How to prove that $\int_{0}^{\infty}{\frac{e^{-nx}}{\sqrt{x}}}\mathrm dx$ exists I am trying to show that the integral $\int_{0}^{\infty}{\frac{e^{-nx}}{\sqrt{x}}}\mathrm dx$ exists ($n$ is a natural number). I tried to use the comparison theorem by bounding from above the integrand by another function which is integrable, but I wasn't successful. 
The same question for this integral as well : $\int_{0}^{\infty}x^2e^{-nx}\mathrm dx$ 
Any help is much appreciated? Thanks!

Regarding $\int_{0}^{\infty}e^{-nx}/\sqrt{x} dx$, the tail is not a problem because for $x > 1$ we have
$$0 < \frac{e^{-nx}}{\sqrt{x}} < e^{-nx}$$
and the right hand side is integrable.
The only other cause for concern is the singularity at $x=0$. But this is also not a problem, because for all $x > 0$, we have $e^{-nx} \leq 1$, and so
$$0 < \frac{e^{-nx}}{\sqrt{x}} \leq \frac{1}{\sqrt{x}}$$
and the right hand side is improperly integrable on $[0,1]$.
