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I have been able to prove the identity $$\int_{0}^{1} \frac{f(x)}{f(x)+f(1-x)} \, dx = \frac{1}{2}$$ for any continous $f:[0,1]\to[0,\infty)$ for which the integrand is defined, with calculus, but I would like to know if there is an intuitive explanation of the identity.

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    $\begingroup$ Should that say "for any continuous $f:[0,1]\to[0,\infty)$ for which the integral is defined"? I'm thinking that $f(x)=1$ (for example) doesn't play nicely in this integrand. Other examples include any function that is constant over some interval $\left(\frac 12-\delta,\frac 12 + \delta\right)$ where $\delta>0.$ $\endgroup$
    – David K
    Oct 28, 2014 at 18:08
  • $\begingroup$ doesn't that integrand goes to $\infty$ when $x = \frac 12$? Or at the very least it is not defined there, is it? $\endgroup$
    – Ant
    Oct 28, 2014 at 21:57
  • $\begingroup$ @Ant Oops! Sorry about that. Should have been + in the denominator. $\endgroup$
    – slo
    Oct 29, 2014 at 7:32
  • $\begingroup$ @Ant: it is claimed that the integral is $\frac12$ for any $f$ for which the integral is defined. If $f\left(\frac12\right)=0$ and $f'\left(\frac12\right)$ exists, then L'Hospital says that $$\lim_{x\to1/2}\frac{f(x)}{f(x)-f(1-x)} =\frac{f'\left(\frac12\right)}{f'\left(\frac12\right)+f'\left(\frac12\right)}$$ Or we can accept the (fairly drastic) change by the OP. $\endgroup$
    – robjohn
    Oct 29, 2014 at 7:50
  • $\begingroup$ @robjohn yes, so we should suppose $f(1/2)=0$ and then eliminate the discontinuity. (By the way, shouldn't there be $f'$ instead? ) $\endgroup$
    – Ant
    Oct 29, 2014 at 7:56

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Hint There is symmetry at play. Make the change of variables $x\to 1-x$, then sum. The integrand, call it $g(x)$, has the property that $g(1-x)=1-g(x)$.

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    $\begingroup$ Well, yes. But that is how I proved it. Can you possibly give a geometric explaination? $\endgroup$
    – slo
    Oct 28, 2014 at 16:32
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    $\begingroup$ @slo Yes. Given a function $g$, plot $g(1-x)$ and $1-g(x)$. Then you'll see what's going on. There's no magical wonder going on here, really. $\endgroup$
    – Pedro
    Oct 28, 2014 at 22:39
  • $\begingroup$ Oh, now you made me feel stupid. It makes sense now. Thanks! $\endgroup$
    – slo
    Oct 29, 2014 at 7:41
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Consider \begin{align} I = \int_{0}^{1} \frac{f(x)}{f(x)-f(1-x)} \, dx \end{align} and let $x = 1-t$ to obtain \begin{align} I &= \int_{1}^{0} \frac{f(1-t)}{f(1-t) - f(t)} \, (-1) dt \\ &= - \int_{0}^{1} \frac{f(1-t)}{f(t) - f(1-t)} \, dt. \end{align} Now adding the two integrals yields \begin{align} 2 I &= \int_{0}^{1} \frac{f(x) - f(1-x)}{f(x) - f(1-x)} \, dx = \int_{0}^{1} \, dt = 1 \end{align} and hence \begin{align} \int_{0}^{1} \frac{f(x)}{f(x)-f(1-x)} \, dx = \frac{1}{2}. \end{align}

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  • $\begingroup$ This would work if it were not for the fact that the denominator of the integrand is $0$ for $t=\frac12$. The OP has corrected this; however, there have been a number of answers recently where divergence of the integral has been ignored, so I felt this should be mentioned. $\endgroup$
    – robjohn
    Oct 29, 2014 at 7:45
  • $\begingroup$ Although I made a mistake, I believe if we instead consider the Cauchy principal value of $\int_{0}^{1} \frac{f(x)}{f(x)-f(1-x)}dx$, the proof is correct. $\endgroup$
    – slo
    Oct 29, 2014 at 18:20
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$$\int_0^1\frac{f(x)}{f(x)-f(1-x)}dx=\frac{1}{2}\int_0^1\frac{f(x)}{1-\frac{1}{2x}}\left(\frac{x-(1-x)}{f(x)-f(1-x)}\right)\frac{dx}{x}$$

My fuzzy intuition is that $\frac{x-(1-x)}{f(x)-f(1-x)}$ is a crude approximation to the the reciprocal of the derivative. Multiplying $f$ by an approximation to the reciprocal of the derivative gives an approximation of the $x$ value. That cancels with the $x$ in the denominator. So you are integrating the rational function $\frac{1}{1-\frac{1}{2x}}$ from $0$ to $1$. Normally, you would get $2\log(2)$, but the error of the approximations above is precisely enough to scale this down to $1$.

I get the sense that this is related to logarithmic differentiation, but I cannot make it precise.

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It's because of mirror-image symmetry around 1/2.

There is a beautiful illustration that clarifies it, but this is my first posting and I don't know how to draw graphics in a post.

I'll try to explain in words: For every point x on the left side of 1/2 (i.e. in the interval [0,1/2)) there is a corresponding point 1-x on the right side of 1/2 (i.e. in the interval (1/2,1]).

You'll notice that your integrand f(x)/(f(x)+f(1-x)) has the nice property that whatever value y it takes at x, it takes on the value 1-y at its corresponding mirror-image point 1-x.

(You can work this out easily enough by substituting (1-x) in for x and doing the algebra.)

So, we're taking the integral of the y values over the interval [0,1/2), and then we're taking the integral of 1-y (those same y values!) over the interval (1/2,1]. You can see that whatever the y values sum to on the left side gets canceled out by whatever their negatives sum to on the right side.

The only thing which remains is the 1 over the interval (1/2,1]. That results in an integral of 1*(1/2) = 1/2.

That's it.

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The square $[0,1]\times [0,1]$ is a union of two sets $\{y \le g(x)\}$ and $\{y \ge g(x)\}$. If $g$ has the property $g(1-x)=1-g(x)$ then the symmetry with respect to the center of the square takes one set into the other, hence they are congruent and have the same area $1/2$.

(assume $f>0$ so $g(x) = \frac{f(x)}{f(x)+f(1-x)} \in (0,1)$)

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