This question is inspired by these two:

Upon reading these two, I realized there might be a geometric way to compute the error function, by a variation on Gauss' famous integral trick for evaluating $\int_{-\infty}^{\infty} e^{-x^2} dx$. Here's the idea:

For convenience, define $E(t) = \int_0^t e^{-y^2} dy$. This is a rescaled $\text{erf}$, which is helpful to eliminate some obnoxious constants in what follows. We have $$E(t)^n = \left(\int_0^t e^{-y^2} dy\right)^n = \left(t\int_0^1 e^{-t^2 y^2} dy\right)^n = t^n \int_{[0,1]^n} e^{-t^2 ||\vec{y}||^2} d^n y$$ where the last expression is an integral over the n-dimensional unit hypercube. As with the usual Gauss trick, we will change variables to spherical coordinates, so the last integral becomes $$t^n \int_0^{\sqrt{n}} e^{-t^2 r^2} A_n(r) dr$$ where $A_n(r)$ is the measure of a sphere of radius $r$ intersected with the unit hypercube, i.e. $A_n(r) := \left| S^{n-1} \cap [0,1]^n\right|$. So we have this interesting expression for the (rescaled) error function, $$\boxed{E(t) = t\left(\int_0^{\sqrt{n}} e^{-t^2 r^2} A_n(r) dr\right)^{1/n}}\tag{1}$$

Now here's where the second link above comes in: The radial mass distribution of an n-cube is approximately Gaussian, due to a miracle of the central limit theorem. In particular, $$A_n \xrightarrow{d} \sqrt{\frac{2\sqrt{\mu}}{\pi \sigma^2}} \exp\left( -\frac{\left(r-\sqrt{n\mu}\right)^2}{\sigma^2/2\sqrt{\mu}} \right)\tag{2}$$ where $\mu=1/3$ and $\sigma = 4/45$ (these are the mean and variance of $X^2$, where $X$ is uniform on $[0,1]$, as explained in the link), and where $\xrightarrow{d}$ denotes weak convergence. If you naively just plug in this expression in place of $A_n$ above, you can evaluate the resulting integral in terms of $E(x)$ and elementary functions. In the limit $n\rightarrow \infty$ the right hand side of (1) would then simplify down to something like $t\exp\left(\frac{-\mu t^2}{1+t^2\sigma^2/2\mu}\right)$, which is of course nonsense.

The problem is of course that weak convergence is not good enough to evaluate this integral. (By the Berry-Esseen theorem, we expect the deviation of $A_n$ from a Gaussian to be $\mathcal{O}(n^{-1/2})$ uniformly across its domain, whereas the integral coming from (2) vanishes exponentially with $n$, so it gets swamped by the error terms as $n\rightarrow \infty$.) So the next thing I tried was an Edgeworth expansion. This allows us to add corrections to the central limit theorem, and also to control the error in the convergence. Unfortunately, as with the naive version above, the error terms in an Edgeworth expansion are always polynomial, whereas the integrals vanish exponentially. So this is also insufficient.

So after realizing this, I decided to back up and try something different. Both the central limit theorem and Edgeworth expansions ultimately come from manipulating the characteristic function (Fourier transform of the distribution function), so perhaps analyzing (1) in Fourier space is the way to go? Note that (1) has the form of an inner product between $e^{-t^2 r^2}$ and the radial mass distribution of a cube $A_n(r)$. So by Parseval, we can evaluate it as an inner product of their Fourier transforms.

Some observations in this direction:

  • Let $V_n(r) := \left| B_r \cap [0,1]^n\right|$ be the volume of the intersection of a ball $B_r$ of radius $r$ centered at the origin and the unit hypercube. Then $V_n(r)$ is the (cumulative) distribution function corresponding to the density $A_n(r)$, and $A_n(r) = \frac{d}{dr} V_n(r)$.
  • $V_n(r)$ has the probabilistic interpretation $\mathbb{P}\left(\sum_{i=1}^n X_i^2 \leq r^2\right)$, where $X_i$ are i.i.d. uniform on $[0,1]$.
  • $X_i^2$ as above has density $\rho(x) = \frac{1}{2\sqrt{x}}$ on the unit interval and $\rho(x) = 0$ elsewhere.
  • $V_n(\sqrt{s}) = \mathbb{P}\left(\sum_{i=1}^n X_i^2 \leq s\right)$, which is the distribution function of $\sum_{i=1}^n X_i^2$. The density for this random variable is $\rho^{*n}$, the convolution of $\rho$ with itself $n$ times. Hence $\rho^{*n}(s) = \frac{d}{ds} V_n(\sqrt{s}) = A_n(\sqrt{s}) \frac{1}{2\sqrt{s}}$.

Putting this all together, the original integral of (1) can be written $$ \begin{align*} \int_0^{\sqrt{n}} e^{-t^2 r^2} A_n(r) dr & = \int_0^{n^{1/4}} e^{-t^2 s} A_n(\sqrt{s}) \frac{1}{2\sqrt{s}} ds \\ & = \int_0^{n^{1/4}} e^{-t^2 s} \rho^{*n}(s) ds \\ & = \langle e^{-t^2 s} , \rho^{*n}(s)\rangle_{L^2} \\ & = \langle \mathcal{F}[e^{-t^2 s}\Theta(s)] , \mathcal{F}[\rho(s)]^n\rangle_{L^2} \\ \end{align*} $$ where $\Theta(s)$ is a Heaviside step function.

The Fourier transform of $e^{-t^2 s}\Theta(s)$ is $\int_0^{\infty} e^{-t^2 s - 2\pi i k s} ds = \frac{-i}{2\pi}\frac{1}{k-it^2/2\pi},$ and the Fourier transform of $\rho(s)$ is $$\int_0^1 e^{-2\pi i k s} \frac{1}{2\sqrt{s}} ds = \int_0^1 e^{-2\pi i k y^2}dy = \frac{E(\sqrt{2\pi ik})}{\sqrt{2\pi ik}}$$ by analytic continuation, if you like. So by Parseval we have $$ \int_0^{\sqrt{n}} e^{-t^2 r^2} A_n(r) dr = \int_{-\infty}^{\infty} \frac{i}{2\pi} \frac{1}{k-\frac{-it^2}{2\pi}} \left(\frac{E(\sqrt{2\pi ik})}{\sqrt{2\pi ik}}\right)^n $$ Closing the contour in the lower half plane, we have a pole at $k=-it^2/2\pi$, and so we can evaluate the integral and find $$ E(t) = t\left( -2\pi i \frac{i}{2\pi} \left(\frac{E(t)}{t}\right)^n \right)^{1/n} = E(t) $$ So it's trivial.

So in the end, it seems I just found a very complicated way of proving that $\text{erf}(t)=\text{erf}(t)$. This is kind of disappointing, given that we seemed to have used several non-trivial facts in the process. My questions are:

I want to throw this idea to the wind and see if anyone else can find a better use for it. Is there any way to patch up the above analysis so as to learn something non-trivial about e.g. the error function or the mass distribution of a hypercube?

If not, is there an intuitive way of seeing why the above analysis is bound to give a trivial answer?

(Also, in case it's not clear, what I might have hoped for in the above analysis was e.g. a non-trivial relationship between $E(t)$ and $E$ evaluated at some other value. So a functional equation or something.)



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