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.

You mean 'how do I find an antiderivative'? – Alex Mar 08 '15 at 17:29 — Alex, Mar 08 '15 at 17:29
Or rather, "how do I calculate this sum ?" – Olórin Mar 08 '15 at 17:30 — Olórin, Mar 08 '15 at 17:30

score 18 · Accepted Answer · answered Mar 08 '15 at 17:30

18

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 $$I=\frac12.$$

answered Mar 08 '15 at 17:30

Olivier Oloa

120,989

Thx, the only thing I don't understand is how you can set u=1-x without getting a minus in front of the integral because of the du/dx=-1 – netwon1227 Mar 08 '15 at 17:43
@netwon1227 Recall that $\int_a^b=-\int_b^a$. So the minus sign disappears: $\int_1^0=-\int_0^1$. Thanks. – Olivier Oloa Mar 08 '15 at 17:44
I am really sorry, but I don't get the second line: I=\int_0^1\frac{f(1-x)}{f(x)+f(1-x)}dx.. You have changed the variable: u \to1-x. Shouldn't it be I=-\int_0^1\frac{f(1-x)}{f(x)+f(1-x)}dx because du/dx=-1 ? – netwon1227 Mar 08 '15 at 17:54
1

@netwon1227 Don't forget to switch $0$ and $1$, so you rather obtain $$-\int_1^0\frac{f(1-x)}{f(x)+f(1-x)}dx $$ which gives $$\int_0^1\frac{f(1-x)}{f(x)+f(1-x)}dx $$ Right? – Olivier Oloa Mar 08 '15 at 17:56
@netwon1227 Observe that $1-0=1$ and $1-1=0$ ... – Olivier Oloa Mar 08 '15 at 18:04
1

You are right. I don't know why I didn't get it. Thank you, very helpful answer. – netwon1227 Mar 08 '15 at 18:16
You are very welcome! – Olivier Oloa Mar 08 '15 at 18:17

score 0 · Answer 2 · answered Mar 08 '15 at 17:32

0

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

answered Mar 08 '15 at 17:32

Michael Hardy

1

score 0 · Answer 3 · answered Apr 08 '15 at 10:58

0

HINT:

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

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

answered Apr 08 '15 at 10:58

lab bhattacharjee

274,582

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

3 Answers3

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