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?

  • $\begingroup$ Some $\TeX$ hints: Use \exp instead of exp to keep the function name from being interpreted as individual variables whose symbols get italicized. If there is no predefined command sequence, e.g. for $\operatorname{frac}$, use \operatorname{frac}. Displayed equations should be in double dollar signs (as opposed to single dollar signs) to allow the proper font sizes for displayed equations to be selected. You can right-click on any $\TeX$ output you see on this site and select "Show Source" to see how it's done. $\endgroup$
    – joriki
    Oct 15, 2011 at 9:40
  • $\begingroup$ You might consider the substitution $u=\log t$, then it turns into a convolution on the $\mathbb{R}$. However, I can not see how that would help - also I think Phira might be right. $\endgroup$ Oct 15, 2011 at 13:50
  • 1
    $\begingroup$ Why it is obvious that a solution does exist ? $\endgroup$
    – Sasha
    Oct 15, 2011 at 16:06

2 Answers 2


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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