# Evaluate $\int_0^\infty\left\{ \frac{1}{t} \right\}^{kn}t^{s-1}dt$ for positive integers $n,k$ and $0<\Re s<1$

Denoting $\{x\}$ the fractional part function, and $k\geq 1$ and $n\geq 1$ integers, I've interested in a closed-form for the integral in this

Question. What's about closed-form for the integral $$\int_0^\infty\left\{ \frac{1}{t} \right\}^{kn}t^{s-1}dt,$$ for $0<\Re s<1$? Many thanks.

I don't know if this integral is well known. See below what is my motivation (do calculations, perhaps artificious from an integral representation of the Riemann Zeta function), and this attempt:

First, the integral is convergence for integers $k\geq 1$ and $n\geq 1$, because using absolute convergence one gets from the inequality (and the appendix) that our integral is convergent for $$\left| \int_0^\infty\left\{ \frac{1}{t} \right\}^{kn}t^{s-1}dt\right|\leq \int_0^\infty\left\{ \frac{1}{t} \right\}t^{\Re s-1}dt,$$ when $0<\Re s<1$.

Secondly, following the hints of solutions in  for some integrals that likes to the integrand in the Question (see  if you want in the section References in the appendix), I believe that I obtain $$\int_0^\infty\left\{ \frac{1}{t} \right\}^{kn}t^{s-1}dt=\sum_{j=0}^\infty\int_0^1\frac{u^{nk}}{(u+j)^{s+1}}du,$$ when $0<\Re s<1$.

As, I've said I'm interested in compute closed-forms for such integral and after try more calculations from the approach in the appendix (that interchange some summation index to get an identity, if it is feasible). I understand that if you answer previous Question you satisfy an answer for this post, but feel free if you want add some remarks with the purpose to get such identities that I evoke.

We know the formula (11) here in MathWorld with reference to , that is the Mellin transform of this fractional part function $$\left\{ \frac{1}{t} \right\} =\operatorname{frac}\left(\frac{1}{t}\right).$$ Then since $0\leq \left\{ \frac{1}{t} \right\}<1$ I've computed with Möbius inversion formula to get $$\left\{ \frac{1}{t} \right\}= \sum_{n=1}^\infty\sum_{k=1}^\infty\frac{\mu(n)}{kn}\left\{ \frac{1}{t} \right\}^{kn},$$ and after combining with the cited Mellin transform by absolute convergence, for $0<\Re s<1$ $$-\frac{\zeta(s)}{s}=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{\mu(n)}{kn}\int_0^\infty\left\{ \frac{1}{t} \right\}^{kn}t^{s-1}dt.$$

References:

 Balazard, M. and Saias, E. The Nyman-Beurling Equivalent Form for the Riemann Hypothesis. Expos. Math. 18, 131-138 (2000).

 Furdui, Limits, Series, and Fractional Part Integrals, Problems in Mathematical Analysis. Springer (2013). (I say hints to Problems likes 2.14 and 2.21.)

• All users, if there are some mistakes in my claims or calculations explain those in your answer. If my result is right I am asking what's about a closed-form for such expression. – user243301 Nov 4 '16 at 10:22
• I've been thinking on this problem a bit the last few days. I don't claim that Mobius Transforms are my strong point, nor is any sort of work related to the Riemann Zeta. That being said, I'll continue thinking and playing around with this. Above all, I recommend editing your post. You have two questions there, and I would split them up. Leave your main one here, and the edit will bump the question to the front page. Then, make another post with the second question. – Brevan Ellefsen Nov 6 '16 at 20:24
• Many thanks for your advice, @BrevanEllefsen – user243301 Nov 7 '16 at 9:07
• Playing with formulas you don't understand is the exact opposite of doing some mathematics. So why do you think that $\left\{ \frac{1}{t} \right\}= \sum_{n=1}^\infty\sum_{k=1}^\infty\frac{\mu(n)}{kn}\left\{ \frac{1}{t} \right\}^{kn}$ – reuns Nov 8 '16 at 16:49

## 3 Answers

Let $$F_k(s) = s\int_0^\infty \{x\}^k x^{-s-1}dx = s \int_0^\infty \left\{\frac1t\right\}^k t^{s-1}dt$$

Show that $$\int_0^x \{t\}^kdt = \frac{\lfloor x \rfloor +\{x\}^{k+1}}{k+1}$$

And that (where it converges) $$\zeta(s) = s\int_0^\infty \lfloor x \rfloor x^{-s-1}dx = \frac{s}{s-1}-s\int_1^\infty \{x\} x^{-s-1}dx, \qquad\quad F_1(s) = -\zeta(s)$$

