Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
Problem. Evaluate $\displaystyle F(n, k):=\int_{0}^{1} \frac{1-\{1/x\}^n}{1-\{1/x\}^k}dx$ where $n$, $k$ are positive integers. ($\{x\}=x-\lfloor x\rfloor$) Someone proposed this interesting problem in AoPS forum. Although I tried various kinds of methods, I couldn't get any closed form of $F$. However, I could obtain an useful result; $$F(n, k)=\int_{0}^{1}\frac{1-t^n}{1-t^k}\psi^{(1)}(t+1)dt \quad\cdots (1)$$ ($\psi^{(1)}$ denotes Trigamma function.) How to get (1)... Substitute $u:=1/x$, then $$F(n, k)=\int_{1}^{\infty}\frac{1}{u^2}\frac{1-\{u\}^n}{1-\{u\}^k}du=\sum_{m\ge 1}\left(\int_{m}^{m+1}\frac{1}{u^2}\frac{1-(u-m)^n}{1-(u-m)^k}du\right)$$ If we make a substitution $t:=u-m$, we get $$\begin{aligned}F(n, k)&=\sum_{m\ge 1}\left(\int_{0}^{1}\frac{1}{(t+m)^2}\frac{1-t^n}{1-t^k}dt\right)=\int_{0}^{1}\frac{1-t^n}{1-t^k}\sum_{m\ge 1}\frac{1}{(t+m)^2}dt\\&=\int_{0}^{1}\frac{1-t^n}{1-t^k}\psi^{(1)}(t+1)dt\end{aligned}$$ So (1) is true. Do you have some ideas for calculating (1) to get a closed (or simplified) form? Thanks. |
|||||||
|
