What is the value of $\int_0^\infty \frac{\{ax^2\}-\{ax\}}{x \log x}dx$?

Question

I am also interested in the following integral: $\int_0^\infty \frac{\{a\cdot4^x\}-\{a\cdot2^x\}}{x} dx$. Both should be identical, after a change of variable. Also, this one: $\int_0^\infty \frac{\{2ax\}-\{ax\}}{x}dx$. The brackets represent the fractional part function. I tried to compute them on WolframAlpha, it seems to converge to $(\log 2)/2$ on a small interval, say $[0, 10]$ but beyond small values, it is a mystery.

Answer 1

For $a > 1,a \not \in \mathbb{Z}$ and $\Re(s) \in (0,1)$

$$F(s)= pv(\int_0^\infty \frac{\{ ax\}}{\log x} x^{-s-1}dx)$$ $$F'(s)=-\int_0^\infty \{ ax\}x^{-s-1}dx=-a^s\int_0^\infty \{ y\}y^{-s-1}dy=\frac{a^s\zeta(s)}{s}\\ F(s)=F(1/2)+\int_{1/2}^s \frac{a^z\zeta(z)}{z}$$

$$G(s)=pv(\int_0^\infty \frac{\{ ax^2\}-\{ax\}}{\log x} x^{-s-1}dx)=pv(\int_0^\infty \frac{\{ ax^2\}}{\log x} x^{-s-1}dx)-pv(\int_0^\infty \frac{\{ ax\}}{\log x} x^{-s-1}dx)\\=pv(\int_0^\infty \frac{\{ at\}}{\log t^{1/2}} t^{-s/2-1/2}dt^{1/2})-pv(\int_0^\infty \frac{\{ ax\}}{\log x} x^{-s-1}dx)= F(s/2)-F(s)$$

$$G(0) = \lim_{s\to 0} F(s/2)-F(s)= \lim_{s\to 0} \int_{s/2}^s\frac{a^z\zeta(z)}{-z} = -\log(2)\zeta(0)= \frac{\log 2}2$$