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.
 A: Put
$$I:=\frac12\int_{-\infty}^\infty e^{-x^2}dx =\int_0^\infty e^{-x^2}dx\implies I^2=\frac14\int_{-\infty}^\infty e^{-x^2}dx\int_{-\infty}^\infty e^{-y^2}dy=$$
$$=\frac14\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(x^2+y^2)}dxdy=\frac14\int_0^\infty\int_0^{2\pi}re^{-r^2}d\theta\,dr=$$
$$=\frac14\left.2\pi\left(-\frac12\right)e^{-r^2}\right|_0^\infty=\frac\pi4\implies I=\frac{\sqrt\pi}2$$
A: I have two ways to derive it.  The simpler one requires multi-variate calculus.  The more complicated approach uses "differentiation under the integral sign."
Since you don't know multivariate calc, I will do the second.
$F(t) = \int_0^{\infty} \dfrac{e^{-t^2(1+x^2)}}{(1+x^2)} dx\\
\frac {dF}{dt} = \int_0^{\infty} -2t e^{-t^2(1+x^2)} dx\\
e^{-t^2}\int_0^{\infty} -2te^{-(tx)^2} dx\\
u = tx, du = t dx\\
e^{-t^2}\int_0^{\infty} -2e^{-u^2} du\\$
$\frac {dF}{dt} = -2e^{-t^2} I$
With $I$ being the our goal.
$\int_0^t \frac {dF}{ds} ds=-2I \int_0^t e^{-s^2} ds\\
F(t) - F(0) = -2I \int_0^t e^{-s^2} ds $ 
As $t$ goes to infinity:
$F(\infty) - F(0) = -2I^2$ 
$F(0) =$$ \int_0^{\infty} \dfrac{1}{(1+x^2)} dx\\
 \tan^{-1}(\infty) = \frac{\pi}{2}$
$F(\infty) =0$
$-2I^2 = -\frac{\pi}{2}\\
I = \frac{\sqrt{\pi}}{2}$
A: It's useless outside of this one specific integral (and its obvious variants), but here's a trick due to Poisson:
\begin{align*}
\left(\int_{-\infty}^\infty dx\; e^{-x^2}\right)^2
  &= \int_{-\infty}^\infty \int_{-\infty}^\infty  \;dx\;dy\; e^{-x^2}e^{-y^2} \\
  &= \int_{-\infty}^\infty \int_{-\infty}^\infty  \;dx\;dy\; e^{-(x^2 + y^2)} \\
  &= \int_0^{2\pi} \!\!\int_0^\infty \;r\,dr\;d\theta\; e^{-r^2} \\
  &= \pi e^{-r^2}\Big\vert_{r=0}^\infty \\
  &= \pi,
\end{align*}
switching to polar coordinates halfway through. Thus the given integral is $\frac{1}{2}\sqrt{\pi}$.
A: Seems appropriate to address this: any proof that the error function is not elementary is really, really, really difficult. It is the main example in An Introduction to Differential Algebra by Irving Kaplansky.
https://en.wikipedia.org/wiki/Elementary_function
A: We assume that $X\sim\,N(0,1)$ 
$$\int_{-\infty}^{+\infty}f_X(x)dx=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}e^{-\frac{x^2}{2}}dx=1$$
as a result
$$2\int_{0}^{+\infty}e^{-\frac{x^2}{2}}dx=\sqrt{2\pi}\implies\,2\sqrt{2}\int_{0}^{+\infty}e^{-u^2}du=\sqrt{2\pi}$$
therefore
$$\int_{0}^{+\infty}e^{-u^2}du=\frac{\sqrt{\pi}}{2}$$
