What is the general form of Powers of Polylogarithms?

  • MHB
  • Thread starter alyafey22
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In summary, the conversation discussed various integrals involving the polylogarithm function and their generalizations. It was mentioned that a general formula can be found for the integral \int^1_0 \frac{\mathrm{Li}_n(x) \mathrm{Li}_m(x)}{x}\, dx, which is related to the function \mathscr{H}(\alpha ,\beta). The conversation also touched on the G-Barnes function and the use of Euler sums to evaluate certain polylog integrals.
  • #1
alyafey22
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MHB
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We should consider the general form

\(\displaystyle L^m_n (a) = \int^1_0 x^{a-1} \mathrm{Li}_{\, n}(x)^m \, dx\)​

This is NOT a tutorial , any attempts or comments are always welcomed.
 
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  • #2
The case \(\displaystyle m=1\) and \(\displaystyle a>0 \in \mathbb{R}\) is fairly straightforward. Additionally, if we generalize your integral to\(\displaystyle \text{Li}_{p,q}(m, n; a, z)=\int_0^zx^{a-1} (\text{Li}_p(x))^m (\text{Li}_q(x))^n\,dx\)Then the special cases\(\displaystyle \text{Li}_{p,q}(1, 1; 0, z)=\int_0^z \frac{ \text{Li}_p(x) \text{Li}_q(x) }{x}\,dx\)

require nothing more than repeated integration by parts, provided \(\displaystyle p+q \in 2\mathbb{N}+1\), since\(\displaystyle \frac{d}{dx} \text{Li}_p(x)=\frac{\text{Li}_{p-1}(x)}{x}\)So, for example, we have:\(\displaystyle (01) \quad \int_0^z \frac{ \text{Li}_1(x) \text{Li}_2(x) }{x}\,dx= \frac{1}{2} \text{Li}_2(z)^2\)

\(\displaystyle (02) \quad \int_0^z \frac{ \text{Li}_2(x) \text{Li}_3(x) }{x}\,dx= \frac{1}{2} \text{Li}_3(z)^2\)

\(\displaystyle (03) \quad \int_0^z \frac{ \text{Li}_3(x) \text{Li}_4(x) }{x}\,dx= \frac{1}{2}
\text{Li}_4(z)^2\)Integrating (02) by parts also gives:\(\displaystyle (04) \quad \int_0^z \frac{ \text{Li}_1(x) \text{Li}_4(x) }{x}\,dx= \text{Li}_2(z) \text{Li}_4(z) - \frac{1}{2} \text{Li}_3(z)^2\)Similarly, an integration by parts of (03) gives\(\displaystyle (05) \quad \int_0^z \frac{ \text{Li}_2(x) \text{Li}_5(x) }{x}\,dx= \text{Li}_3(z) \text{Li}_5(z) - \frac{1}{2} \text{Li}_4(z)^2\)as well as\(\displaystyle (06) \quad \int_0^z \frac{ \text{Li}_1(x) \text{Li}_6(x) }{x}\,dx= \)

\(\displaystyle \text{Li}_2(z) \text{Li}_6(z)- \text{Li}_3(z) \text{Li}_5(z) + \frac{1}{2} \text{Li}_4(z)^2\)
 
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  • #3
Hey DW , thanks for your input (which I always wait for). I think the generalization you put is interesting to look at.

I was playing a little bit with these integrals and found interesting to look at the infinite sum

\(\displaystyle \mathscr{C}(\alpha , k) =\sum_{n\geq 1}\frac{1}{n^{\alpha}(n+k)}\,\,\, ; \,\,\,\,\mathscr{C}(1, k)=\frac{H_k}{k} \)

We can find a general formula to find the sum

\begin{align}
\mathscr{C}(\alpha , k) &=\sum_{n\geq 1}\frac{1}{k\, n^{\alpha-1}}\left( \frac{1}{n}-\frac{1}{n+k}\right)\\ &= \frac{1}{k}\zeta(\alpha)-\frac{1}{k}\mathscr{C}(\alpha-1 , k)\\ &= \frac{1}{k}\zeta(\alpha)-\frac{1}{k^2}\zeta(\alpha-1)+\frac{1}{k^2}\mathscr{C}(\alpha-2 , k)\\ &= \sum_{n=1}^{\alpha-1}(-1)^{n-1}\frac{\zeta(\alpha-n+1)}{k^n}+(-1)^{\alpha-1}\frac{H_k}{k^\alpha}
\end{align}

Hence we have the general formula

\(\displaystyle \mathscr{C}(\alpha , k) = \sum_{n=1}^{\alpha-1}(-1)^{n-1}\frac{\zeta(\alpha-n+1)}{k^n}+(-1)^{\alpha-1}\frac{H_k}{k^\alpha}\)

Dividing by \(\displaystyle \frac{1}{k^{\beta}}\) and summing w.r.t to $k$ we have

\(\displaystyle \sum_{k\geq 1}\frac{\mathscr{C}(\alpha , k)}{k^{\beta}} = \sum_{n=1}^{\alpha-1}(-1)^{n-1}\zeta(\alpha-n+1)\zeta(\beta+n)+(-1)^{\alpha-1}\sum_{k\geq 1}\frac{H_k}{k^{\alpha+\beta}}\)

Now we use that

\(\displaystyle \sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)\)

Hence we have

\(\displaystyle \sum_{n=1}^\infty \frac{H_n}{n^{\alpha+\beta}}= \left(1+\frac{{\alpha+\beta}}{2} \right)\zeta({\alpha+\beta}+1)-\frac{1}{2}\sum_{k=1}^{{\alpha+\beta}-2}\zeta(k+1)\zeta({\alpha+\beta}-k)\)

