Definite Integral $\int_0^1 \left \{\frac{(-1)^{\lfloor 1/x \rfloor}}{x} \right\}\, dx$

Question

The curly brackets mean 'FractionalPart' which, I believe, is defined as {${x}$}$=x-\lfloor x \rfloor$ where $x \in \mathbb{R}$.

The conjecture I have found but can not prove is that the definite integral has a closed form of $A+B*ln(C/\pi)$ for positive integers A,B and C. The graph of the integrand gets 'very wild' near x=0 and I have come up with values from -.14478 to 3.928 - which means I can't get a handle on it at all. Any help or guidance to material that could teach me new techniques would also be appreciated. Values of A=B=C=1 does give me a result close to my negative estimate.

Also, using telescoping sum tricks, $\displaystyle \int_0^1(-1)^{\lfloor 1/x \rfloor} dx=1-2\ln(2)$ , so the conjecture seems to have a similar format. The denominator of 'x' in the integrand, this time, has me really puzzled.

Do the curly braces in your integrand have any particular meaning, or are they just for grouping? — hmakholm left over Monica, Apr 04 '16 at 21:55
@Henning Makholm -They mean 'FractionalPart' - thanks for the alert. I added clarification to the question. — Bob Kadylo, Apr 04 '16 at 22:09
Please do not copy other people's stuffs from other places and claiming it to be your own — GohP.iHan, Apr 05 '16 at 07:48
@GohP.iHan - How is my statement "The conjecture I have found but can not prove" - 'copying and claiming it to be my own' ?? The word "found" clearly implies it already existed somewhere else. I never said "My conjecture is: " - that would be plagiarism. — Bob Kadylo, Apr 06 '16 at 20:36

Jack D'Aurizio · Accepted Answer · 2016-04-04T22:25:44.053

We have that $\left\lfloor\frac{1}{x}\right\rfloor = n$ for any $x\in \left(\frac{1}{n+1},\frac{1}{n}\right]$. The integral of $\frac{1}{x}$ over such interval is $\log\left(1+\frac{1}{n}\right)$, hence:

$$ \int_{0}^{1}(-1)^{\left\lfloor\frac{1}{x}\right\rfloor}\frac{dx}{x}=\sum_{n\geq 1}(-1)^n \log\left(1+\frac{1}{n}\right)=\color{red}{\log\frac{2}{\pi}} $$

since the middle term is just the logarithm of Wallis' product. If we add a fractional part to the integrand function, things become a little more complicated, but not that much.

$$\begin{eqnarray*} \int_{0}^{1}\left\{(-1)^{\left\lfloor\frac{1}{x}\right\rfloor}\frac{1}{x}\right\}\,dx &=& \sum_{n\geq 1}\int_{\frac{1}{n+1}}^{\frac{1}{n}}\left\{\frac{(-1)^n}{x}\right\}\,dx\\ &=&\int_{\frac{1}{2}}^{1}\left(2-\frac{1}{x}\right)+\int_{\frac{1}{3}}^{\frac{1}{2}}\left(\frac{1}{x}-2\right)\,dx+\ldots\end{eqnarray*}$$ hence:

$$\begin{eqnarray*} \int_{0}^{1}\left\{(-1)^{\left\lfloor\frac{1}{x}\right\rfloor}\frac{1}{x}\right\}\,dx &=& \log\frac{2}{\pi}+\sum_{k=1}^{+\infty}\left(\frac{2k}{2k(2k-1)}-\frac{2k}{2k(2k+1)}\right)\\&=&\log\frac{2}{\pi}+\sum_{k\geq 1}\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)\\&=&\color{red}{1+\log\frac{2}{\pi}}.\end{eqnarray*}$$

There is much to learn in what you have shared here ! – Bob Kadylo Apr 05 '16 at 00:02 — Bob Kadylo, Apr 05 '16 at 00:02

Definite Integral $\int_0^1 \left \{\frac{(-1)^{\lfloor 1/x \rfloor}}{x} \right\}\, dx$

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