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!

Adding a few things to the good answers you already received, using the change of variable suggested by TeeJay, you obtain $$\int{\frac{e^{-nx}}{\sqrt{x}}}dx=\frac{\sqrt{\pi } \text{erf}\left(\sqrt{n} \sqrt{x}\right)}{\sqrt{n}}$$ from which $$\int_{0}^{\infty}{\frac{e^{-nx}}{\sqrt{x}}}dx=\frac{\sqrt{\pi }}{\sqrt{n}}$$ provided $n \gt 0$.
