On some extra questions after the solution of a cauchy problem. I have the Cauchy problem 
$$ \begin{cases}  y' = 2xy + \frac{1}{\sqrt{x}} \\y(1) = 0 \end{cases} $$
I solved this problem and I obtain
$$y(x) = e^{x^2} \int_{1}^x e^{-t^2} \frac{1}{\sqrt{t}} dt $$
Now I am tasked in 


*

*finding the maximum domain in which this is defined and call it $I$

*find $\lim_{x \rightarrow \inf(I)} y(x)$ and find  $\lim_{x \rightarrow \sup(I)} y(x)$

*Given $f(x) = e ^{- x^2} y(x)$  find the cardinality of the set $\{f^{-1}(\lambda) \}$ where $\lambda \in f(I)$. 


I think the answer to point three is that for all $\lambda$ the cardinality of the set $\{f^{-1}(\lambda) \}$  is one because $$e^{-t^2} \frac{1}{\sqrt{t}}> 0 \forall t$$
and so the integral function would be strictly increasing.
I am doubtful on the maximum domain though, how would I find the max domain? is $ 0$ a problem since I have $\sqrt{t}$ at the denominator?
 A: 
(1) boils down to: for which values of $\mathrm x$ is the integral convergent?

To avoid complex numbers and division by $0$, $\mathrm x$ must be greater than $0$. Making the change of variable $\mathrm{u=\sqrt t}$ we can turn the integral into (leaving out the constant factor of $2$)
$$\mathrm{
\int_1^{\sqrt x}{e^{-u^4}}\,du
}$$
Now we can compare the above integral for arbitrarily large values of $\mathrm x$ with the Gaussian integral: 
$$
\mathrm{
\Big{|}\int_1^{\infty}{e^{-u^4}}\,du\Big{|}<\int_{-\infty}^{\infty}{e^{-u^2}}\,du={\sqrt\pi}
}$$
We can compare the integral for arbitrariliy small values of $\mathrm x$ (which are still greater than $0$) with the integral of unity:
$$\mathrm{
\Big{|}\int_1^0e^{-u^4}\,du\Big{|}<\int_0^11\,du=1
}$$
So we have $\mathrm{I=(0,\infty)}$.

(2)

The integral is finite in the closure of $\mathrm I$. But $\mathrm{e^{x^2}}$ is unbounded in $\mathrm I$.
$$\mathrm{
\lim_{x\to\sup(I)}y(x)=\infty
}$$
According to Weierstrass M-test $$\mathrm{
\left\{\sum_{k=0}^n{(-u^4)^k\over k!}\right\}_n
}$$
converges uniformly to $\mathrm{e^{-u^4}}$ for $\mathrm{u\in[0,1]}$ which enables us to do the following maipulations.
$$\mathrm{
\int_0^1e^{-u^4}\,du=\sum_{k=0}^\infty(-1)^k\int_0^1{u^{4k}\over k!}\,dk=\sum_{k=0}^\infty{(-1)^k\over(4k+1)k!}:=K
}$$
And therefore
$$\mathrm{
\lim_{x\to \inf(I)}y(x)=-2K
}$$

(3)

$$\mathrm{
f(x)=2\int_1^{\sqrt x}e^{-u^4}\,du\implies f'(x)={e^{-x^2}\over\sqrt x}>0\;\forall\;x\in I
}$$
Therefore $\mathrm{f}$ is strictly increasing and hence is injective in $I$. Consequently for $\mathrm{\lambda\in f(I)}$ there is a unique $\mathrm{x:f(x)=\lambda}$ i.e. $\mathrm{\#(f^{-1}(\lambda))=1}$. 
