Finding $\lim_{x \to \infty}\int_0^x{e^{-x^2+t^2}}\,dt$ If we aren’t able to solve the integral $\int e^{-x^2}\,dx$, then how is it possible to find the $\lim_{x \to \infty}\int_0^x{e^{-x^2+t^2}}\,dt$? This was given to me by my prof, and I asked him multiple times if it was able to be solved. He said yes, but I’m just not getting it. Any thoughts?

More generally,
consider
$I(f)
=\lim_{x \to \infty}
\frac1{f(x)}\int_0^x f(t)dt
$
where $f(x) \to \infty$.
I'll try to find
a condition on $f$
such that
$I(f) = 0$.
By L'Hopital's rule,
$I(f)
=\lim_{x \to \infty}
\frac1{f(x)}\int_0^x f(t)dt
=\lim_{x \to \infty}
\frac{f(x)}{f'(x)}
$
if this limit exists.
Therefore,
if
$\lim_{x \to \infty}
\frac{f(x)}{f'(x)}
=0
$,
then
$I(f) = 0$.
If
$f(x) = e^{x^a}$
with
$a > 1$
(the OP's case is $a=2$),
then
$f'(x)
=ax^{a-1}e^{x^a}
$,
so
$\frac{f(x)}{f'(x)}
=\frac1{ax^{a-1}}
\to 0
$
as $x \to \infty$.
