TITLE: Evaluating $\int_{0}^{1}\cdots\int_{0}^{1}\bigl\{\frac{1}{x_{1}\cdots x_{n}}\bigr\}^{2}\:\mathrm{d}x_{1}\cdots\mathrm{d}x_{n}$
QUESTION [65 upvotes]: Here is my source of inspiration for this question.
I suggest to evaluate the following new one.

$$
I_{n}:= \int_0^1 \! \cdots \! \int_0^1 
\left\{\frac{1}{x_1x_2  \cdots x_n}\right\}^{2} \:\mathrm{d}x_1\,\mathrm{d}\,x_2 \cdots \mathrm{d}x_n 
$$
where $\left\{x\right\}=x-\lfloor x\rfloor$ denotes the fractional part of $x$ throughout.

(a) Could you prove the result below?

$$
I_2= \int_0^1\!\!\int_0^1 
\left\{\frac{1}{x y}\right\}^2 \:\mathrm{d}x \,\mathrm{d}y = 1-\gamma+\dfrac{\gamma^2}{2}-\dfrac{\pi^2}{24}+\ln(2\pi)-\dfrac{\ln^2(2\pi)}{2}.
$$
where $\gamma$ denotes the Euler-Mascheroni constant.

(b) Could you find a general formula for the multiple fractional integral $I_n$?
Please, this is a challenge problem.
Thanks.

REPLY [28 votes]: Here is an approach.

Theorem 1. Let $s$ be a complex number such that $-1<\Re{s}<1$, $s\neq 0$.
Then 
  $$
\int_{0}^{1}  x^{s}\left\{1/x\right\}^{2}\:\mathrm{d}x = -\frac{2\zeta(s)}{s(1+s)}-\frac{\zeta(1+s)}{1+s}-\frac{1}{1-s} \tag1
$$ where $\left\{x\right\}=x-\lfloor x\rfloor$ denotes the fractional part of $x$ and where $\zeta$ denotes the Riemann zeta function.

Proof. Let us assume first that $\Re{s}>1$. We have
\begin{align}
\displaystyle \int_{0}^{1} x^{s}\left\{1/x\right\}^{2}\:\mathrm{d}x  & = \sum_{k=1}^{\infty}
\displaystyle \int_{1/(k+1)}^{1/k} x^{s}\left\{1/x\right\}^{2} \:\mathrm{d}x \\
& = \sum_{k=1}^{\infty} \displaystyle \int_{k}^{k+1} (x-k)^{2} \frac{\mathrm{d}x}{x^{s+2}}  \\
& = \sum_{k=1}^{\infty} \displaystyle \int_{0}^{1}\frac{x^{2}}{(x+k)^{s+2}}\mathrm{d}x \\
& = \sum_{k=1}^{\infty} \displaystyle \int_{0}^{1}\left(\frac{1}{(x+k)^{s}}-\frac{2k}{(x+k)^{s+1}}+ \frac{k^{2}}{(x+k)^{s+2}}\right)\mathrm{d}x \\
& = \sum_{k=1}^{\infty} \displaystyle  \left. \left( -\frac{1}{(s-1)} \frac{1}{(x+k)^{s-1}} +\frac{2}{s}\frac{k}{(x+k)^{s}} 
-\frac{k^{2}}{(s+1)(x+k)^{s+1}}   \right) \right|_{0}^{1}\\
 & = -\frac{1}{1-s} -\frac{2\zeta(s)}{s(1+s)}-\frac{\zeta(1+s)}{1+s}. 
\end{align}
Since the integral on the left hand side is clearly analytic for $\Re s >-1$ and since all singularities on the right hand side  (including $s=0$) are removable, we may extend the preceding identity by analytic continuation to obtain $(1)$. 
$\,\Box$
A first application of $(1)$.

Example. $$
I_{2}= \int_{0}^{1}\!\!\int_{0}^{1} 
\left\{\frac{1}{x y}\right\}^{2} \:\mathrm{d}x \mathrm{d}y = 1-\gamma+\dfrac{\gamma^2}{2}-\dfrac{\pi^2}{24}+\ln(2\pi)-\dfrac{\ln^2(2\pi)}{2} \tag2
$$
  where $\left\{x\right\}=x-\lfloor x\rfloor$  denotes the fractional part of $x$ and $\gamma$ is the Euler constant.

