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

Question

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 [2] for some integrals that likes to the integrand in the Question (see [2] 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.

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We know the formula (11) here in MathWorld with reference to [1], 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:

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

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

Answer 1

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. $$\boxed{F_k(s) = \frac{k}{s-1} F_{k-1}(s-1)-\zeta(s)= -\sum_{m=0}^{k-1} \zeta(s-m)\prod_{l=0}^{m-1} \frac{k-l}{s-l-1}}$$

Answer 2

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 $$ f_n(s)=-\sum_{k=0}^{n-1}\frac{\binom{n}{k}}{\binom{s}{k}}\frac{\zeta(s-k)}{s-k}\tag{6} $$

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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 $0\lt\mathrm{Re}(s)\lt n$.

Answer 3

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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}

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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}

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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.