Here is my source of inspiration for this question.

I suggest to evaluate the following

new one.In:=∫10⋯∫10{1x1x2⋯xn}2dx1dx2⋯dxn

where {x}=x−⌊x⌋ denotes thefractional partof x throughout.(a) Could you prove the result below?

I2=∫10∫10{1xy}2dxdy=1−γ+γ22−π224+ln(2π)−ln2(2π)2.

where γ denotes theEuler-Mascheroniconstant.(b) Could you find a general formula for the multiple fractional integral In?

Please, this is a challenge problem.

Thanks.

**Answer**

For any α>0 and n∈Z+. Let In(α) be the n-dimensional integral

In(α)=∫[0,1]n{1/n∏i=1xi}αn∏i=1dxi

It is clear all these In(α)∈(0,1). As a result, the corresponding generating function

I(α,z)=∞∑k=0Ik+1(α)zk

converges absolutely for any |z|<1.

**We will first assume z is small and positive.**

For any given n, substitute xi by e−ti for i=1,…,n. Let t=n∑i=1ti and x=et. We have

In(α)=1(n−1)!∫∞0{et}αe−ttn−1dt

and hence

I(α,z)=∫∞0{et}αe(z−1)tdt=∫∞1{x}αxz−2dx=∞∑k=1∫10xα(x+k)2−zdx

In the last expression, if we replace the denominator (x+k)2−z by its integral representation

1(x+k)2−z=1Γ(2−z)∫∞0s1−ze−(x+k)sds

We get

I(α,z)=1Γ(2−z)∞∑k=1∫10xα[∫∞0s1−ze−(x+k)sds]dx=1Γ(2−z)∫10xα[∫∞0s1−ze−xses−1ds]dx=1Γ(2−z)∫∞0s1−zes−1[∫10xαe−xsdx]ds=1Γ(2−z)∫∞0s1−zes−1[1sα+1∫s0xαe−xdx]ds

Up to this step, we are using the assumption z is small and positive. A consequence of this assumption is all the terms involved in above steps are non-negative numbers.

The replacement of the denominator by its integral representation, the switching order of summation and integrations are automatically valid.

To proceed further, we need to split what's inside the square bracket in last expression of (∗1). If we do that, we will notice there are terms that are no longer non-negative. Furthermore, if z remains to be small and positive, some of the terms simply diverge.

To bypass this obstacle, we will use the fact as long as ℜz<1, the last expression

of (∗1) defines an analytic function in z. Instead of sticking with z is small and positive,

**we will switch our assumption to z is real and sufficiently negative**

After we work out what the last expression of (∗1) really are, we will analytic continue

the result back to small and positive z.

For integer α>0, we have

I(α,z)=1Γ(2−z)∫∞0s−(α+z)es−1[−e−xα∑k=0α!(α−k)!xα−k]s0ds=1Γ(2−z)∫∞0s−(α+z)es−1[α!−e−sα∑k=0α!(α−k)!sα−k]ds=α!Γ(2−z)∫∞0[s−(α+z)−1es−1α−1∑k=0s−(k+z)(α−k)!]e−sds

Since ∫∞0sβ−1es−1e−sds=Γ(β)(ζ(β)−1), we find

I(α,z)=α!Γ(2−z)[Γ(1−α−z)−α−1∑k=0Γ(1−k−z)ζ(1−k−z)−1(α−k)!]

In particular, when α=2, this leads to

I(2,z)=2!Γ(2−z)[Γ(−1−z)−12Γ(1−z)(ζ(1−z)−1)−Γ(−z)(ζ(−z)−1)]

Since Γ(1+w)=wΓ(w), this can be simplified as

I(2,z)=2z(1−z2)−11−z(ζ(1−z)−1)+2z(1−z)(ζ(−z)−1)=−1z+1−11−zζ(1−z)+2z(1−z)ζ(−z)

We know ζ(w) is analytic over the whole complex plane except at w=1.

It has a simple pole with residue 1 there. Using this info, we know RHS(∗2) may have poles at z=0 and ±1.

- We don't care what happens at z=1.
- Near z=−1, RHS(∗2) ∼−1z+1+O(1)+(−1+O(z+1))(1−z−1+O(1))=O(1).
- Near z=0, RHS(∗2) ∼O(1)−(1+O(z))(1−z+O(1))+1z(2ζ(0)+O(z))=O(1) again.

Combine these, we know RHS(∗2) is analytic at z=−1 and 0. There is nothing

from extending its validity to the unit disk |z|<1 where I(α,z) coincides

with its definition as a generating function. As a result,

\bbox[4pt,border:1px solid blue;]{

I_n(2) = (n-1)!\frac{d^{n-1}}{dz^{n-1}}\left[-\frac{1}{z+1} - \frac{1}{1-z}\zeta(1-z) + \frac{2}{z(1-z)}\zeta(-z) \right]_{z=0}

}\tag{*3}

If we throw RHS(*2) to WA and ask for its Taylor expansion, WA returns

I(2,z) =

(-1-\gamma + \log(2\pi))

+ z \left(1 - \gamma+ \frac{\gamma^2}{2}-\frac{\pi^2}{24} + \log(2\pi) - \frac{\log^2(2\pi)}{2}\right)

+ O(z^2)

This implies

\begin{align}

I_1(2) &= -1-\gamma + \log(2\pi)\\

I_2(2) &= 1 - \gamma+ \frac{\gamma^2}{2}-\frac{\pi^2}{24} + \log(2\pi) - \frac{\log^2(2\pi)}{2}

\end{align}

as expected.

**Attribution***Source : Link , Question Author : Olivier Oloa , Answer Author : achille hui*