Let $n\in \mathbb N.$
$I_n= \int_{-n}^{\infty} e^{-t^2}t^2 dt.$
Question: Can we compute $I_n$? Where $I_n$ converges as $n\to \infty$?
My try: Put $t^2=y. $ So $I_n= \int_{n^2}^{\infty} e^{-y} y \frac{1}{2t} dy.$ My confusion is: can we write $t=\sqrt{y}$ as $t$ could be negative!! And so how should I handle this?
I know: $\int_0^{\infty} e^{-t}t^{n-1} dt = \Gamma(n).$
What do you mean exactly by "can we compute $I_n$"? Are you asking if the integral converges or if you can find a closed form expression for it using elementary functions (like polynomials, exponentials, etc).
If you are asking if the integral converges, the answer is yes. If you are asking if there is a closed form expression for it, I strongly believe the answer is no, unless you accept using the Gauss function, $erf(x)$, in your solution:
$erf(x)=\frac{2}{\pi} \int_0^x e^{-t^2} dt$.
Moreover, $\lim_{n \to \infty}I_n=\frac{\sqrt{\pi}}{2}$.