2

This is a puzzle I made up for your amusement.

Which special number, $a$, is such that:

$$\int\limits_{-\infty}^\infty a^{x^2} dx = \int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty a^{x^2+y^2} dx dy = \int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty a^{x^2+y^2+z^2} dx dy dz = \cdots$$

Community
  • 1
zooby
  • 4,343
  • From where did you get this puzzle? Also have you tried anything? – Zacky Nov 17 '19 at 23:00
  • @Nyssa I made it up for your ammusement. – zooby Nov 17 '19 at 23:22
  • This is an interesting puzzle (and has garnered some interesting answers), but Math.SE isn't really the place for such things. Note that the description of the puzzle tag itself says specifically "If the answer is known to you please do not use this tag to 'riddle' other users, but rather to ask about the correctness of a possible solution or ways to extend and improve an existing solution." ... Puzzle or not, when you do know an answer (especially when you have invented the question), please say so. This helps readers know, for instance, not to waste effort explaining concepts to you. – Blue Nov 18 '19 at 15:30

3 Answers3

2

If $L=\int\limits_{-\infty}^\infty a^{x^2} dx$, then your first equation requires that $L=L^2$. So either $L=0$ or $L=1$.

Since $a^{x^2}\geq0$, the only way for $L$ to be $0$ is if $a=0$.

If $L=1$, then $$\begin{align} &\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty a^{x^2+y^2}\, dx\, dy = 1\\ \implies&\int\limits_{0}^{2\pi}\int\limits_{0}^\infty a^{r^2}\, r\,dr\, d\theta = 1\\ \implies&2\pi\int\limits_{0}^\infty a^{r^2}\, r\,dr = 1\\ \implies&\lim_{t\to\infty}\pi\left[\frac{1}{\ln(a)}a^{r^2}\right]_0^{t} = 1\\ \implies&\lim_{t\to\infty} a^{t^2} = \frac{\ln(a)}{\pi}+1 \end{align}$$

The limit on the left is only defined for $a\leq1$, and it is either $1$ (if $a=1$) or $0$. $a=1$ is not a solution since it makes the original integral infinite. So $$0=\frac{\ln(a)}{\pi}+1$$ which implies $a=e^{-\pi}$.

2'5 9'2
  • 54,717
  • Very nice explanation: why $L=L^2$? – Sebastiano Nov 17 '19 at 23:22
  • Good explanation using polar coordinates. – zooby Nov 17 '19 at 23:32
  • It's because $\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty a^{x^2+y^2} dx dy=\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty a^{x^2}a^{y^2} dx dy=\int\limits_{-\infty}^\infty a^{y^2}\left(\int\limits_{-\infty}^\infty a^{x^2} dx\right) dy=\left(\int\limits_{-\infty}^\infty a^{x^2} dx\right)\left(\int\limits_{-\infty}^\infty a^{y^2} dy\right)=\left(\int\limits_{-\infty}^\infty a^{x^2} dx\right)\left(\int\limits_{-\infty}^\infty a^{x^2} dx\right)=L^2$. – 2'5 9'2 Nov 17 '19 at 23:41
2

If the first integral is $I$, the second is $I^2$, the third is $I^3$ and so on. To have the equality work we can either have $I=0$ or $I=1$. If $a=0$ we have $I=0$ if we ignore the problem that the integrand is undefined at the origin. As changing one point does not change the integral, we can just define the integrand to be any value we want at the origin and handwave that away. Otherwise, we need to find $a$ so that the integral is $1$. Alpha tells me that $$\int_0^\infty a^{x^2}dx=\frac {\sqrt \pi}{2\sqrt{-\log(a)}} \text{ for } 0 \lt a \lt 1$$ We want the integral to be $\frac 12$, so $$\frac {\sqrt \pi}{2\sqrt{-\lg(a)}}=\frac 12\\ \sqrt{-\log (a)}=\sqrt \pi\\ \log(a)=-\pi\\ a=e^{-\pi}\approx 0.0432139$$

Ross Millikan
  • 374,822
1

Well, some would probably argue that 0 isn't a number, but 1 also seems to work.

Morten Kruse Pedersen
  • 11
  • $1$ does not work because the integrals are infinite. $0$ has the problem that the integrand is not defined at the origin. I believe we can accept that, but we need to say something about it. – Ross Millikan Nov 17 '19 at 23:22
  • But we get $\infty = \infty = \infty = \dots$ for lots of values of $a$. – GEdgar Nov 17 '19 at 23:23
  • I didn't think of those. I guess any number where $a>1$ diverges so all integrals would be infinite! There is another number between 0 and 1 that also works. – zooby Nov 17 '19 at 23:31