2
$\begingroup$

This question already has an answer here:

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.

$\endgroup$

marked as duplicate by user228113, Jack, Hans Lundmark, egreg, colormegone Jun 7 '16 at 22:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Can you do double integrals? $\endgroup$ – DonAntonio Jun 7 '16 at 18:57
  • $\begingroup$ your question has an issue of the variable $x$ there. $\endgroup$ – Jack Jun 7 '16 at 18:58
  • 1
    $\begingroup$ Where did you see that the answer is $\sqrt{\pi}/4$? It should be $\sqrt{\pi}/2.$ $\endgroup$ – tilper Jun 7 '16 at 19:03
  • 5
    $\begingroup$ Going back close to the beginning of this site: math.stackexchange.com/questions/9286/… Many of the approaches there do not use multiple integration. $\endgroup$ – colormegone Jun 7 '16 at 19:09
  • 1
    $\begingroup$ user, I put an answer about that part. $\endgroup$ – Will Jagy Jun 7 '16 at 20:33
10
$\begingroup$

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 $\frac{1}{2}\sqrt{\pi}$.

$\endgroup$
  • 3
    $\begingroup$ Ah, good to know that this trick is due to Poisson, didn't know that. Great answer. $\endgroup$ – Mathematician 42 Jun 7 '16 at 19:09
  • $\begingroup$ Lovely trick. Thanks. $\endgroup$ – user242756 Jun 7 '16 at 19:21
  • $\begingroup$ Oddly enough, the modern approach of choosing an appropriate contour and using residue calculus appears to only have been worked out (or at least publicized) by Polya in the late 1940s. Integrating $f(z) = e^{\pi i z^2} / \sin \pi z$ over the parallelogram with vertices $\pm \frac{1}{2} \pm R e^{\pi i/4}$ as $R \to \infty$ gives the result. $\endgroup$ – anomaly Jun 7 '16 at 19:30
3
$\begingroup$

Put

$$I:=\frac12\int_{-\infty}^\infty e^{-x^2}dx =\int_0^\infty e^{-x^2}dx\implies I^2=\frac14\int_{-\infty}^\infty e^{-x^2}dx\int_{-\infty}^\infty e^{-y^2}dy=$$

$$=\frac14\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(x^2+y^2)}dxdy=\frac14\int_0^\infty\int_0^{2\pi}re^{-r^2}d\theta\,dr=$$

$$=\frac14\left.2\pi\left(-\frac12\right)e^{-r^2}\right|_0^\infty=\frac\pi4\implies I=\frac{\sqrt\pi}2$$

$\endgroup$
  • 1
    $\begingroup$ to cover the quadrant $[0,\infty)\times[0,\infty)$ only is necessary to take the angle $[0,\frac{\pi}{2}]$ $\endgroup$ – janmarqz Jun 7 '16 at 19:22
  • 1
    $\begingroup$ @janmarqz Thank you, edited. $\endgroup$ – DonAntonio Jun 7 '16 at 20:36
2
$\begingroup$

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}$

$F(\infty) =0$

$-2I^2 = -\frac{\pi}{2}\\ I = \frac{\sqrt{\pi}}{2}$

$\endgroup$
  • $\begingroup$ Yes, I missed cleaning up some stuff. I think I have it all correct right now. $\endgroup$ – Doug M Jun 7 '16 at 20:57
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $$\int_{0}^{+\infty}e^{-u^2}du=\frac{\sqrt{\pi}}{2}$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.