6
$\begingroup$

This morning I've calculated the case $k=2$ of an integral involving the fractional part function $ \left\{ x \right\} $, this $$-\int_0^1\log\left( \left\{ x^{-1/k} \right\} \right)dx,\tag{1}$$ in terms of well-known constants and the series $\sum_{n=1}^\infty\frac{\log (n+1)}{n^2}$.

Question. Is it known from the literature the calculation $(1)$ for integers $k\geq 2?$ Or in other case can you calculate a different case for example $k=3$? I am saying to calculate in terms of constants or series, as I did with my example $k=2$. Thanks in advance.

$\endgroup$
6
$\begingroup$

$$\begin{eqnarray*}\mathfrak{I}_k &=& -\int_{0}^{1}\log\left(\{x^{-1/k}\}\right)\,dx=-\int_{0}^{1}kz^{k-1}\log\left(\{z^{-1}\}\right)\,dz\\&=&-k \int_{1}^{+\infty}\frac{\log\left(\{u\}\right)}{u^{k+1}}\,du=-k\sum_{n\geq 1}\int_{n}^{n+1}\frac{\log(u-n)}{u^{k+1}}\,du\\ &=&-k\sum_{n\geq 1}\int_{0}^{1}\frac{\log u}{(u+n)^{k+1}}\,du=\frac{(-1)^{k}}{(k-1)!}\int_{0}^{1}\log(u)\,\psi^{(k)}(u+1)\,du\\&=&\int_{0}^{1}\left[\frac{\psi^{(k-1)}(u+1)}{(k-1)!}-(-1)^k\zeta(k)\right]\frac{du}{u}\end{eqnarray*}$$ in the $k=3$ case, by integration by parts, leads to $$ \begin{eqnarray*}\mathfrak{I}_3 &=& -\frac{1}{2}+\zeta(3)+\int_{0}^{1}\left[\frac{1}{2}\psi'(u+1)+u\zeta(3)-\frac{1}{2}\zeta(2)\right]\frac{du}{u^2} \\&=&\frac{3\,\zeta(3)-\zeta(2)}{2}+\int_{0}^{1}\left[\psi(u+1)+\gamma-\zeta(2)u+\zeta(3)u^2\right]\frac{du}{u^3}\\&=&\gamma-\zeta(2)+\tfrac{11}{6}\zeta(3)+3\int_{0}^{1}\left[\log\Gamma(u+1)+\gamma u-\frac{\zeta(2)}{2}u^2+\frac{\zeta(3)}{3}u^3\right]\frac{du}{u^4}\end{eqnarray*}$$ where $$ \log\Gamma(u+1)+\gamma u -\frac{\zeta(2)}{2}u^2+\frac{\zeta(3)}{3}u^3= \sum_{m\geq 1}\left[\frac{u}{m}-\frac{u^2}{2m^2}+\frac{u^3}{3m^3}-\log\left(1+\frac{u}{m}\right)\right] $$ produces: $$\boxed{\mathfrak{I}_3 = \color{blue}{\frac{3\,\zeta(3)-\zeta(2)}{2}+\sum_{m\geq 1}\frac{\log(m+1)-\log m}{m^3}}}\tag{A}$$ By Frullani's theorem, the last series is related with $\int_{0}^{1}\frac{(1-u)\,\text{Li}_3(u)}{u\log u}\,du $.
We may also write $(A)$ in the following form: $$ \mathfrak{I}_3 = \frac{3\zeta(3)-\zeta(2)}{2}+\log(2)+\sum_{h\geq 0}\frac{(-1)^h}{h+1}\left[\zeta(4+h)-1\right]\tag{B} $$ which is better suited for numerical evaluation purposes.
The shown approach admits a straightforward generalization: for any $k\geq 2$, $\mathfrak{I}_k$ is given by a linear combination of $\zeta(2),\ldots,\zeta(k)$ and the series $\sum_{m\geq 1}\frac{\log(m+1)-\log(m)}{m^k}$, associated with $\int_{0}^{1}\frac{(1-u)\,\text{Li}_k(u)}{u\log u}\,du $ and $\sum_{h\geq 0}\frac{(-1)^h}{h+1}\left[\zeta(h+k+1)-1\right]$.

$\endgroup$
  • $\begingroup$ Many thanks and congratulations for such nice result (A), and (B) that has a great mathematical beauty. I am going to need some hours to read and understand your calculations. $\endgroup$ – user243301 Oct 25 '17 at 19:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy