Evaluating ∫10⋯∫10{1×1⋯xn}2dx1⋯dxn\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}

Here is my source of inspiration for this question.

I suggest to evaluate the following new one.

In:=1010{1x1x2xn}2dx1dx2dxn
where {x}=xx denotes the fractional part of x throughout.

(a) Could you prove the result below?

I2=1010{1xy}2dxdy=1γ+γ22π224+ln(2π)ln2(2π)2.
where γ denotes the Euler-Mascheroni constant.

(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 nZ+. Let In(α) be the n-dimensional integral

In(α)=[0,1]n{1/ni=1xi}αni=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 eti for i=1,,n. Let t=ni=1ti and x=et. We have

In(α)=1(n1)!0{et}αettn1dt

and hence
I(α,z)=0{et}αe(z1)tdt=1{x}αxz2dx=k=110xα(x+k)2zdx
In the last expression, if we replace the denominator (x+k)2z by its integral representation

1(x+k)2z=1Γ(2z)0s1ze(x+k)sds

We get

I(α,z)=1Γ(2z)k=110xα[0s1ze(x+k)sds]dx=1Γ(2z)10xα[0s1zexses1ds]dx=1Γ(2z)0s1zes1[10xαexsdx]ds=1Γ(2z)0s1zes1[1sα+1s0xαexdx]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 z<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(2z)0s(α+z)es1[exαk=0α!(αk)!xαk]s0ds=1Γ(2z)0s(α+z)es1[α!esαk=0α!(αk)!sαk]ds=α!Γ(2z)0[s(α+z)1es1α1k=0s(k+z)(αk)!]esds
Since 0sβ1es1esds=Γ(β)(ζ(β)1), we find
I(α,z)=α!Γ(2z)[Γ(1αz)α1k=0Γ(1kz)ζ(1kz)1(αk)!]

In particular, when α=2, this leads to
I(2,z)=2!Γ(2z)[Γ(1z)12Γ(1z)(ζ(1z)1)Γ(z)(ζ(z)1)]
Since Γ(1+w)=wΓ(w), this can be simplified as
I(2,z)=2z(1z2)11z(ζ(1z)1)+2z(1z)(ζ(z)1)=1z+111zζ(1z)+2z(1z)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))(1z1+O(1))=O(1).
  • Near z=0, RHS(2) O(1)(1+O(z))(1z+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

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