How we can solve that $\lim _{_{x\rightarrow \infty }}\int _0^x\:e^{t^2}dt$? How we can solve that $\lim _{_{x\rightarrow \infty }}\int _0^x\:e^{t^2}dt$ ?
P.S: This is my method as I thought:
$\int _0^x\:\:e^{t^2}dt>\int _1^x\:e^tdt=e^x-e$ which is divergent, so all your answers, helped me to think otherwise, maybe my method help something else :D
 A: Since
$$
e^{x^2}=1+x^2+\frac{x^4}{2!}+\dotsb
$$
we have that, for $x\ge0$, $e^{x^2}\ge1+x^2$. So
$$
\int_{0}^{x}e^{t^2}\,dt\ge\int_{0}^x(1+t^2)\,dt=x+\frac{x^3}{3}
$$
Can you finish?
A: This function diverges extremely fast. Notably, $e^{t^2}$ is monotone increasing with limit $\infty$ as $t \to \infty$. Thus your integral diverges (and it get very, very large very, very quickly).
A: $$e^{t^2}> e^t\text{ from 1 to $\infty$, and the part from 0 to 1 is finite.}$$ and the integral$ \int_1^\infty e^t $ diverges. Therefore, by the comparison test, it diverges too.
A: $$
\begin{align}
\int_0^xe^{t^2}\,\mathrm{d}t
&\ge\int_0^x\frac tx\,e^{t^2}\,\mathrm{d}t\\
&=\frac1{2x}\left(e^{x^2}-1\right)
\end{align}
$$
As $x\to\infty$, the function on the right goes to $\infty$ extremely fast.
A: Or just use $e^{x^2} \ge 1$ on $[0,\infty)$ to see $\int_0^x e^{t^2}dt \ge x \to \infty.$
A: It is vary simple:
$$y(x)=\int_0^xe^{t^2}dt=\frac{1}{2} \sqrt{\pi } \text{erfi}(x)$$
So:
$$\lim_{x \rightarrow+\infty}y(x)=+\infty$$
