I would like to know how to calculate this integral n check its nature :
$$ A= \int_1^\infty \frac{e^{-t}}{1+t^{2}} dt . $$ I think I figured it out , we can do it with Equivalence like this (since it is positive): $$ \frac{e^{-t}}{1+t^{2}} \leq \frac{1}{1+t^{2}} And \frac{1}{1+t^{2}} \simeq \frac{1}{t^{2}} $$
Now I know $$ \frac{e^{-t}}{1+t^{2}} $$ is positive n tends to $0$ so the nature of $A$ is the nature of : $$ \Sigma \frac{e^{-t}}{1+t^{2}} \simeq \Sigma \frac{1}{t^{2}}$$ wich we know converges (the Riemann serie)