For any given function $f\colon [0,1]\to\Bbb R$, what is $\int_0^1\frac{f(x)}{f(x)+f(1-x)}dx$?

Question

I have a general function assuming the following integral does exist $$\int_0^1\frac{f(x)}{f(x)+f(1-x)}dx.$$ How do I solve it? I have tried to split it up from $0$ to $0.5$ and from $0.5$ to $1$, but I don't know what to do next.

Thanks for help.

Answer 1

Hint. Set $\displaystyle I=\int_0^1\frac{f(x)}{f(x)+f(1-x)}dx$. By the change of variable $x \to1-x$ you get that $$ I=\int_0^1\frac{f(1-x)}{f(x)+f(1-x)}dx. $$ Then observe that $$ I+I=\int_0^1\frac{f(x)}{f(x)+f(1-x)}dx+\int_0^1\frac{f(1-x)}{f(x)+f(1-x)}dx=\int_0^1\frac{f(x)+f(1-x)}{f(x)+f(1-x)}dx=1 $$ giving easily $$I=\frac12.$$

Answer 2

$$ u = 1-x,\qquad x = 1-u $$ $$ \int_0^1 \frac{f(x)\,dx}{f(x)+f(1-x)} = \int_1^0 \frac{f(1-u)(-du)}{f(1-u)+f(u)} = \int_0^1\frac{f(1-x)\,dx}{f(1-x)+f(x)}. $$ What the sum of the first and last integrals is, is clear.

Answer 3

HINT:

Use $I=\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$

$I+I=\int_a^bf(x)\ dx+\int_a^bf(a+b-x)\ dx=\int_a^b[f(x)+f(a+b-x)]\ dx$