Find $a$ such that $\int_{-\infty}^\infty a^{x^2}dx=\int_{-\infty}^\infty\int_{-\infty}^\infty a^{x^2+y^2}dxdy=\cdots$ This is a puzzle I made up for your amusement.


Which special number, $a$, is such that:
$$\int\limits_{-\infty}^\infty a^{x^2} dx = \int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty a^{x^2+y^2} dx dy = \int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty a^{x^2+y^2+z^2} dx dy dz = \cdots$$

 A: 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 $a=e^{-\pi}$.
A: 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 
$$\frac {\sqrt \pi}{2\sqrt{-\lg(a)}}=\frac 12\\
\sqrt{-\log (a)}=\sqrt \pi\\
\log(a)=-\pi\\
a=e^{-\pi}\approx 0.0432139$$
A: Well, some would probably argue that 0 isn't a number, but 1 also seems to work.
