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 $.


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

I read the title and missed the hints. Here is a different way of computing the limit.

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 $$ \lim_{x\to\infty}xe^{-x^2}\int_0^xe^{t^2}\,\mathrm{d}t=\frac12\tag{6} $$


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