EDITED text:

I was trying to get an answer to a slightly different problem, but I posted it here in greater generality, and made a slight mistake in it too. The question actually is, given that $f$ is continuously differentiable, and that $lim_tf'(t)exp(-t^2) = 0$ is $int _-infty^+inftyf'(t)exp(-t^2)dt$ finite?

ORIGINAL text:
Given $lim_tf(t) = 0$ and that $f$ is continuously differentiable can we say $int _-infty^+inftyf(t)dt$ is finite?

Try the following function:

If $|t|leqfracpi2$$$f(t) = cos(t) + 2^frac-14 $$

Otherwise
$$f(t) = frac1t$$

Why is this a counter example? Well, its easy to show this function is continuously differentiable at all points besides $t=pmfracpi2$, and at these points, you can show through some algebra that this function is continuous and differentiable.

But, because of the nature of $frac1t$, this function will have an infinite integral.

Now your question asks for $f'(t)e^-t^2$. We can set the function I found equal to this expression to find the actual $f$ that violates your condition.

Edit

My original example was $f'(t)exp(-t^2) = frac1+1$, which happens to also be sufficient to disprove OP's question.