does $\int_0^{\infty} e^{(-t^2 -1/t^{2})} $ exists as improper riemann integral or lebesgue integral 
$\int_0^{\infty} e^{(-t^2 -1/t^{2})} $ dt

This question is from Apostol's mathematical analysis and the author has given many questions of this type in exercises of Ch-10 . I have done many of them but this is 1 of the question which I was not able to solve.
I used the defination of 10.32 for improper riemann integral and theorem 10.33 for lebesgue integral in most of the questions along with comparison tests for reimann integral but in this question I was not able to use any of the above bcoz of the type of function in the integrand.
So, I need help in how to approach this question.
Thanks!

I will prove it is Lebesgue integrable. Clearly your issue is near $0$, since by comparison with $e^{-t^2}$ you can integrate it on $(\epsilon, \infty)$ for any $\epsilon > 0$ so we will focus on integrating on $(0,1)$.
Using the substitution $s = 1/t$ you get
$$ \int_{0}^1 e^{-t^2 -1/t^2}dt = \int_{1}^{\infty} e^{-s^2 - 1/s^2} s^{-2} ds $$
And now just use comparison test with $e^{-s^2}$.
