What is the integral of $e^{a \cdot x+b \cdot y}$ evaluated over the Koch Curve

Solution 1:

not an answer yet, just some thoughts.

Say our Koch curve $K$ starts at $(0,0)$, ends at $(1,0)$ and the midpoint is at $(1/2, 1/(2\sqrt{3}\;))$. Mark seems to have used this, since his computation with $a=b=1$ agrees with mine.

Self-similarity is described by two maps of the plane to itself: $$ L(x,y) = \left(\frac{x}{2}+\frac{y}{2\sqrt{3}},\frac{x}{2\sqrt{3}}-\frac{y}{2}\right), \\ R(x,y) = \left(\frac{x}{2}-\frac{y}{2\sqrt{3}}+\frac{1}{2},-\frac{x}{2\sqrt{3}}-\frac{y}{2}+\frac{1}{2\sqrt{3}}\right), $$ So $L(K)$ is the left half and $R(K)$ is the right half. Set $K$ is the unique nonempty compact set with $K = L(K) \cup R(K)$. Map $L$ shrinks by factor $1/\sqrt{3}$, reflects in the $x$-axis, rotates by $\pi/6$, and fixes the point $(0,1)$. Map $R$ shrinks by factor $1/\sqrt{3}$, reflects in the $x$-axis, rotates by $-\pi/6$ and fixes the point $(1,0)$.

The measure $\mu$ on $K$ is made up of two parts, which are images of $\mu$ under $L, R$, respectively, with half the measure. That is, for integrable $f$ we have $$ \int_K f\,d\mu = \int_{L(K)} f\,d\mu+\int_{R(K)} f\,d\mu = \frac{1}{2}\int_K f\circ L\,d\mu + \frac{1}{2}\int_K f\circ R\,d\mu $$

Now if we write $$ q(a,b) := \int_K e^{ax+by}d\mu(x,y) $$ the self-similarity shows $$ q(a,b) = \frac{1}{2}q\left(\frac{a}{2}+\frac{b}{2\sqrt{3}}, \frac{a}{2\sqrt{3}}-\frac{b}{2}\right)+\frac{1}{2}\exp\left(\frac{a}{2}+\frac{b}{2\sqrt{3}}\right)q\left(\frac{1}{2}-\frac{b}{2\sqrt{3}},-\frac{a}{2\sqrt{3}}-\frac{b}{2}\right) $$

We could use this recursively to evaluate $q(a,b)$ numerically. At each iteration, the point $(a,b)$ where $q$ should be evaluated moves closer to the origin by factor $1/\sqrt{3}$. We stop when we are "close enough" to $(0,0)$, since we know $q(0,0)=1$. But, of course, at each iteration the number of exponentials we have to evaluate doubles, so it is a slow method.