Integrating by parts $$F_k(s) = s(s+1)\int_0^\infty \frac{\lfloor x \rfloor +\{x\}^{k+1}}{k+1} x^{-s-2}dx = s\frac{\zeta(s+1)+F_{k+1}(s+1)}{k+1}$$ i.e. $$\boxed{F_k(s) = \frac{k}{s-1} F_{k-1}(s-1)-\zeta(s)= -\sum_{m=0}^{k-1} \zeta(s-m)\prod_{l=0}^{m-1} \frac{k-l}{s-l-1}}$$

• interesting mathematical exercice for you : add the domain of convergence to all those formulas – reuns Nov 8 '16 at 16:51
• Many thanks for your answer and advice. You and users as previous moderator are the heart of this site. I will study all details. – user243301 Nov 8 '16 at 19:31
• @user243301 ok so what is the domain of convergence of $\int_0^\infty \{x\}^k x^{-s-1}dx, k \in \mathbb{N}^*$ ? and how do you prove that where it converges $\int_0^\infty \{x\} x^{-s-1}dx = -\zeta(s)$ ? – reuns Nov 9 '16 at 20:52
• $g(x) = \{x\} = \mathcal{O}(1)$ means that $|g(x)| < C$ for some $C$ so that $|g(x) x^{s-1}| < C x^{-Re(s)-1}$ and $\int_1^\infty |\{x\} x^{-s-1}|dx < \int_1^\infty C x^{-Re(s)-1}dx= \frac{C}{Re(s)}$ when $Re(s) > 0$ – reuns Nov 9 '16 at 22:23
• 2nd step (the most important) : proving that $\int_1^\infty |h(x)|dx$ converges $\implies \int_1^\infty h(x)dx$ converges – reuns Nov 9 '16 at 22:27

We can derive a recursion for $f_n$ which is valid for $n\ge2$ and $1\lt s\lt n$: \begin{align} f_n(s) &=\int_0^\infty\left\{\frac1t\right\}^nt^{s-1}\,\mathrm{d}t\\ &=\int_0^\infty\{t\}^nt^{-s-1}\,\mathrm{d}t\\ &=-\frac1s\int_0^\infty\{t\}^n\,\mathrm{d}t^{-s}\\ &=\frac ns\int_0^\infty\{t\}^{n-1}t^{-s}\,\mathrm{d}t-\frac{\zeta(s)}s\\ &=\frac nsf_{n-1}(s-1)-\frac{\zeta(s)}s\tag{1} \end{align} We can explicitly compute $f_1(s)$ for $0\lt s\lt1$ using $(1)$-$(4)$ of this answer: \begin{align} f_1(s) &=\int_0^\infty\{t\}t^{-s-1}\,\mathrm{d}t\\ &=\lim_{L\to\infty}\int_0^L\{t\}t^{-s-1}\,\mathrm{d}t\\ &=\lim_{L\to\infty}-\frac1s\int_0^L\{t\}\,\mathrm{d}t^{-s}\\ &=\lim_{L\to\infty}\left(\frac1s\int_0^Lt^{-s}\,\mathrm{d}t-\frac1s\sum_{k=1}^Lk^{-s}\right)\\ &=\lim_{L\to\infty}\left(\frac1s\frac{L^{1-s}}{1-s}-\frac1s\sum_{k=1}^Lk^{-s}\right)\\ &=-\frac{\zeta(s)}s\tag{2} \end{align} Multiply $(1)$ by $\frac{\Gamma(s+1)}{n!}$ and rearrange to get $$\frac{\Gamma(s+1)}{n!}f_n(s)-\frac{\Gamma(s)}{(n-1)!}f_{n-1}(s-1)=-\frac{\Gamma(s)\zeta(s)}{n!}\tag{3}$$ Summing $(3)$, we get $$\frac{\Gamma(s+1)}{n!}f_n(s)-\frac{\Gamma(s-n+2)}{1!}f_1(s-n+1)=-\sum_{k=0}^{n-2}\frac{\Gamma(s-k)\zeta(s-k)}{(n-k)!}\tag{4}$$ Applying $(2)$ to $(4)$, we get $$\frac{\Gamma(s+1)}{n!}f_n(s)=-\sum_{k=0}^{n-1}\frac{\Gamma(s-k)\zeta(s-k)}{(n-k)!}\tag{5}$$ Simplifying $(5)$ gives $$f_n(s)=-\sum_{k=0}^{n-1}\frac{\binom{n}{k}}{\binom{s}{k}}\frac{\zeta(s-k)}{s-k}\tag{6}$$

Extension by Analytic Continuation

The recursion $(1)$ computes $f_n(s)$ from $f_1(s-n+1)$ and the integral for $f_1(s-n+1)$ in $(2)$ converges when $n-1\lt\mathrm{Re}(s)\lt n$. The integral for $f_n(s)$ in $(1)$ converges for $0\lt\mathrm{Re}(s)\lt n$, is analytic, and agrees with $(6)$ for $n-1\lt\mathrm{Re}(s)\lt n$. By analytic continuation, $(6)$ holds for $0\lt\mathrm{Re}(s)\lt n$.

