Are there any functions $f$ beyond trivial examples where $\int f(x +f(x + f(x +\dotsb)))\,dx$ $= F(x+F(x+F(x +\dotsb))) + C$ for some function $F$?

Question

Basically the title explains most of my question here. Purely out of curiosity (I have no real application), I am wondering if there are any "interesting" functions $f$ where we know of a function $F$ (perhaps $f$'s antiderivative, but not necessarily) such that

$$\int f(x + f(x + f(x + \dotsb)))\,dx = F(x + F(x + F(x + \dotsb))) + C.$$

One obvious case where this is true: $f(x) = 0$, $F(x) = 0$.

Are there any other examples where $f$ is doing any real transformation? Or is such a pair $f$ and $F$ impossible?

Answer 1

Based on others and my comments, one can construct arbitrary solutions to the problem using a Taylor series ansatz. Use the functional equations from the comments \begin{align} \phi(x)&=f(x+\phi(x))\tag{1a}\label{eq:1a}\\ \psi(x)&=F(x+\psi(x))\tag{1b}\label{eq:1b} \end{align} fulfilling $\phi(x) = \psi'(x)$. Insert the Taylor expansions \begin{align} \phi(x)&=\sum_{k=0}^n \frac{\phi^{(k)}(0)}{k!} \, x^k \tag{2a}\\ f(x)&=\sum_{k=0}^n \frac{f^{(k)}(\phi(0))}{k!} \, (x-\phi(0))^k\tag{2b} \end{align} into \eqref{eq:1a} to get \begin{align} f^{(0)}(\phi(0)) &= \phi(0)\\ f^{(1)}(\phi(0)) &= \frac{\phi'(0)}{1+\phi'(0)}\\ f^{(2)}(\phi(0)) &= \frac{\phi''(0)}{(1+\phi'(0))^3} \\ f^{(3)}(\phi(0)) &= \frac{\phi'''(0) (1+\phi'(0))-3 \phi''(0)^2}{(1+\phi'(0))^5} \label{eq:3}\tag{3}\\ &\vdots \end{align} as well as the corresponding coefficients of $F$, \begin{align} F^{(0)}(0) &= 0 \\ F^{(1)}(0) &= \frac{\phi(0)}{1+\phi(0)} \\ F^{(2)}(0) &= \frac{\phi'(0)}{(1+\phi(0))^3} \\ F^{(3)}(0) &= \frac{\phi''(0) (1+\phi(0))-3 \phi'(0)^2}{(1+\phi(0))^5}. \label{eq:4}\tag{4}\\ &\vdots \end{align} For $\phi(0)=0$ we can now require that $F'(x)=a f(x)$ to get two solutions: $\phi(x)\equiv 0$ and \begin{align} \phi^{(1)}(0) &= a-1 \label{eq:5}\tag{5}\\ \phi^{(2)}(0) &= \frac{3a^2(a-1)}{a+1} \\ \phi^{(3)}(0) &= \frac{3 (a-1) a^3 (5 a^3+5 a^2+5 a-4)}{(a+1)^2 (a^2+a+1)} \\ \phi^{(4)}(0) &= \frac{15 (a-1) a^4 (7 a^6+7 a^5+14 a^4-5 a^3+11 a^2-20 a+4)}{(a+1)^3 (a^2+1) (a^2+a+1)} \\ \phi^{(5)}(0) &= \frac{45 (a-1) a^5 p_5(a)}{(a+1)^4 (a^2+1) (a^2+a+1)^2 (a^4+a^3+a^2+a+1)}\\ &\vdots \end{align} with $p_5(a)=(21 a^{13}+63 a^{12}+147 a^{11}+175 a^{10}$${}+231 a^9+104 a^8+75 a^7-109 a^6$${}-101 a^5-189 a^4-59 a^3-32 a^2+68 a-8)$, which reduces to $\phi(x)\equiv 0$ for $a\to 1$.

As the denominator looks familiar to me (from some $q$ series), this solution might be written down in closed form.

Some Mathematics code:

ordfs = 5;
\[Phi]s[x_] = Series[\[Phi][x], {x, 0, ordfs}](*/.\[Phi][0]->0*)
fs[x_] = Series[f[x], {x, \[Phi][0], ordfs}(*/.\[Phi][0]->0*)]
CoefficientList[Normal[\[Phi]s[x] - f[x + \[Phi]s[x]]], x]
sof = FullSimplify[Solve[% == 0, fs[x][[3]] /. Rational[_, _] :> 1]]
\[Psi]s[x_] = Integrate[\[Phi]s[x], x]
ordFs = Length[(\[Psi]s[x] - F[x + \[Psi]s[x]])[[3]]] - 1;
Fs[x_] = Series[F[x], {x, 0, ordFs}]
CoefficientList[Normal[\[Psi]s[x] - F[x+\[Psi]s[x]]], x]
soF = Simplify[Solve[% == 0, Table[Derivative[k][F][0], {k,0,ordFs}]]]

Simplify[Normal[\[Phi]s[x] - f[x + \[Phi]s[x]]] /. sof]
Simplify[Normal[\[Psi]s[x] - F[x + \[Psi]s[x]]] /. soF]

Update 16.12.23:

Noting that the Taylor series of $f(x)$ and $F(x)$ in \eqref{eq:3} and \eqref{eq:4} only differ by an additional derivative order, we can set $\phi(x)=\psi(x)=e^x$ to get $f(x)=F(x)$ (use the integration constant $C=1$). The resulting common series expansion A274447 can be identified as the Lambert W function, also known as ProductLog, such that \begin{align}\label{eq:6}\tag{6} \phi(x)=\psi(x)=e^x \Rightarrow f(x)=F(x)=W(e^x). \end{align} Indeed, this solution fulfills the functional equation \eqref{eq:1a}, as \begin{align} e^x &= f(x + e^x) = W(e^{x+e^x}) \label{eq:7a}\tag{7a}\\ \Leftrightarrow\quad \phi &= f(\ln\phi + \phi) = W(\phi \, e^\phi) \label{eq:7b}\tag{7b}, \end{align} which is the definition of $W$. See also MO:417026.

Update 18.12.23:

Playing around with the large number of identities for Lambert's W, a more general solution combo reads \begin{align} \phi(x) &= e^{a x + b} \label{eq:8a}\tag{8a}\\ \psi(x) &= \phi'(x)=a e^{a x + b} \label{eq:8b}\tag{8b}\\ f(x) &= a^{-1} W(a \, \phi(x)) \label{eq:8c}\tag{8c}\\ F(x) &= a^{-1} W(a \, \psi(x)) \label{eq:8d}\tag{8d}, \end{align} with arbitrary $a,b$.

Answer 2

Since you are just looking for a pair $f$ and $F$, you don't have to solve a functional equation for $\varphi$ and $\psi$. Take for instance $\varphi(x)=\sin(x)$, $\psi(x)=\cos(x)$. Then $I+\varphi$ and $I+\psi$ are both homeomorphisms $\mathbb R\to\mathbb R$, so you can define $f := \varphi\circ(I +\varphi)^{-1}$ and $F:=\psi\circ(I+\psi)^{-1}$.

Rmk For a smooth example (actually, analytic) you may take $\varphi(x)=a\sin(x)$, $\psi(x)=a\cos(x)$ with $|a|<1$, so that now $I+\varphi$ and $I+\psi$ are smooth diffeos. Of course, if you don't insist that the functions be defined on the whole line, there is even more freedom in the choice of $\varphi$.