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!
 A: 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.
A: 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 $[0,1]$.
A: 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 $n \gt 0$.
A: Without giving too much away, there is a way to simply evaluate the integral. It is related to the Gamma function $\Gamma(\frac{1}{2})$ but there is a subtlety involing the variable $x$ in place of $nx$. By performing a variable substitution, you should get a Gaussian integral which can be directly evaluated. For starters, let $u=\sqrt x$ and go from there. 