Proof. We use a change of variable in the given integral:
$$
\begin{align}
\displaystyle u  & = xy \\
v  & = x \\
\end{align}
$$ then the absolute value of the Jacobian determinant and the new domain of integration are $$ \left| \frac{D(u,v)}{D(x,y)}\right|=v, \quad 0<u<v<1, $$ leading to
$$
\begin{align}
I_{2} & = \int_{0}^{1}\!\!\int_{0}^{1} 
\left\{\frac{1}{x y}\right\}^{2} \:\mathrm{d}x \mathrm{d}y 
\\\\ & = \int_{0}^{1} \! \int_{u}^{1} 
\left\{1/u\right\}^2\frac{\mathrm{d}v}{v} \mathrm{d}u
\\\\ & =-\int_{0}^{1}  
\left\{1/u\right\}^{2} \ln u\:  \mathrm{d}u 
\\\\ & =-\left.\frac{d}{ds}\left(\int_{0}^{1}  
u^s \left\{1/u\right\}^{2} \mathrm{d}u\right) \right|_{s=0}
\\\\ & =-\left.\frac{d}{ds}\left(-\frac{1}{1-s} -\frac{2\zeta(s)}{s(1+s)}-\frac{\zeta(1+s)}{1+s}\right) \right|_{s=0}
\\\\ & = 1-\gamma+\dfrac{\gamma^2}{2}-\dfrac{\pi^2}{24}+\ln(2\pi)-\dfrac{\ln^2(2\pi)}{2}
\end{align}
$$
where, for $s$ near $0$, we have used
$$
\begin{align}
& \zeta(1+s) =\frac1s+\gamma-\gamma_1 s +\mathcal{O}(s^2),
\\& \zeta(s) =-\frac12-\dfrac{\ln(2\pi)}{2} s +\left(\dfrac{\gamma^2}{4}-\dfrac{\pi^2}{48}+\ln(2\pi)-\dfrac{\ln^2(2\pi)}{4}+\dfrac{\gamma_1}{2}\right)s^2+\mathcal{O}(s^3).
\end{align}
$$ $\Box$
Now one may observe that $(1)$ has the following consequence.

Theorem 2. Let $n=1,2,3, \cdots.$
Then 
  $$
\frac{(-1)^{n-1}}{(n-1)!} \int_{0}^{1}  
\left\{1/x\right\}^{2} \ln ^{n-1} x\:  \mathrm{d}x = (-1)^{n}+1 -\sum_{k=0}^{n-1}\frac{\gamma_{k}}{k!}+2\sum_{k=0}^{n} \frac{(-1)^k}{k!}\zeta^{(k)}(0) \tag3
$$
  where $\left\{x\right\}=x-\lfloor x\rfloor$  denotes the fractional part of $x$ and where $\gamma_{k}$ denote the Laurent-Stieltjes constants which may be defined by 
  $$
\gamma_{k} = \displaystyle \lim_{N \rightarrow \infty}\left(\sum_{m=1}^{N}\frac{\ln^{k}m}{m}-\frac{\ln^{k+1}N}{k+1} \right).
$$

Proof. Let $s$ be a complex number such that $0<\Re{s}<1$. Using Theorem 1, one may just write 
$$
\int_{0}^{1} x^{s}\left\{1/x\right\}^2 \:\mathrm{d}x 
= \sum_{n=0}^{\infty}\left(\int_{0}^{1} \left\{1/x\right\}^2 \ln^{n}x \:\mathrm{d}x \right)  \frac{s^{n}}{n!},
$$ and one may take into account the standard Laurent series 
$$\begin{align} 
\displaystyle \zeta(1+s) =\frac{1}{s}+\sum_{k=0}^{\infty} (-1)^{k} \frac{\gamma_{k}}{k!} s^{k}
\end{align}$$ to obtain $(3)$ by identifying the Taylor coefficients on both sides of $(1)$. 
$\,\Box$
From $(3)$ we deduce the following result.

