# integral equation

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?

• 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. Oct 15, 2011 at 9:40
• 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. Oct 15, 2011 at 13:50
• Why it is obvious that a solution does exist ? Oct 15, 2011 at 16:06

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?