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Assume that we have a function $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$ and $D$ a constant. How can we solve the following equation for $f$:

$$ \int^{b}_{x_2=a} \int^{b}_{x_1=a} \frac{1}{(f(x_1)+f(x_2))^2}d x_1 d x_2 = D \log(\tfrac{b}{a})$$

where $0.

Thanks.

  • 0
    Are $a$ and $b$ fixed, of supposed to satisfy some conditions?2012-06-07
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    They are fixed positive real numbers.2012-06-07
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    Have you tried letting $F(a,b) = \int_a^b \int_a^b (f(x)+f(y))^{-2} \mathrm dx\, \mathrm dy = D(\log b - \log a)$ and comparing partial deriviatives with respect to $a$ and $b$?2012-06-07
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    I tried it but it doesn't seem to give a solution. We obtain `$ int^{b}_{x=a} (f(x)+f(b))^{-2} dx = D/ (2 b) $` and `$ int^{b}_{x=a} (f(x)+f(a))^{-2} dx = D/ (2 a) $` if I didn't make a mistake somewhere. However using the same method again does simplify them further.2012-06-07
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    @korpeo Could you let us know in what context you found this problem?2012-06-07
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    You can take $D$ to be 1, can't you, by toggling $f$ and $\sqrt Df$.2012-06-07
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    Actually x1 and x2 are random variables and f is a function of them in a problem I am working on. I assumed them to have a uniform(a,b) distribution. The model I am working on resulted in this equation for which I couldn't find a closed-form solution and I think the details of the model is not necessary to solve the equation.2012-06-07
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    @korpeo Thanks. The reason I asked the context since sometimes the context can provide some insight into the problem.2012-06-08
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    Unless I've made an error somewhere, $\frac{\partial^2}{\partial a\partial b} F(a,b) = -2(f(a)+f(b))^{-2}$. But $\frac{\partial^2}{\partial a\partial b} D(\log b - \log a) = 0$, so it seems there is no solution.2012-06-08

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