Given the fuctions $f(x)=e^{x^2}$ and $F(x)=\int_0^xf(t)dt$
Show that the limit of $F(x)=\infty$ as $x\to\infty$.
Here is my thought process
I can see that it looks a lot like a Gaussian integral but reading up on Gaussian integrals didn't really help me.
Here is what I've tried so far
I know $e^{t^2}$ is continuous
$F(x)=\int_0^1f(t)dt+\int_1^xf(t)dt=\int_0^1e^{t^2}dt+\int_1^xe^{t^2}dt$
On
If you have learnt multivariable calculus then following might help.
Let $$I=\int_0^{\infty} e^{x^2} dx$$ Similarly we can write $$I=\int_0^{\infty} e^{y^2} dy$$
Hence $$I^2=\int_0^{\infty}\int_0^{\infty} e^{x^2+y^2} dx dy$$
Now using polar coordinates this simplifies to $$I^2=\int_0^{\frac {\pi}{2}}\int_0^{\infty} e^{r^2} rdr d\theta$$ $$\Rightarrow I^2=\int_0^{\frac {\pi}{2}} d\theta \int_0^{\infty} e^{r^2} rdr$$ Now substitute $r^2=u$ $$\Rightarrow I^2=\int_0^{\frac {\pi}{2}} \frac {d\theta}{2} \int_0^{\infty} e^{u} du$$ $$\Rightarrow I^2=\frac {\pi}{4}\cdot (\infty -1)$$ $$\Rightarrow \lim_{x\to \infty} F(x)=\infty$$
Use the fact that\begin{align}x\geqslant 1\implies F(x)&=\int_0^1e^{t^2}\,\mathrm dt+\int_1^xe^{t^2}\,\mathrm dt\\&\geqslant\int_0^1e^{t^2}\,\mathrm dt+\int_1^xe^t\,\mathrm dt\\&=\int_0^1e^{t^2}\,\mathrm dt+e^x-e.\end{align}