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 ?
 A: $$I=\int_0^\infty\frac{e^{-t}}{\sqrt{t}}\,dt$$
$u=\sqrt{t}\,,dt=2\sqrt{t}du$
$$I=2\int_0^\infty e^{-u^2}\,du=\sqrt{\pi}$$ 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 $I=\sqrt{\pi}$
A: $$\int_0^{+\infty}\frac{e^{-t}}{\sqrt{t}}dt=\int_{0}^{1}\frac{e^{-t}}{\sqrt{t}} \, dt + \int_1^{+\infty}\frac{e^{-t}}{\sqrt{t}} \, dt=A+B$$
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 
$$B=\int_{1}^{+\infty}\frac{e^{-t}}{\sqrt{t}}\,dt\leq\int_1^{+\infty}e^{-t} \, dt < \infty$$
A: As Henry Lee's answer mentions, you get $\sqrt\pi,$ so $2e/3$ is wrong. But if the function is everywhere positive and you computed it correctly and get a finite number rather than $+\infty,$ then it converges.  If it's not everywhere positive, then the meaning of convergence may become problematic. For example,
$$
\lim_{a\,\to\,+\infty}\int_0^a \frac{\sin x} x \, dx = \frac \pi 2
$$
and that is a sort of convergence, but
$$
\int\limits_{\{\,x \,:\, (\sin x)/x \, \ge\,0\}} \frac{\sin x} x\,dx = +\infty \quad\text{ and } \quad \int\limits_{\{\,x \,:\, (\sin x)/x \, < \,0\}} \frac{\sin x} x\,dx = -\infty
$$
and some questions arise about when one should consider the thing convergent.
But one thing to bear in mind is that the question of whether something converges or not is often simpler than the question of what it converges to. Thus
$$
0 < \int_0^1 \frac{e^{-t}}{\sqrt t} \, dt \le \int_0^1 \frac 1 {\sqrt t} \, dt < +\infty
$$
and
$$
0 < \int_1^\infty \frac{e^{-t}}{\sqrt t}\,dt \le \int_1^\infty e^{-t} \, dx = e^{-1} < +\infty
$$
so what you have converges.
A: 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. 
A: If we can calculate a finite value for the integral of course the integral converges but if our goal is simply to check for the convergence we don’t need to calculate it explicitly but we can proceed for example by some quick convergence test. 
For example, in this case we have that
$$ \int^{+\infty}_0\frac{e^{-t}} {\sqrt t} \, dt=\int^{1}_0\frac{e^{-t}} {\sqrt t} \, dt+
\int^{+\infty}_1\frac{e^{-t}} {\sqrt t} \, dt$$
and as $t\to 0^+$
$$\frac{e^{-t}} {\sqrt t} \sim 
\frac{1} {\sqrt t}$$
and as $t\to \infty$
$$\frac{e^{-t}} {\sqrt t} \sim 
\frac{1} {e^t}$$
therefore both integral converge by limit comparison test and therefore the given integral converges.
