Evaluate $\lim_{n \to \infty}\int_{0}^{\infty}e^{-nx}x^{-1/2}dx$

Question

I would like to evaluate $$\lim_{n \to \infty}\int_{0}^{\infty}e^{-nx}x^{-1/2}dx.$$ The purpose of the problem was to show that

(1) $\int_{0}^{\infty}e^{-nx}x^{-1/2}dx$ converges for every natural number $n$

(2) $\lim_{n \to \infty}\int_{0}^{\infty}e^{-nx}x^{-1/2}dx \neq 0$

I used the substitution method where $u = \sqrt{x}$, then $\int_{0}^{\infty}e^{-nx}x^{-1/2}dx$ was reduced to $2\int_{0}^{\infty}e^{-nu^2}du$. When $n = 2^m$ for some natural number $m$, then it seems that $2\int_{0}^{\infty}e^{-nu^2}du = \frac{\sqrt{\pi}}{\sqrt{2}^m}$, but how do I show that $\int_{0}^{\infty}e^{-nx}x^{-1/2}dx$ converges for every natural number $n$ as well as $\lim_{n \to \infty}\int_{0}^{\infty}e^{-nx}x^{-1/2}dx \neq 0$.

Any help will be greatly appreciated.

Answer 1

$$I_n=\int_0^1\frac{e^{-nx}}{\sqrt x}dx+\int_1^\infty\frac{e^{-nx}}{\sqrt x}dx<\int_0^1\frac{dx}{\sqrt x}+\int_1^\infty e^{-nx}dx=2+\frac{e^{-n}}n$$ proves that the integral converges.

With $nx=t^2$, we have

$$I_n=\frac2{\sqrt n}\int_0^{\infty}e^{-t^2}dt,$$ which obviously has the limit $0$.

Answer 2

For $n$ large enough, $$ \int_0^\infty e^{-nx} x^{-1/2} dx < \int_0^\infty x^{-2} dx < \infty, $$ so the dominated convergence theorem allows exchanging the integral and the limit, implying $$ \lim_{n \to \infty} \int_0^\infty e^{-nx} x^{-1/2} dx = \int_0^\infty \left(\lim_{n \to \infty} e^{-nx} x^{-1/2} \right) dx = \int_0^\infty 0 dx = 0. $$

Answer 3

Since $y=nx$ gives $\int_0^\infty x^{-1/2}e^{-nx}dx=n^{-1/2}\int_0^\infty y^{-1/2}e^{-y}dy=\sqrt{\frac{\pi}{n}}$, the limit is $0$. For proof the integral over $y$ is $\sqrt{\pi}$, take your pick.

Answer 4

$$I(n)=\int_0^\infty e^{-nx}x^{-1/2}dx$$ $$u=nx\to dx=\frac{du}{n}$$ $$I(n)=\int_0^\infty e^{-u}\left(\frac un\right)^{-1/2}\frac{du}{n}=n^{-1/2}\int_0^\infty u^{-1/2}e^{-u}du=n^{-1/2}\Gamma(1/2)=n^{-1/2}\sqrt{\pi}$$ now we can say: $$\lim_{n\to\infty}I(n)=\sqrt{\pi}\lim_{n\to\infty}n^{-1/2}=0$$