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I was wondering why $\pi$ shows up in the equation for the area under a Gaussian: $$ \int_{-\infty}^{\infty}e^{-ax^2} ~dx= \sqrt{\frac{\pi}{a}} $$ This isn't intuitive to me, as I think of $\pi$ as being intrinsically linked with rotation in 2D or higher, and this function lives in 1D.

Looking at a derivation of this result, it seems that $\pi$ shows up because $e^{-a(x^2+y^2)}$ is rotationally symmetric in 2D. (This lets us calculate the square of the above integral by converting to polar coordinates). So, the area under $e^{-ax^2}$ is related to $\pi$ because its higher dimensional cousin $e^{-ax^2}e^{-ay^2}=e^{-a(x^2+y^2)}$ is rotationally symmetric in 2D.

So, $e^{-ax^2}$ is 2D rotationally symmetric in spirit, even though it lives in 1D. This idea of understanding something about a function by using a higher dimensional symmetry seems interesting. I wonder how many functions have this sort of hidden rotational symmetry. Explicitly, what are the functions $f: \mathbb{R} \to \mathbb{R}$ so that $f_{2D}: \mathbb{R}^2 \to \mathbb{R}$ [defined by $f_{2D}(x,y) = f(x)f(y)$] is rotationally symmetric about the z-axis?

It seems that these functions form a group under multiplication (at least as long as they don't take on zero values). The neutral element is $f(x) = 1$ and taking the reciprocal plays the role of the inverse. To see this, note that if some 2D function $h: \mathbb{R}^2 \to \mathbb{R}$ is rotationally symmetric, then so $1/h$. For multiplicative closure, note that if $h_1$ and $h_2$ are 2D rotationally symmetric functions, then so is their product. These 2D properties then induce the desired properties in 1D. Can we find the generators of this group - the basic building blocks that allow us to construct all the other "secretly rotationally symmetric" 1D functions?

Edit Upon reflection, I realized that any rotationally symmetric 2D function $h: \mathbb{R}^2 \to \mathbb{R}$ has to be of the form $h(x,y) = H(x^2 + y^2)$ for some function $H: \mathbb{R} \to \mathbb{R}$. This is because the value of the rotationally symmetric function is totally determined by its distance from the origin. So, for $f$ to be a "secretly rotationally symmetric" 1D function, it is sufficient and necessary that $f_{2D}(x,y) = f(x)f(y) = H(x^2 + y^2)$ for some function $H$.

Edit 2 We can also see that $f$ has to be even (if $f$ doesn't take on zero values). We need $f_{2D}(-x,y) = f_{2D}(x,y) \implies f(-x)f(y) = f(x)f(y) \implies f(-x) = f(x)$. So, if $f$ is smooth we can expand it as a Taylor series without any odd powers: $$ f(x) = f(0) + \frac{f''(0)}{2!}x^2 + \dots + \frac{f^{(2n)}(0)}{(2n)!}x^{2n} + \dots $$ This means we can write $f(x)$ in terms of a function of $x^2$, say $F(x^2)$. The condition described in Edit 1 then becomes $F(x^2)F(y^2) = H(x^2 + y^2)$. In the special case that $F = H$, we get $F(x^2)F(y^2) = F(x^2 + y^2)$, which is equivalent to $F(x)F(y) = F(x + y)$. [In the case of the Gaussian, we have $F(x^2) = e^{-ax^2}$ or equivalently $F(x) = e^{-ax}$.]

So, under these conditions, it seems sufficient to have $f(x) = F(x^2)$ with $F$ satisfying $F(x)F(y) = F(x + y)$. It is interesting that there appears to be some link between rotational symmetry and this structure preserving condition moving from addition to multiplication. Might there be additional generator functions beyond these?

David Egolf
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  • This might help: https://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-limits-k-1-infty-frac1k2-basel-pro – Alex Jun 18 '20 at 21:16
  • And: https://math.stackexchange.com/questions/39802/why-does-123-cdots-frac112 – Alex Jun 18 '20 at 21:16
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    Note $F(x^2)F(0)=H(x^2)$, so either $F(0)=0$ so $H=0$ and $F=0$ or $F(x^2)=F(0)G(x^2)$, where $G(x+y)=G(x)G(y)$. In the second case $G(n/m)^m=G(1)^n$ for $n,m\in \mathbb{Z}, m\neq 0$ so if $G$ is continuous it must have the form $e^{\lambda x}$. Thus the general solution has the form $f(x)=Ae^{\lambda x^2}$. – tkf Jun 18 '20 at 23:45
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    Also you do not need to restrict $f$ to being smooth. If $f(-x)=f(x)$ then $f=F(x^2)$ for some $F$ by definition, so the Taylor series argument is redundant. Further you have $f(-x)=f(x)$ as long as $f\neq0$ - you do not need to restrict $f$ to being a non-zero function. So this really is all solutions. – tkf Jun 19 '20 at 00:05
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    Before I saw your superior solution, I was going to suggest $$\frac d {d\theta}f(r\cos(\theta))f(r\sin(\theta))=0$$ which simplifies to $$\frac {f'(x)}{xf(x)}=c$$ which has general solution $f(x)=Ae^{c/2x^2}$. – tkf Jun 19 '20 at 00:12
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    So the group you mentioned is $C_2 \times \mathbb{R}\times \mathbb{R}$, where $\mathbb{R}$ is regarded as a group with respect to addition. I am not posting this as an answer as you did almost all the work. – tkf Jun 19 '20 at 00:13

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