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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 $F(x)$ :

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$$\int_{-\infty}^{+\infty} F(f(x)) – F(x) dx = 0$$

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.

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- Integral $I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}\mathrm dx=\frac{10\pi^3}{81 \sqrt 3}$

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.

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