Theorem 3. Let $n=1,2,3, \cdots.$ Set
  $$
I_{n}:= \int_{0}^{1} \! \cdots \! \int_{0}^{1} 
\left\{\frac{1}{x_{1}x_{2}  \cdots x_{n}}\right\}^{2} \:\mathrm{d}x_{1}\mathrm{d}x_{2} \cdots \mathrm{d}x_{n}. 
$$
  Then 
  $$
I_{n}=   (-1)^{n}+1 -\sum_{k=0}^{n-1}\frac{\gamma_{k}}{k!}+2\sum_{k=0}^{n} \frac{(-1)^k}{k!}\zeta^{(k)}(0) \tag4
$$
  where $\left\{x\right\}=x-\lfloor x\rfloor$  denotes the fractional part of $x$ and where $\gamma_{k}$ denote the Laurent-Stieltjes constants defined by 
  $$
\gamma_{k} = \displaystyle \lim_{N \rightarrow \infty}\left(\sum_{m=1}^{N}\frac{\ln^{k}m}{m}-\frac{\ln^{k+1}N}{k+1} \right).
$$

Proof. We reduce $ I_{n} $ to a one dimensional integral
by making the change of variables 
$$
\begin{align}
\displaystyle u_{1}  & = x_{1} \\
u_{2}  & = x_{1}x_{2} \\
 & \vdots  \\
u \, & = x_{1}x_{2} \cdots x_{n} .
\end{align}
$$
The initial domain of integration is transformed into $ \displaystyle   0<u< \cdots <u_{2}< u_{1}<1 $ 
with a Jacobian equal to $\displaystyle u_{1}  \cdots u_{n-1}$. Hence
$$
\begin{align}   
I_{n} = \displaystyle \int_{0}^{1} \! \int_{u}^{1} \cdots \! \int_{u_{2}}^{1} 
\left\{1/u\right\}^{2} \: \frac{\mathrm{d}u_{1} \cdots \mathrm{d}u_{n-1}}{u_{1}  \cdots u_{n-1}}  \mathrm{d}u 
\end{align}
$$ then integrating out successively $\displaystyle u_{1}, u_{2}, \dots ,u_{n-1}$ gives 
$$  
I_{n} = \frac{(-1)^{n-1}}{(n-1)!}\displaystyle \int_{0}^{1}  
\left\{1/u\right\}^{2} \ln ^{n-1} u\:  \mathrm{d}u 
$$ and we are directly led to $(4)$ by $(3)$.
$\Box$
Here is a general formula answering an open problem from Ovidiu Furdui (2006) on this class of multiple integrals involving the fractional part.

Theorem  (O. Oloa, general case, 18th Aug 2014). Let $n=1,2,3...$ and let $p=1,2,3....$ 
Set
  $$
\begin{align}  
I_{n,p} :=\int_{0}^{1} \! \cdots \! \int_{0}^{1} 
\left\{\frac{1}{x_{1}  \cdots x_{n}}\right\}^{p} \:\mathrm{d}x_{1} \cdots \mathrm{d}x_{n}
\end{align}
$$ where $\left\{x\right\}=x-\lfloor x\rfloor$  denotes the fractional part of $x$. 
Then
  $$
\begin{align}
\displaystyle I_{n,1} :=\int_{0}^{1} \! \cdots \! \int_{0}^{1} 
\left\{\frac{1}{x_{1}  \cdots x_{n}}\right\} \:\mathrm{d}x_{1} \cdots \mathrm{d}x_{n} = 1-\sum_{k=0}^{n-1}\frac{\gamma_{k}}{k!},
\end{align}
$$ and, for $p=2,3,4,...,$ 
  $$
\begin{align} 
I_{n,p} = \frac{(-1)^{n}}{(p-1)^{n}}- 1-\sum_{k=0}^{n-1}\frac{\gamma_{k}}{k!}
- \frac{p!}{n!}\sum_{\ell=0}^{p-2}\frac{(-1)^{\ell}}{(p-\ell-1)!}\left.\left(\! \frac{\zeta(s-\ell)}{(1+s)(1-s)_{\ell}}\! \right)^{(n)}\right|_{s=0}
\end{align}
$$ where $\gamma_{k}$ are the Laurent Stieltjes constants and where $\zeta$ denotes the Riemann zeta function.