Given the equation $g(x) - \int_0^y e^{t^2}\,{\rm d}t = 0$, with $g\colon \Bbb R \to \Bbb R$ of class ${\cal C}^\infty$, show that for each $x \in \Bbb R$ there is a unique $y = y(x)$ that solves the equation and that this association is also ${\cal C}^\infty$.
I managed to solve this problem by copying the argument given in an example before in the textbook, but I think that this result is very counter-intuitive. I'll put here what I've made, first.
Let $x_0 \in \Bbb R$ be arbitrary and define $p(y) = \frac{\sqrt{\pi}}{2}\,{\rm erfi}(y) - g(x_0),$ where ${\rm erfi}$ is the imaginary error function. Since $p'(y) = e^{y^2} > 0$, $p$ is strictly increasing and since $\lim_{y \to \pm \infty}{\rm erfi}(y) = \pm\infty$, the same goes for $p$ and there is a unique root $f(x_0)$ for $p$. Now define $F\colon \Bbb R^2 \to \Bbb R$ by $$F(x,y) = g(x) - \int_0^y e^{t^2}\,{\rm d}t.$$Since $g$ is ${\cal C}^\infty$, $F$ is too, and $F(x_0,f(x_0))=0$. Now: $$\frac{\partial F}{\partial y}(x_0,f(x_0)) = -e^{f(x_0)^2} \neq 0,$$and by the Implicit Function Theorem, there is an interval $I$ containing $x_0$ and $y: I \to \Bbb R$ of class ${\cal C}^\infty$ such that $F(x,y(x)) = 0$ for all $x \in I$, and each $y(x)$ is uniquely determined in these conditions. Since $F(x_0, f(x_0)) = 0$, we have that $f(x_0) = y(x_0), $ so that the solution is unique and depends ${\cal C}^\infty$ n $x$, as wanted.
My problem with this is that we're probably wandering into complex territory without noticing. Take for example, $g(x) = -1$ for all $x$. Then $g$ is ${\cal C}^\infty$ and $$ g(x) - \int_0^ye^{t^2}\,{\rm d}t \leq -1 < 0, \quad \forall\,x,y \in \Bbb R,$$since $e^{t^2} > 0$ for all $t$, so we don't have any solution!
Can someone explain what is happening?
I don't think there is a problem. $f(y) = \int_0^y e^{t^2}\,{\rm d}t$ is strictly increasing and continuous on $\Bbb R$ with $$ \lim_{y \to -\infty} f(y) = -\infty, \lim_{y \to +\infty} f(y) = +\infty \, $$ and therefore a bijective mapping from $\Bbb R$ onto $\Bbb R$.
$f$ is of class ${\cal C}^\infty$ and $f'$ is nowhere zero, therefore the inverse mapping $f^{-1}$ is ${\cal C}^\infty$ as well.
$y(x) = f^{-1}(g(x))$ is the solution of $g(x) - \int_0^y e^{t^2}\,{\rm d}t = 0$. If $g(x)$ is negative then $y = y(x)$ is negative which is no contradiction, as then $$ \int_0^ye^{t^2}\,{\rm d}t = - \int_y^0e^{t^2}\,{\rm d}t < 0 \, . $$