# Functional equation $f(x) + f\left(1-\frac{1}{x}\right) = \tan^{-1}(x)$ and definite integral

Let $$f(x)$$ be a function $$f :\mathbb{R}\to \mathbb{R}$$ such that $$f(x) + f\left(1-\frac{1}{x}\right) = \tan^{-1}(x)$$ for all real $$x$$ except $$0$$.

Find $$\int_0^1f(x)\ \mathrm dx$$.

My approach till now:

Put $$x = \frac{1}{x}$$ in the functional equation and consider the domain of integration $$(0,1)$$ such that $$\tan^{-1}\frac{1}{x} = \cot^{-1}(x)$$ and add the original functional equation and the resulting equation after the substitution to get:

$$f(x) + f(1-x) = \frac{\pi}{2} - f\left(\frac{1}{x}\right) - f\left(1-\frac{1}{x}\right)$$ and integrate both sides from $$0$$ to $$1$$.

Let $$I = \int_0^1f(x)\ \mathrm dx$$, then LHS of above functional equation becomes $$2I$$. Now I am not able to evaluate the RHS, some $$\frac{\ln(2)}{2}$$ term always creeps up and doesn't gets cancelled and its not even in the answer.

• Couple things to note about this function. It can't be continuous (but it most likely is almost everywhere) and we have the following limits: $$\lim_{x\to1^+} f(x) = \lim_{x\to0^-} = -\lim_{x\to\infty} f(x) =-\frac{\pi}{8}$$ $$\lim_{x\to1^-} f(x) = \lim_{x\to0^+} = -\lim_{x\to-\infty} f(x) = \frac{3\pi}{8}$$ Mar 31 '20 at 11:39
• That's if this function exists at all Mar 31 '20 at 11:44
• @NinadMunshi yes its apparently not continous at x = 1 atleast since f(1) = $\frac{\pi}{4}$ Mar 31 '20 at 12:11
• $f(1)\neq \frac{\pi}{4}$ where did you get that? Mar 31 '20 at 12:11
• ohh sorry my bad...... i need to work out the limits at x=1 first myself again Mar 31 '20 at 12:12

Let $$g(x)=1-\frac1x$$. Then $$g(g(x))=\frac1{1-x}$$ and $$g(g(g(x)))=x$$.

Thus, $$f(x)+f(g(x))=\tan^{-1}(x)\tag1$$ $$f(g(x))+f(g(g(x)))=\tan^{-1}(g(x))\tag2$$ $$f(g(g(x)))+f(x)=\tan^{-1}(g(g(x)))\tag3$$ Since $$2f(x)=(1)-(2)+(3)$$, we get $$f(x)=\frac12\left(\tan^{-1}(x)-\tan^{-1}(g(x))+\tan^{-1}(g(g(x)))\right)\tag4$$ As mentioned in comments, this function is not continuous:

What is interesting is that the derivative of this function is continuous (if we subtract $$\frac\pi2$$ from $$f$$ in $$[0,1]$$ to counter the jump discontinuities):

In any case, \begin{align} &\int_0^1f(x)\,\mathrm{d}x\\ &=\frac12\left(\int_0^1\tan^{-1}(x)\,\mathrm{d}x-\int_0^1\tan^{-1}(g(x))\,\mathrm{d}x+\int_0^1\tan^{-1}(g(g(x)))\,\mathrm{d}x\right)\tag5\\ &=\frac12\left(\int_0^1\tan^{-1}(x)\,\mathrm{d}x-\int_{-\infty}^0\tan^{-1}(x)\,\mathrm{d}g(g(x))+\int_1^\infty\tan^{-1}(x)\,\mathrm{d}g(x)\right)\tag6\\ &=\scriptsize\frac12\left(\frac\pi4-\color{#C00}{\int_0^1\frac{x}{1+x^2}\,\mathrm{d}x}+\color{#090}{\int_{-\infty}^0\frac1{(1-x)\left(1+x^2\right)}\,\mathrm{d}x}+\frac\pi2-\color{#00F}{\int_1^\infty\frac{x-1}{x\left(1+x^2\right)}\,\mathrm{d}x}\right)\tag7\\ &=\frac12\left(\frac\pi4-\color{#C00}{\frac{\log(2)}2} +\color{#090}{\frac\pi4} +\frac\pi2-\color{#00F}{\frac{\pi-2\log(2)}4}\right)\tag8\\[3pt] &=\frac{3\pi}8\tag9 \end{align} Explanation:
$$(5)$$: apply $$(4)$$
$$(6)$$: apply $$g$$ and $$g\circ g$$ to get $$\tan^{-1}(x)$$ in each integral
$$(7)$$: integrate by parts
$$(8)$$: evaluate the integrals by partial fractions
$$(9)$$: simplify

To long for a comment. (I'm afraid if my calculation be a little wrong. I'm in a hurry situation, sorry)

By your notations, we have $$2I=\frac{\pi}{2}-\int_0^1f(\frac{1}{x})dx-\int_0^1f(1-\frac{1}{x})dx.$$ Take $$\dfrac{1}{u}=1-\dfrac{1}{x}$$, we have $$x=\dfrac{u}{u-1}$$, $$dx=-\dfrac{du}{(u-1)^2}$$ and $$\int_0^1f(1-\frac{1}{x})dx=-\int_0^\infty\frac{f(1/u)}{(u-1)^2}du$$ On the other hand, by letting $$x=\dfrac{1}{u}$$, we have $$\int_0^1f(\frac{1}{x})dx=\int_1^\infty\frac{f(u)}{u^2}du=\int_0^\infty \frac{f(u-1)}{(u-1)^2}du.$$