I have to prove that there exists a $ \delta > 0$ and a function $h \in C^1(]-\delta, \delta[)$ with the property $\forall x \in ] -\delta, \delta[ : h(x) + h(x) \int_0^{h(x)} e^{-t^2} \, dt = x$. $\\$
After that I have to calculate $h'(0)$. $\\$
My idea is to use the implicit function theorem: $\\$
Let $F(x,y) = y + y \int_0^y e^{-t^2} \, dt -x $. $\\$
$F(x,y) = 0$ if $x = y = 0. $ $\\$
Now I have to show that $\dfrac{ \partial F (x,y)}{\partial y} \neq 0$. I don't know how to derivate my function as I have a y as a border. Furthermore it is not easy to calculate the integral... $\\$
And how can I calculate $h'(0)$? For that my idea would be to rearrange F(x,y) after y and derivate my function. But there is again the problem with the integral... Thanks!
You are seeking a local inverse to:
$$f(y)=y+y\int_0^y e^{-x^2}\,dx$$
Now:
$$f’(y)=1+ye^{-y^2}+\int_0^y e^{-x^2}dx$$
So $f’(0)=1.$ Also, $f’(y)$ is continuous, so $f’(y)>0$ for $y\in I=(-\epsilon,+\epsilon),$ for some $\epsilon>0.$ So $f$ is strictly increasing on that interval and has an inverse on the range of $I.$
Let $\delta_1=-f(-\epsilon/2)$ and $\delta_2=f(\epsilon/2).$ Then $-\delta_1<f(0)<\delta_2.$ Let $\delta=\min(\delta_1,\delta_2).$
Then $J=(-\delta,\delta)\subset f(I).$ So there is an inverse $h$ of $f$ on $J.$
Now, $$\begin{align}\lim_{x\to x_0}\frac{h(x)-h(x_0)}{x-x_0}&=\lim_{x\to x_0} \frac{h(x)-h(x_0)}{f(h(x))-f(h(x_0))}\\ &=\lim_{y\to h(x_0)} \frac{y-h(x_0)}{f(y)-f(h(x_0))}\\&=\frac{1}{f’(h(x_0))} \end{align}$$
This means $h’$ is defined on $J$ and $h’(0)=1.$