Show that $ f(x) = e^{x^2/2} \int^{\infty}_{x} e^{-t^2/2} dt $ is decreasing in $] 0, \infty [$ and find its limit

Question

We have $ f(x) = e^{x^2/2} \int^{\infty}_{x} e^{-t^2/2} dt $

1. Show that $f(x)$ is decreasing over $] 0, \infty [$,

2. Find $ \lim_{x \to \infty} f(x) $.

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I had few attempts though none convinced me about $f(x)$ being decreasing. I think I make a serious mistake already at the beginning.

Thank you for helping me

Answer 1

Consider

$$f(x)=e^{x^2/2}\int_x^{\infty}e^{-t^2 /2}dt=\int_x^{\infty}e^{-(t^2-x^2) /2}dt$$

and performing the change of variable $t-x=w$ your function reads

$$f(x)=\int_0^{\infty}e^{-w(w+2x) /2}dw=\int_0^{\infty}e^{-w^2/2-w\,x}dw$$

Then your questions

1.- $\displaystyle f'(x)=-\int_0^{\infty}e^{-w^2/2-w\,x}w\,dw<0$

2.-$\displaystyle \lim_{x\to\infty} f(x)=\lim_{x\to\infty}\int_0^{\infty}e^{-w^2/2-w\,x}dw=\int_0^{\infty}\lim_{x\to\infty}e^{-w^2/2-w\,x}dw=0$

Answer 2

$ e^{-x^{2}/2}=\int_x^{\infty} te^{-t^{2}/2}dt >x\int_x^{\infty} e^{-t^{2}/2}dt $. From this can you conclude that $xf(x) <1$? Once you do this you can easily see that $f'(x)=xf(x)-1 <0$. So $f$ is decreasing and $f(x) \to 0$ as $ x\to \infty$.