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Find f(t) such that:

$\log (x-1)=\int_1^{\infty } \log \left(1-f(t)^{-x}\right) \, dt$

I am not familiar with solving integral equations. I was thinking of expressing logarithm inside integral with series, then moving integral inside this sum, but this leads me nowhere.

Do you have any idea? I do not expect there should be some nice formula for f(t) but at least some numerical method.

azerbajdzan
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  • differentation w.r.t. $x$ is an idea – tired Dec 25 '16 at 20:47
  • So after differentiation I got:$$\frac{1}{x-1}=\int_1^{\infty } \frac{\log (f(t))}{f(t)^x-1} , dt$$ – azerbajdzan Dec 25 '16 at 20:53
  • isn't there a $f'(t)$ missing? – tired Dec 25 '16 at 20:56
  • Should it be $$\frac{1}{x-1}=\int_1^{\infty } \frac{\frac{x f'(t)}{f(t)}+\log (f(t))}{f(t)^x-1} , dt$$To tell you truth I am not sure how to differentiate inside integral. I considered f(t) is independent of x. – azerbajdzan Dec 25 '16 at 21:04
  • Can you find $f(t)$ such that $0=\displaystyle\int_{1}^{\infty}\lg\left(1-\frac{1}{f(t)^2}\right)dt$? – drzbir Dec 25 '16 at 23:43
  • No, for x=2 there is no solution, because integral of always negative function can not be equal zero. But it does not mean, that for x>2 there is no solution, or does it? – azerbajdzan Dec 26 '16 at 11:15
  • I am just trying to simplify the problem. Try for $x=3$, maybe you will see a pattern. – drzbir Dec 26 '16 at 16:29

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