If $f(x) + f(1 - 1/x) = \ln|x| $ find $\int_0^1 f(x) dx $

Question

This is a question from a practice workbook for a college entrance exam.

Let $A$ denote the set of all real numbers other than 0 and 1. A function $f:A \to R$ satisfies $f(x) + f(1 - \frac{1}{x}) = \ln|x|, \forall x \in A$ then the value of $\int_0^1f(x)dx$ is ?

I tried adjusting the integral $I = \int_0^1f(x)dx$ into a different form by using the substitution $x = 1 - \frac{1}{t}$

This gives $ I = -\int_1^\infty f(1 - \frac{1}{t})\frac{dt}{t^2}$ I expected a form with similar limits so I could add them and apply the functional equation but am unsure of how to proceed from here.

Answer 1

The thing to notice here is that if $1-\frac 1x $ is composed with itself twice, it gives the identity function, $x$. Starting with $$f(x) +f\left ( 1-\frac 1x \right) =\ln |x| $$ transform $x\to 1-\frac 1x$: $$f\left( 1-\frac 1x \right)+f\left(\frac{1}{1-x}\right) =\ln \left | 1-\frac 1x \right | $$ Do the same transformation again: $$f\left( \frac{1}{1-x} \right) +f(x) = \ln \left | \frac{1}{1-x} \right | $$ This gives you three equations in three variables. Solve for $f(x)$ : $$f(x) =\frac 12 \ln \left | \frac{x^2}{(x-1)^2} \right | = \ln |x| -\ln |x-1| $$ Now you may do the integral without any trouble.