# 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. ]