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!

The integrand is less than $1/\sqrt{x}$, so that $0<\int_\varepsilon^1 \frac{e^{-nx}dx}{\sqrt{x}}<\int_0^1 \frac{dx}{\sqrt{x}}=2$, and $\int_0^1 \frac{e^{-nx}dx}{\sqrt{x}}$ exists. As to $\int_1^\infty \frac{e^{-nx}dx}{\sqrt{x}}$, the integrand rapidly decreases to $0$ as $x\rightarrow\infty$, in particular, $0<\int_1^N \frac{e^{-nx}dx}{\sqrt{x}}<\int_1^\infty \frac{dx}{x^2}=1$ is sufficient.
