Proving convergence of improper integral, g(t) I want to prove convergence of the improper integral,

$$\int_1^\infty g(t) dt$$
Where $g(t)$ is given by: $g(t)=(\frac{1-e^{-2t^2}}{t^2})$ , $t\neq0$

I have thought about using different tests, but I keep getting stuck at the fact that as t approaches zero from the right hand side, the function value increases towards infinity, so I can't figure out a limit from the right hand side.  

I hope you can help. 

Best regards,
Christoffer
$0&lt;\frac{1−e^{−t^2}}{t^2} &lt; \frac{1}{t^2}$ which is integrable on $[1,∞)$, so by comparison, your integral converges.
If you look at the integrand, it can be rewritten as 
$$
f(t) = 2 \int_{0}^{1} e^{-2t^2x} dx
$$
So the whole expression (with $\varepsilon \to \infty$)
$$
L = 2 \int_{1}^{\varepsilon}\int_{0}^{1}e^{-2xt^2}dxdt = 2\int_{0}^{1}\int_{1}^{\varepsilon} e^{-2xt^2}dt dx
$$
Use the $\texttt{erf}$ function for the second integral:
$$
L = \sqrt{2} \int_{0}^{1}\frac{1}{\sqrt{x}}[\texttt{erf}(\varepsilon) - \texttt{erf}(1)]
$$
Now, compute the simple integral, and then take the limit as $\varepsilon \to \infty$. Can you handle from here?