# The integral $\int_0^\infty e^{-t^2}dt$ [duplicate]

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.

• Can you do double integrals? Jun 7, 2016 at 18:57
• your question has an issue of the variable $x$ there.
– user9464
Jun 7, 2016 at 18:58
• Where did you see that the answer is $\sqrt{\pi}/4$? It should be $\sqrt{\pi}/2.$
– user307169
Jun 7, 2016 at 19:03
• Going back close to the beginning of this site: math.stackexchange.com/questions/9286/… Many of the approaches there do not use multiple integration. Jun 7, 2016 at 19:09
• user, I put an answer about that part. Jun 7, 2016 at 20:33

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

• Ah, good to know that this trick is due to Poisson, didn't know that. Great answer. Jun 7, 2016 at 19:09
• Lovely trick. Thanks.
– user242756
Jun 7, 2016 at 19:21
• 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. Jun 7, 2016 at 19:30

Put

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

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

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

• to cover the quadrant $[0,\infty)\times[0,\infty)$ only is necessary to take the angle $[0,\frac{\pi}{2}]$ Jun 7, 2016 at 19:22
• @janmarqz Thank you, edited. Jun 7, 2016 at 20:36

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) =$$\int_0^{\infty} \dfrac{1}{(1+x^2)} dx\\ \tan^{-1}(\infty) = \frac{\pi}{2} F(\infty) =0 -2I^2 = -\frac{\pi}{2}\\ I = \frac{\sqrt{\pi}}{2} • Yes, I missed cleaning up some stuff. I think I have it all correct right now. Jun 7, 2016 at 20:57 Seems appropriate to address this: any proof that the error function is not elementary is really, really, really difficult. It is the main example in An Introduction to Differential Algebra by Irving Kaplansky. https://en.wikipedia.org/wiki/Elementary_function We assume that X\sim\,N(0,1)$$\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}$\$