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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.

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  • $\begingroup$ Fractional part, sorry. $\endgroup$
    – Vincent Granville
    Mar 15, 2019 at 2:34
  • $\begingroup$ Is $a > 0$? What have you tried? $\endgroup$
    – JavaMan
    Mar 15, 2019 at 2:49
  • $\begingroup$ Yes, $a > 0$. Not sure if it changes anything, whether $a$ is rational, irrational, or a normal number. It would probably if this was a sum (which I am interested in too) instead of an integral. On the grand scheme of things, I am trying to find an integral formula, for a number $a$, such that if the formula is satisfied, then the digits of $a$ in base 2 have a 50% proportion of zero's. Note that for some number, for instance $a=0.1001111000000001111111111111111...$ the proportion of zero's or one's does not even exists in the first place. $\endgroup$
    – Vincent Granville
    Mar 15, 2019 at 3:03
  • $\begingroup$ Are you familiar with Frullani's Integral? $\endgroup$
    – Ryan Goulden
    Mar 25, 2019 at 23:04
  • $\begingroup$ Yes Ryan, at least a little bit. See my article on this topic (it might even provide a generalization) at dsc.news/2HDYKkp. $\endgroup$
    – Vincent Granville
    Mar 26, 2019 at 1:22

1 Answer 1

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

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

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

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

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    $\begingroup$ Nice way to get around the separate integrals diverging at $s=0$. $\endgroup$
    – marty cohen
    Mar 15, 2019 at 3:59
  • $\begingroup$ Isn't last limit $\lim_{s\rightarrow 0}$ evaluated as $\frac12 \log 2$? This is due to the integration with respect to $z$. $\endgroup$
    – Sungjin Kim
    Mar 25, 2019 at 22:40
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    $\begingroup$ In the second line, we have $$\zeta(s) = \frac{s}{s-1}-\int_{0+}^{\infty} \frac{ \{ x \} }{x^{s+1}} dx, \ \ \Re(s)>0.$$ This would change the expression just a little bit. It might be that through the Cauchy's principal value. In any case, I think there needs a justification in your answer. $\endgroup$
    – Sungjin Kim
    Mar 26, 2019 at 1:00
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    $\begingroup$ @i707107 For $\Re(s)> 1$ then for $\Re(s)>0$ it is $\zeta(s) =\sum_{n=1}^\infty s\int_n^\infty x^{-s-1}dx=s\int_1^\infty \lfloor x \rfloor x^{-s-1}dx= \frac{s}{s-1}-s\int_1^{\infty} \{x\}x^{-s-1} dx$ which is $= -s \int_0^\infty \{x\}x^{-s-1}dx$ for $\Re(s) \in (0,1)$ (and the functional equation of $\zeta(s)$ follows from the Fourier series of $\{x\}$) $\endgroup$
    – reuns
    Mar 26, 2019 at 1:34
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    $\begingroup$ Okay, I see. In my expression, it must be $$\frac{s}{s-1}-\int_1^{\infty} \frac{ \{ x\} }{x^{s+1}} dx, \ \ \Re(s)>0.$$ I was confused $0+$ with $1$. Now, with $\int_{0+}^{\infty}$, everything makes sense. $\endgroup$
    – Sungjin Kim
    Mar 26, 2019 at 1:37

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