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Given the integral equation

$\exp(x)-1=\int_0^{\infty} \frac{\mathrm dt}{t}\operatorname{frac}\left(\frac{ \sqrt x}{\sqrt t}\right) f(t)\;,$

where $\operatorname{frac}$ denotes the fractional part of a number, $ \operatorname{frac}(x)= x-\lfloor x\rfloor$.

My questions are:

  1. Can we deduce from this integral equation that $ f(x)= O(x^{1/4+\epsilon}) $ for some positive $\epsilon$?

  2. Can we solve this integral by the Hilbert-Schmidt method?

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    Why it is obvious that a solution does exist ?2011-10-15

2 Answers 2

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You cannot conclude anything for the asymptotics of the integrand, because the function can have very high, very narrow peaks that contribute almost nothing to the integral.

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The integral equation has the form $\int_{0}^{\infty}\frac {dt} {t} f\left(x/t\right)g\left(t\right)=h\left(x\right)$ The left hand side is known as the Mellin convolution. Defining the Mellin transforms: $F\left(s\right)=\int_{0}^{\infty}dt t^{s-1}f\left(t\right)$ $G\left(s\right)=\int_{0}^{\infty}dt t^{s-1}g\left(t\right)$ $H\left(s\right)=\int_{0}^{\infty}dt t^{s-1}h\left(t\right)$ Your integral equation implies that $F\left(s\right)G\left(s\right)=H\left(s\right)$ See this Mellin convolution and Mellin transform. Does this help?