4

This question is inspired by the formula $$\displaystyle\int_0^1{\left\lfloor{1\over x}\right\rfloor}^{-1}\!\!dx={1\over2^2}+{1\over3^2}+{1\over4^2}+\cdots = \zeta(2)-1,$$ see for instance this question.

It appears that for any integer $n\geq 0$, we have

$$\int_0^1 \lfloor x^{-1} \rfloor^{-1} x^n dx = \frac{1}{n+1}(\zeta(2)+\zeta(3) + \dots + \zeta(n+2) ) - 1.$$

I believe that Yiorgos' answer could be generalized to prove the above formula. Can anyone pull it off?

Side question: If $n$ is replaced by a complex number $s$, for which values of $s$ does the integral converge? (We interpret $x^s$ as $\exp(x\log s)$, where $\log s$ is the principal branch of the complex log.) If the above formula is true, then for any integer $n\geq 0$, have $$\int_0^1 \lfloor x^{-1} \rfloor^{-1} ((n+1)x^n-nx^{n-1}) dx = \zeta(n+2) - 1.$$ Is it true that for complex value of $s$ for which the integral converges, we have $$\int_0^1 \lfloor x^{-1} \rfloor^{-1} ((s+1)x^s-sx^{s-1}) dx = \zeta(s+2)-1?$$

Community
  • 1
Bruno Joyal
  • 54,711

2 Answers2

4

Letting $t=x^{-1}$ so $dx=-t^{-2}dt$ this integral can be rewritten as:

$$\begin{align}\int_{1}^\infty \lfloor t\rfloor^{-1} t^{-(n+2)}dt &= \sum_{k=1}^\infty \frac{1}k\int_{k}^{k+1}t^{-(n+2)}dt \\&= \frac{1}{n+1}\sum_{k=1}^\infty \frac{1}{k}\left(k^{-(n+1)}-(k+1)^{-(n+1)}\right) \end{align}$$ That's pretty much the exact same as the proof for $n=0$.

Now, $$\begin{align} \frac{1}{k}(k+1)^{-(n+1)}&= \frac{1}{k(k+1)}(k+1)^{-n}\\ &=\left(\frac{1}{k}-\frac{1}{k+1}\right)(k+1)^{-n} \\ &=\frac{1}{k}(k+1)^{-n}- (k+1)^{-(n+1)}\\ &=\dots\text{ induction applied here on }\frac1k(k+1)^{-n}\dots\\ &=\frac{1}{k}-\frac{1}{k+1}-\frac{1}{(k+1)^2}-\cdots-\frac{1}{(k+1)^{n+1}} \end{align}$$

So $$\frac{1}{k}\left(k^{-(n+1)}-(k+1)^{-(n+1)}\right) = \left(\sum_{j=2}^{n+1} \frac{1}{(k+1)^j}\right) + \frac{1}{k^{n+2}}+\frac{1}{k+1}-\frac1k$$

This yields the result you want.

The side question (at least when $s\neq -1$) essentially asks about when:

$$\sum_{k=1}^\infty \frac{1}{k}\left(k^{-(s+1)}-(k+1)^{-(s+1)}\right)$$ converges. When $s$ is real and $s>-1$, the terms are positive, and bounded above by the sequence $k^{-(s+1)}-(k+1)^{-(s+1)}$, which is a telescoping sequence, so it converges.

Thomas Andrews
  • 177,126
3

To answer the side question.

Assume that $ \Re s >1$. Letting $t=x^{-1}$ so $dx=-t^{-2}dt$ the initial integral can be rewritten as:

$$\begin{align}\int_{1}^\infty \lfloor t\rfloor^{-1} t^{-(s+2)}dt &= \sum_{k=1}^\infty \frac{1}k\int_{k}^{k+1}t^{-(s+2)}dt \\&= \frac{1}{s+1}\sum_{k=1}^\infty \frac{1}{k}\left(k^{-(s+1)}-(k+1)^{-(s+1)}\right)\\&= \frac{1}{s+1}\left(\sum_{k=1}^\infty \frac{1}{k^{s+2}}-\sum_{k=1}^\infty \frac{1}{k(k+1)^{s+1}}\right) \end{align}$$ Thus $$ \int_0^1 \lfloor x^{-1} \rfloor^{-1} (s+1)\:x^s \:\mathrm dx = \zeta(s+2)-\sum_{k=1}^\infty \frac{1}{k(k+1)^{s+1}}, \quad \Re s >1 \tag1 $$ and we have $$ \int_0^1 \lfloor x^{-1} \rfloor^{-1} s\:x^{s-1} \:\mathrm dx = \zeta(s+1)-\sum_{k=1}^\infty \frac{1}{k(k+1)^{s}}, \quad \Re s >2. \tag2 $$ Then we may substract $(2)$ from $(1)$, using $$ \sum_{k=1}^\infty \frac{1}{k(k+1)^{s}}-\sum_{k=1}^\infty \frac{1}{k(k+1)^{s+1}}=\sum_{k=1}^\infty \frac{1}{k(k+1)^{s}}\left(1-\frac{1}{(k+1)}\right)=\zeta(s+1)-1 $$ to get, by analytic continuation, $$\int_0^1 \lfloor x^{-1} \rfloor^{-1} ((s+1)x^s-sx^{s-1}) dx = \zeta(s+2)-1, \quad \Re s >0.$$

Olivier Oloa
  • 120,989