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Let $f: \mathbb R\rightarrow \mathbb R$ satisfying \begin{equation*} f(x) + f\bigg(1-{1\over x}\bigg) = \arctan x\end{equation*} for all $x\neq 0$. I want to evaluate \begin{equation*} \int_0^1 f(x) dx \end{equation*}

I know that $\tan\bigg(f(x) + f\bigg(1-{\cfrac 1x}\bigg)\bigg) = x$ and $-\pi/2 < f(x) + f\bigg(1-{1\over x}\bigg) < \pi/2$, but none of this seems useful yet. Since $x\neq 0$, the answer should be the solution to \begin{equation*} \lim_{n\rightarrow 0}\int_n^1f(x)dx\end{equation*} But again, I'm still unsure where to go from here. Any hint would be appreciated.

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  • $\begingroup$ Try playing with $x$ in your initial equation. $\endgroup$ Feb 22, 2023 at 0:02
  • $\begingroup$ @ericforman, I have, one of my results was $f(0) = \pi/4 + f(1)$ $\endgroup$
    – Lex_i
    Feb 22, 2023 at 0:07
  • $\begingroup$ @ericforman Oh wait, hold up. $1-1/x > 0$ only when $x > 1$. But our domain is $(0, 1]$, so does that mean $f(1-1/x)$ is not defined on $(0, 1]$, so we treat it as zero? $\endgroup$
    – Lex_i
    Feb 22, 2023 at 0:12
  • $\begingroup$ Try to see how a substitution involving $x$ would help you get a more helpful expression. Seeing as the expression includes $\arctan x$ some substitutions should come to mind. $\endgroup$ Feb 22, 2023 at 0:19
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    $\begingroup$ One possible hint: I recognize $1 - \frac{1}{x}$ as a rational function which gives the identity if you compose it with itself three times. $\endgroup$ Feb 22, 2023 at 0:22

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