# $\lim\limits_{x \to +\infty}x\, e^{-x^2}\int_{0}^{x}e^{t^2}dt$

Hello every one Please I need your help for the 3rd question, I tried but i fail every time.

for every real $x$, we put $f(x)=e^{-x^2}\int_{0}^{x}e^{t^2}dt$.

1. Show that $f$ is odd of class $C^{\infty}$ on $\mathbb{R}$.

2. Show that $f$ is a solution of the functional equation $y'+2xy=1$.

3. Prove that $\lim\limits_{x \to +\infty}2xf(x)=1$.

thanks.

• L'Hôpital with $e^{x^2}$ in the denominator should do the trick. – Py42 Apr 14 '14 at 9:55

First, note that \begin{align} xe^{-x^2}\int_0^xe^{t^2}\,\mathrm{d}t &\ge e^{-x^2}\int_0^xte^{t^2}\,\mathrm{d}t\\ &=\frac12\tag{1} \end{align} Next, note that \begin{align} xe^{-x^2}\int_0^{x-1}e^{t^2}\,\mathrm{d}t &\le xe^{-x^2}\int_0^{x-1}e^{(x-1)t}\,\mathrm{d}t\\ &\le\frac{ex}{x-1}e^{-2x}\tag{2} \end{align} Therefore, \begin{align} xe^{-x^2}\int_0^xe^{t^2}\,\mathrm{d}t &=xe^{-x^2}\int_{x-1}^xe^{t^2}\,\mathrm{d}t +xe^{-x^2}\int_0^{x-1}e^{t^2}\,\mathrm{d}t\\ &\le xe^{-x^2}\int_{x-1}^x\frac{t}{x-1}e^{t^2}\,\mathrm{d}t +\frac{ex}{x-1}e^{-2x}\\ &=\frac12\frac{x}{x-1}\left(1-e^{-2x+1}\right) +\frac{ex}{x-1}e^{-2x}\tag{3} \end{align} Taking the limit of $(3)$ yields $$\lim_{x\to\infty}xe^{-x^2}\int_0^xe^{t^2}\,\mathrm{d}t\le\frac12\tag{4}$$ Therefore, $(1)$ and $(4)$ imply $$\lim_{x\to\infty}xe^{-x^2}\int_0^xe^{t^2}\,\mathrm{d}t=\frac12\tag{6}$$