exponential function - The integral $int_0^infty e^{-t^2}dt$

Me and my highschool teacher have argued about the limit for quite a long time.

We have easily reached the conclusion that integral from $0$ to $x$ of $e^{-t^2}dt$ has a limit somewhere between $0$ and $\pi/2$, as we used a little trick, precisely the inequality $e^t>t+1$ for every real $x$. Replacing $t$ with $t^2$, inversing, and integrating from $0$ to $x$, gives a beautiful $\tan^{-1}$ and $\pi/2$ comes naturally.

Next, the limit seemed impossible to find. One week later, after some google searches, i have found what the limit is. This usually spoils the thrill of a problem, but in this case it only added to the curiosity. My teacher then explained that modern approaches, like a computerised approximation, might have been applied to find the limit, since the erf is not elementary. I have argued that the result was to beautiful to be only the result of computer brute force.

After a really vague introduction to fourier series that he provided, i understood that fourier kind of generalised the first inequality, the one i have used to get the bounds for the integral, with more terms of higher powers.

To be on point: I wish to find a simple proof of the result that the limit is indeed $\sqrt\pi/2$, using the same concepts I am familiar with. I do not know what really Fourier does, but i am open to any new information.

Thank you for your time, i appreciate it a lot. I am also sorry for not using proper mathematical symbols, since I am using the app.

Answer

It's useless outside of this one specific integral (and its obvious variants), but here's a trick due to Poisson:

$$\begin{align*} \left(\int_{-\infty}^\infty dx\; e^{-x^2}\right)^2 &= \int_{-\infty}^\infty \int_{-\infty}^\infty \;dx\;dy\; e^{-x^2}e^{-y^2} \\ &= \int_{-\infty}^\infty \int_{-\infty}^\infty \;dx\;dy\; e^{-(x^2 + y^2)} \\ &= \int_0^{2\pi} \!\!\int_0^\infty \;r\,dr\;d\theta\; e^{-r^2} \\ &= \pi e^{-r^2}\Big\vert_{r=0}^\infty \\ &= \pi, \end{align*}$$
switching to polar coordinates halfway through. Thus the given integral is $\frac{1}{2}\sqrt{\pi}$.