Vtyjkyuk

Given a function $f$ which is continuously differentiable over the real line, and given that $displaystyle lim_to inftyf(t)exp(-t^2) = 0$ how does one prove that $displaystyle int_-infty^inftyf'(t)exp(-t^2)textdt$ is finite?

I think you can't. Just looking at the positive side of your integral, consider
$$f:tmapsto frace^t^2sqrtt+1$$
You do have $lim_tto+infty f(t)e^-t^2=0$, but
$$f'(t)e^-t^2 = frac2tsqrtt+1-frac1sqrtt+1t+1 sim_tto+infty 2sqrt t$$
which is not integrable on $[0,+infty[$.

The result is wrong as stated

By integration by parts for $a >0$
$$int_- a^a f^prime(t) e^-t^2 dt = underbraceleft[f(t)e^-t^2right]_-a^a_A+2underbraceint_- a^a tf(t) e^-t^2 dt_B $$

$A$ converges to zero as $a to infty$ by hypothesis.

Now take a function $f$ that vanishes for $xle 1$ and such that $int_1^infty tf(t) e^-t^2 dt$ diverges. For example $$f : x mapsto t^alpha e^t^2$$ with $-2 <alpha <0$. This will provide a counterexample as $B$ diverges for $a to infty$ while $f$ satisfies the hypothesis $displaystyle lim_to inftyf(t)exp(-t^2) = 0$.

This cannot be true for all functions $f(t)$. Consider $f(t)=fracsin tte^t^2$. This is continuously differentiable on the real line, and $f'(t)e^-t^2=2sin t-fracsin tt^2+fraccos tt$. The $sin t$ part makes the integral oscillate indefinitely and thus diverge.

More explicitly,

$int^infty_-infty(2sin t-fracsin tt^2+fraccos tt)dt=2int^infty_-inftydtsin t+[fracsin tt]^infty_-infty=2int^infty_-inftydtsin t $ -(DNE)