# Continued integrals

Many of the most awesome formulas you meet when getting into mathematics are continued fractions like

$\Phi = 1+\frac{1}{1+\frac{1}{1+\frac{1}{\ldots}}}$

$2 = \sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}}$.

What about nested/continued integrals? Here is a simple one:

$e^x=1+x+\int x+\left(\int x+\left(\int x+\left(\ldots\right)dx\right)dx\right)dx$.

The way to see this is to recognize that the x in the first integral is going to integrate to $x^2/2$, the x in the second will be integrated twice $\int x^2/2 dx = x^3/3!$, and so on.

In general additive integrals of this kind turn into sums (assuming convergence, handwave, handwave…):

$I(x)=\int f(x)+\left(\int f(x)+\left(\int f(x)+\left(\ldots\right)dx\right)dx\right)dx = \sum_{n=1}^\infty \int^n f(x) dx$.

On the other hand, $I'(x)=f(x)+I(x)$.

So if we insert $f_k(x)=\sin(kx)$ we get the sum $I_k(x)=-\cos(kx)/k-\sin(kx)/k^2+\cos(kx)/k^3+\sin(x)/k^4-\cos(kx)/k^5-\ldots$. For $x=0$ we end up with $I_k(0)=\sum_{n=0}^\infty 1/k^{4n+2} - \sum_{n=0}^\infty 2/k^{4n+1}$. The differential equation has solution $I_k(x)=ce^x-\sin(kx)/(k^2+1) - k\cos(kx)/(k^2+1)$. Setting $k=0$ the integral is clearly zero, so $c=0$. Tying it together we get:

$\sum_{n=0}^\infty 1/k^{4n+2}-\sum_{n=0}^\infty 1/k^{4n+1}=-k/(k^2+1)$.

Things are trickier when the integrals are multiplicative, like $I(x)=\int x \int x \int x \ldots dx dx dx$. However, we can turn it into a differential equation: $I'(x)=x I(x)$ which has the well known solution $I(x)=ce^{x^2/2}$. Same thing for $f_k(x)=\sin(kx)$, giving us $I_k(x)=ce^{-\cos(kx)/k}$. Since we are running indefinite integrals we get those pesky constants.

Plugging in $f(x)=1/x$ gives $I(x)=cx$. If we set $c=1$ we get the mildly amusing and in retrospect obvious formula

$x=\int \frac{\int \frac{\int \frac{\ldots}{x} dx}{x} dx}{x} dx$.

We can of course mess things up further, like $I(x)=\int\sqrt{\int\sqrt{\int\sqrt{\ldots} dx} dx} dx$, where the differential equation becomes $I'^2=I$ with the solution $I(x)=(1/4)(c^2 + 2cx + x^2)$. A surprisingly simple solution to a weird-looking integral. In a similar vein:

$2\cot^{-1}(e^{c-x})=\int\sin\left(\int\sin\left(\int\sin\left(\ldots\right)dx\right) dx\right) dx$

$-\log(c-x)=\int \exp\left(\int \exp\left(\int \exp\left(\ldots \right) dx \right) dx \right) dx$

$1/(c-x)=\int \left(\int \left(\int \left(\ldots \right)^2 dx \right)^2 dx \right)^2 dx$

And if you want a real mind-bender, use the Lambert W function:

$I(x)=\int W\left(\int W\left(\int W\left(\ldots \right) dx \right) dx \right) dx$, then $x=\int_1^{I(x)}1/W(t) dt + c$.

(that is, you get an implicit but well defined expression for the (x,I(x)) values. With Lambert, the x and y axes always tend to switch place).

[And yes, convergence is handwavy in this essay. I think the best way of approaching it is to view the values of these integrals as the functions invariant under the functional consisting of the integral and its repeated function: whether nearby functions are attracted to it (or not) under repeated application of the functional depends on the case. ]

# Mathematical anti-beauty

Browsing Mindfuck Math I came across a humorous Venn diagram originally from Spikedmath.com. It got me to look up the Borwein integral. Wow. A kind of mathematical anti-beauty.

“As we all know”, $sinc(x)=sin(x)/x$ for $x\neq 0$ and defined to be 1 for $x=0$. It is not that uncommon as a function. Now look at the following series of integrals:

$\int_0^{\infty} sinc(x) dx = \pi/2$,

$\int_0^{\infty} sinc(x) sinc(x/3) dx = \pi/2$ ,

$\int_0^\infty sinc(x) sinc(x/3) sinc(x/5) dx = \pi/2$.

The pattern continues:

$\int_0^\infty sinc(x) sinc(x/3) sinc(x/5) \cdots sinc(x/13) dx = \pi/2$.

And then…

$\int_0^\infty sinc(x) sinc(x/3) sinc(x/5)$ $\cdots sinc(x/13) sinc(x/15) dx$
$=\frac{467807924713440738696537864469}{935615849440640907310521750000}\pi$

It is around 0.499999999992646 $\pi$ – nearly a half, but not quite.

What is going on here? Borwein & Borwein give proofs, but they are not entirely transparent. Basically the reason is that 1/3+1/5+…1/13 < 1, while 1/3+1/5+…1/13 + 1/15 > 1, but why this changes things is clear as mud. Thankfully Hanspeter Schmid has a very intuitive explanation that makes what is going on possible to visualize. At least if you like convolutions.

In any case, there is something simultaneously ugly and exciting when neat patterns in math just ends for no apparent reason.

Another good example is the story of the Doomsday conjecture. Gwern tells the story well, based on Klarreich: a certain kind of object is found in dimension 2, 6, 14, 30 and 62… aha! They are conjectured to occur in all  $2^n-2$ dimensions. A branch of math was built on this conjecture… and then the pattern failed in dimension 254. Oops.

It is a bit like the opposite case of the number of regular convex polytopes in different dimensions: 1, infinity, 5, 6, 3, 3, 3, 3… Here the series start out crazy, and then becomes very regular.

Another dimensional “imperfection” is the behaviour of the volume of the n-sphere: $V_n(r)=\frac{\pi^{n/2}r^n}{\Gamma(1+n/2)}$

The volume of a unit sphere increases with dimension until $n \approx 5.25$, and then decreases. Leaving the non-intuitiveness of why volumes would shrink aside, the real oddness is that the maximum is for a non-integer dimension. We might argue that the formula is needlessly general and only the integer values count, but many derivations naturally bring in the Gamma function and hence the possibility of non-integer values.

Another association is to this integral problem: given a set of integers $x_i$, is the integral $\int_0^\pi \prod_i \sin(x_i \theta) d\theta = 0$ ? As shown in Moore and Mertens, this is NP-complete. Here the strangeness is that integrals normally are pretty well behaved. It seems absurd that a particular not very scary trigonometric integral should require exponential work to analyze. But in fact, multivariate integrals are NP-hard to approximate, and calculating the volume of a n-dimensional polytope is actually #P-complete.

We tend to assume that mathematics is smoother and more regular than reality. Everything is regular and exceptionless because it is generated by universal rules… except when it isn’t. The rules often act as constraints, and when they do not mesh exactly odd things happen. Similarly we may assume that we know what problems are hard or not, but this is an intuition built in our own world rather than the world of mathematics. Finally, some mathematical truths maybe just are. As Gregory Chaitin has argued, some things in math are irreducible; there is no real reason (at least in the sense of a comprehensive explanation) for why they are true.

Mathematical anti-beauty can be very deep. Maybe it is like the insects, rot and other memento mori in classical still life paintings: a deviation from pleasantness and harmony that adds poignancy and a bit of drama. Or perhaps more accurately, it is wabi-sabi.