What are the functions verifying:
$$\int_0^{\infty} f(t) \cos(2\pi xt)=\lambda \frac{1}{x} f(\frac{1}{x})$$
With $\lambda$ a constant ? (Functions $x^{-\alpha}$ with $0<\alpha<1$ are solutions but can we find other solutions?)
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$\begingroup$ Well, it is still not quite integrable... In any case, the equation is linear in $f$, so any linear combination of solutions also does the job. In fact, $\alpha$ can be complex, and `linear combinations' can be replaced by integrals. This brings us to the Mellin transform. In general one can expect that the solution can be found by taking the Mellin transform of both sides; when I have more time, I will try to write down the details. $\endgroup$– Mateusz KwaśnickiCommented Apr 11, 2018 at 20:35
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1$\begingroup$ Some more thoughts: After (formally) taking the Mellin transform of both sides, the equation takes form $(2\pi)^{-s}\Gamma(s)\cos(\pi s/2)\mathcal{M}f(1-s)=\lambda \mathcal{M}f(1-s)$, which is ill-posed and suggests that the only solutions are linear combinations of $x^{s-1}$, where $s$ is a (complex) solution of the equation $(2\pi)^{-s}\Gamma(s)\cos(\pi s/2)=\lambda \mathcal{M}f(1-s)$. There is an infinite family of solutions; when $\lambda=\tfrac{1}{2}$, infinitely many of them seem to lie on $\operatorname{Re}s=\tfrac{1}{2}$, and in general they appear to concentrate around this line. $\endgroup$– Mateusz KwaśnickiCommented Apr 11, 2018 at 21:56
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