# Putnam, A3 (2016)

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.

• Try playing with $x$ in your initial equation. Feb 22, 2023 at 0:02
• @ericforman, I have, one of my results was $f(0) = \pi/4 + f(1)$ Feb 22, 2023 at 0:07
• @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? Feb 22, 2023 at 0:12
• 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. Feb 22, 2023 at 0:19
• 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. Feb 22, 2023 at 0:22