# Interesting integral related to floor function

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.

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Is $\{y\}$ the floor of $y$? – Ron Gordon Feb 23 '13 at 1:31
Yes. $\{x\}=x-\lfloor x\rfloor$ – hunminpark Feb 23 '13 at 13:16
{x} denotes the fractional part of x, not the floor (greatest integer function) of x: {x} = x - floor(x). – PolyaPal Feb 23 '13 at 13:40