Recently I came across this tweet by @Keyneqq which includes an integral with a Sierpinski Triangle pattern. I think the tweet was meant as a bit of a joke, but I was curious if the integral could be evaluated in a reasonable way.
In order to make sense of it, I assumed that the integrand is $x$ for all of the integrals, and that the pattern goes on forever. After some analysis, I found the following sequence of functions which, as $n\rightarrow\infty$, generates the entire fractal:
$$f_0(a,b)=\int_{\int_{\int_{\int_{b}^{1}xdx}^{\int_{0}^{1}xdx}xdx}^{0}xdx}^{\int_{1}^{\int_{\int_{0}^{1}xdx}^{\int_{0}^{a}xdx}xdx}xdx}xdx$$
$$f_{n}(a,b)=f_{n-1}(f_{n-1}(a, 0),f_{n-1}(1, b))$$
FYI, $f_0$ resolves to this polynomial:
$$f_0(a,b)=\frac{1}{32768}\left[(a^{16}-4a^{12}-122a^8+252a^4+3969)-(b^{16}-8b^{14}+24b^{12}-32b^{10}+16b^8)\right]$$
The value we’re looking for is then $$\lim_{n\rightarrow\infty}f_n(1,0)$$
I wrote a Python script to calculate a few iterations, it appears to converge very quickly to a value of around $0.121126$. This seems to be a promising solution, but of course a Python script is not a proof of convergence. Is there a way to prove the convergence of this series that, as far as I can tell, can only be defined recursively?
Additional information
Here's the code for those interested:
from decimal import Decimaldef f(n, a, b): if n == 0: return (((Decimal(a ** 16) - 4 * a ** 12 - 122 * a ** 8 + 252 * a ** 4 + 3969) - (b ** 16 - 8 * b ** 14 + 24 * b ** 12 - 32 * b ** 10 + 16 * b ** 8)) / 32768) return f(n - 1, f(n - 1, Decimal(a), 0), f(n - 1, 1, Decimal(b)))
print(f(5, 1, 0))
At the limit,
$$f_\infty(1,0) = f_\infty(f_\infty(1,0),f_\infty(1,0))\qquad f_\infty\equiv\lim_{n\rightarrow\infty}f_n$$
Taking $m=f_\infty(1,0)$ we then have
$$m = f_\infty(m,m)$$
So the desired value $m$ is some sort of fixed point of the function. Using the value $m=0.121126$ this is verified by the script. In fact it appears that $f(a,a)$ converges to that same value of $0.121126$ for all $|a|<2.089$.
