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

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.

• Try the substitution $y=nx$. – Lord Shark the Unknown May 31 at 19:55
• But that limit is zero. – eyeballfrog May 31 at 19:58

$$I_n=\int_0^1\frac{e^{-nx}}{\sqrt x}dx+\int_1^\infty\frac{e^{-nx}}{\sqrt x}dx<\int_0^1\frac{dx}{\sqrt x}+\int_1^\infty e^{-nx}dx=2+\frac{e^{-n}}n$$ 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 $$0$$.

• Thank you very much for the answer I appreciate it. – James Jun 5 at 12:21

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 $$\lim_{n \to \infty} \int_0^\infty e^{-nx} x^{-1/2} dx = \int_0^\infty \left(\lim_{n \to \infty} e^{-nx} x^{-1/2} \right) dx = \int_0^\infty 0 dx = 0.$$

• Thank you for the insight as well as for the editing my mistakes in the question – James Jun 5 at 12:21

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.

• Thank you very much for the explanation. – James Jun 5 at 12:22

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

• Thank you very much for the answer – James Jun 5 at 12:22