 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 $I=\sqrt{\pi}$
