Earlier today there was a question asking, for any $f$ from $[0,1] \rightarrow \Bbb R,$ how to evaluate the following integral,
$$ \int_0^1 \frac{f(x)}{f(x)+f(1-x)} \ dx. $$
Link here. The question implicitly assumed that the function was 'sufficiently nice' (integrable and perhaps even continuous) and arrived at the conclusion that it must equal $1/2.$ This got me thinking about the conditions required for this to occur, so my question is as follows,
Under what conditions for $f$ is the above function Riemann integrable?
My Thoughts
Since the composition of continuous functions is continuous, it follows that given the function is always integrable if $f$ is continuous (or piecewise continuous on finitely many intervals) and $f(x)+f(1-x) \neq 0.$
With a bit of googling, I came across the following result: If $f$ is integrable and $g$ is continuous, then the composition $g \circ f$ is integrable. Hence setting $g(x)=1/x,$ we have that $1/f$ is integrable if $f$ is either positive or negative everywhere.
Since the product of two integrable functions is integrable, it follows that if $f$ is integrable and $f(x) + f(1-x)$ is positive or negative everywhere, then the integrand exists.
Further, if there are finitely many intervals $I_1, \dots, I_n$ partitioning $[0,1]$ s.t. $f(x)+f(1-x)$ is either positive or negative everywhere on each interval, then the function is integrable on each interval, so it is therefore integrable on $[0,1].$
This is pretty good, but can this be further improved? In particular, the remaining cases that haven't been considered are,
- $f$ is not integrable.
- $f(x)+f(1-x)$ 'jumps' between positive and negative intervals infinitely many times (so when the second condition doesn't hold).
Background
I've had a first course in analysis, so I know the definition of a Riemann integrable in terms of Daurbox sums along with basic facts like integrable functions being bounded, continuous functions being integrable, sums and products of integrable functions being integrable, etc along with the fundamental theorem of calculus.
(Feel free to ignore/remove this part if it's off topic)
Finally as a related but somewhat off-topic question, if it is integrable, does it always equal $1/2?$ I've only seen a proof of integration by substitution when $f$ is continuous, so I'm not convinced that the manipulation holds in general.
