Let $a_n > 0$ and $b_n$ real.
Let $f(x)= x - \sum_{i=0}^{\infty} \dfrac {a_n}{x+b_n}$
Now apparently for every function $F(x)$ :
$$\int_{-\infty}^{+\infty} F(f(x)) - F(x) dx = 0$$
If the integral converges.
Why is this true ?
I considered contour integrals and substitution but found no proof.
I assume there is a substitution such that we get
$$\int_{-\infty}^{+\infty} F(f(x)) - F(x) dx =\int_{+\infty}^{-\infty} F(f(u)) - F(u) du$$
(from which the equality to $0$ follows)
I also assume this can be proved by an argument principle.
And I assume there is a much simpler way to prove this.
Not a full answer but perhaps the first step to a simple full answer.
I found a proof for the case $F$ is continuous and $f(x) = x - \frac{1}{x}$.
And indeed it uses some simple substitutions.
We may write \begin{align} \int_{-\infty}^{\infty}f\left(x-\frac{1}{x}\right)dx&=\int_{0}^{\infty}f\left(x-\frac{1}{x}\right)dx+\int_{-\infty}^{0}f\left(x-\frac{1}{x}\right)dx=\\ &=\int_{-\infty}^{\infty}f(2\sinh T)\,e^{T}dT + \int_{-\infty}^{\infty}f(2\sinh T)\,e^{-T}dT=\\ (collecting\space terms ) &=\int_{-\infty}^{\infty}f(2\sinh T)\,2\cosh T\,d T=\\ &=\int_{-\infty}^{\infty}f(x)\,dx. \end{align} To go from the first to the second line, we make the substitution $x=e^{T}$ in the first integral and $x=-e^{-T}$ in the second one.
The simplicity of this proof and its substitutions imho seems to suggest we can generalize this to a simple full proof by substitutions methods.