• Many thanks for your answer. You and users as user1952009 are the heart of this site. I will study all details. – user243301 Nov 8 '16 at 19:30
• I don't understand your 1st step – reuns Nov 8 '16 at 19:56
• and (when $n \in \mathbb{N}^*$) the result is a finite sum, as I wrote in my answer – reuns Nov 8 '16 at 19:57
• The first step is substituting $\frac1t=k+u$ where $k$ is a non-negative integer and $u\in[0,1)$ – robjohn Nov 8 '16 at 20:33
• @robjohn Did you read my answer ? – reuns Nov 8 '16 at 21:23

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$\ds{\int_{0}^{\infty}\braces{1 \over t}^{kn}t^{s - 1}\,\dd t:\ ?.\qquad k,n \in \mathbb{N}_{\ \geq\ 0}\,,\quad\Re\pars{s} \in \pars{0,1}}$.

\begin{align} \int_{0}^{\infty}\braces{1 \over t}^{kn}t^{s - 1}\,\dd t & \,\,\,\stackrel{t\ \mapsto\ 1/t}{=}\,\,\, \int_{\infty}^{0}\braces{t}^{kn}\,\pars{1 \over t}^{s - 1} \pars{-\,{\dd t \over t^{2}}} = \int_{0}^{\infty}\braces{t}^{kn}\, t^{-s - 1}\,\dd t \\[5mm] & = \int_{0}^{\infty}\pars{t - \left\lfloor t\right\rfloor}^{kn}\, t^{-s - 1}\,\dd t \end{align}

With $\ds{N \in \mathbb{N}_{\ >\ 0}}$: \begin{align} &\int_{0}^{N}\pars{t - \left\lfloor t\right\rfloor}^{kn}\, t^{-s - 1}\,\dd t \\[5mm] = &\ \int_{0}^{1}\pars{t - 0}^{kn}t^{-s - 1}\,\dd t + \int_{1}^{2}\pars{t - 1}^{kn}\, t^{-s - 1}\,\dd t + \cdots + \int_{N - 1}^{N}\pars{t - N + 1}^{kn}\, t^{-s - 1}\,\dd t \\[5mm] = &\ \int_{0}^{1}t^{kn}t^{-s - 1}\,\dd t + \int_{0}^{1}t^{kn}\, \pars{t + 1}^{-s - 1}\,\dd t + \cdots + \int_{0}^{1}t^{kn}\, \pars{t + N - 1}^{-s - 1}\,\dd t \\[5mm] = &\ \int_{0}^{1}t^{kn}\sum_{k = 0}^{N - 1}{1 \over \pars{k + t}^{s + 1}}\,\dd t \end{align}
When $\ds{N \to \infty}$, it becomes: \begin{align} \int_{0}^{\infty}\braces{1 \over t}^{kn}t^{s - 1}\,\dd t & = \int_{0}^{1}t^{kn}\,\zeta\pars{s + 1,t}\,\dd t\label{1}\tag{1} \end{align} where $\ds{\zeta\pars{a,b}}$ is the Hurwitz Zeta function.

So far, there are many representations of the Hurwitz Zeta Function which can be 'inserted' in the above result \eqref{1} to yield an expression as a series but a 'closed form' doesn't seem feasible at this time.

In particular, CAS doesn't help at all, either.

• Many thanks also to you for your great answers. Your aproach in very nice. – user243301 Nov 9 '16 at 9:23
• What do you mean with a closed-form ? – reuns Nov 9 '16 at 16:25
• @user243301 Thanks. You're welcome. – Felix Marin Nov 9 '16 at 19:55
• @user1952009 As usual, it means something which can be expressed in terms of 'elementary functions' $\left(~\exp, \ln, \ldots~\right)$ and/or 'special' funtions like $\Gamma, \zeta, \mathrm{Bessel}\, \ldots$ or something else. – Felix Marin Nov 9 '16 at 19:58
• @FelixMarin as usual :D did you read other answers ? You end up with an integral of $\zeta(s,a)$ whereas the answer (when $k \in \mathbb{N}^*$) is just a finite sum of $\zeta(s-m)$ – reuns Nov 9 '16 at 20:06