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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:
Can we deduce from this integral equation that $ f(x)= O(x^{1/4+\epsilon}) $ for some positive $\epsilon$?
Can we solve this integral by the Hilbert-Schmidt method?
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.
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?
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