And the generalization is the following formula

\begin{align}
\sum_{k\geq 1}\frac{\mathscr{C}(\alpha , k)}{k^{\beta}} &= \sum_{n=1}^{\alpha-1}(-1)^{n-1}\zeta(\alpha-n+1)\zeta(\beta+n) -\frac{1}{2}\sum_{n=1}^{{\alpha+\beta}-2}(-1)^{\alpha-1}\zeta(n+1)\zeta({\alpha+\beta}-n)\\ &+(-1)^{\alpha-1}\left(1+\frac{{\alpha+\beta}}{2} \right)\zeta({\alpha+\beta}+1)\end{align}

Interestingly we also have that

\(\displaystyle \sum_{k\geq 1}\frac{\mathscr{C}(\alpha , k)}{k^{\beta}}=\sum_{k\geq 1}\frac{\mathscr{C}(\beta, k)}{k^{\alpha}}\)

Generally we will use the symbol

\(\displaystyle \mathscr{H}(\alpha,\beta)=\sum_{k\geq 1}\frac{\mathscr{C}(\alpha , k)}{k^{\beta}}\)

So we have the symmetric property

\(\displaystyle \mathscr{H}(\alpha,\beta) = \mathscr{H}(\beta,\alpha)\)

\(\displaystyle \mathscr{H}(1,\beta)= \left(1+\frac{{\beta}}{2} \right)\zeta({\beta}+1)-\frac{1}{2}\sum_{k=1}^{{\beta}-2}\zeta(k+1)\zeta({\beta}-k)\)

Wasn't that fun , yes it was. I'll have to check my work because I might have some errors.
 
  • #4
ZaidAlyafey said:
Hey DW , thanks for your input (which I always wait for). I think the generalization you put is interesting to look at.

Thanks Zaid! The feeling's mutual. Very much so. :D
ZaidAlyafey said:
Wasn't that fun , yes it was. I'll have to check my work because I might have some errors.
Certainly was... ;)Incidentally, I saw a paper some years ago on arXiv that dealt with Polylog Integrals of the exact same type as I posted above, but where \(\displaystyle p+q \in 2\mathbb{N}
\)Can't seem to find it now, but the author used Euler Sums to evaluate them (nudge, nudge, wink, wink... ;) )
 
  • #5
DreamWeaver said:
Incidentally, I saw a paper some years ago on arXiv that dealt with Polylog Integrals of the exact same type as I posted above, but where \(\displaystyle p+q \in 2\mathbb{N}
\)Can't seem to find it now, but the author used Euler Sums to evaluate them (nudge, nudge, wink, wink... ;) )

Really , that would be interesting if you can find it. Generally, I can find a general formula for the integral

\(\displaystyle \int^1_0 \frac{\mathrm{Li}_n(x) \mathrm{Li}_m(x)}{x}\, dx \)

which is related to \(\displaystyle \mathscr{H}(\alpha ,\beta)\). Will post that when I have time. Currently I read about the G-Barnes function which is really interesting to play with.
 
  • #6
ZaidAlyafey said:
Really , that would be interesting if you can find it. Generally, I can find a general formula for the integral

\(\displaystyle \int^1_0 \frac{\mathrm{Li}_n(x) \mathrm{Li}_m(x)}{x}\, dx \)

which is related to \(\displaystyle \mathscr{H}(\alpha ,\beta)\). Will post that when I have time. Currently I read about the G-Barnes function which is really interesting to play with.
I'll keep trying to find that old paper, and will let you know if I have any luck.Likewise, I look forward to your general solution to

\(\displaystyle \int^1_0 \frac{\mathrm{Li}_n(x) \mathrm{Li}_m(x)}{x}\, dx \)

as and when you have time. (Yes)
 
  • #7
We find series and integral representations of \(\displaystyle \mathscr{H}(\alpha,\beta)\)

\(\displaystyle \mathscr{H}(p,q)=\sum_{k\geq 1}\frac{\mathscr{C}(p,k)}{k^{q}}= \sum_{k\geq 1}\sum_{n\geq 1}\frac{1}{k^{q}n^{p}(n+k)}\)

Also we have

\(\displaystyle \mathscr{H}(p,q)= \int^1_0 \frac{\mathrm{Li}_p(x)\,\, \mathrm{Li}_q(x)\, }{x}\, dx \)

So we have the following

\begin{align}
\mathscr{H}(p,q)=\int^1_0 \frac{\mathrm{Li}_p(x)\,\, \mathrm{Li}_q(x)\, }{x}\, dx&= \sum_{n=1}^{p-1}(-1)^{n-1}\zeta(p-n+1)\zeta(q+n) -\frac{1}{2}\sum_{n=1}^{{p+q}-2}(-1)^{p-1}\zeta(n+1)\zeta({p+q}-n)\\ &+(-1)^{p-1}\left(1+\frac{{p+q}}{2} \right)\zeta({p+q}+1)\end{align}

Now for the special case \(\displaystyle p=q\) we have

\begin{align}
\mathscr{H}(q,q)=\int^1_0 \frac{ \mathrm{Li}_q(x) ^2 \, }{x}\, dx&= \sum_{n=1}^{q-1}(-1)^{n-1}\zeta(q-n+1)\zeta(q+n) -\frac{1}{2}\sum_{n=1}^{{2q}-2}(-1)^{q-1}\zeta(n+1)\zeta({2q}-n)\\ &+(-1)^{q-1}\left(1+q \right)\zeta(2q+1)\end{align}

We have for the special case \(\displaystyle q=2\)

\(\displaystyle \mathscr{H}(2,2)=\int^1_0 \frac{ \mathrm{Li}_2(x) ^2 \, }{x}\, dx=2\zeta(3)\zeta(2)-3\zeta(5)\)
 
  • #8
An interesting form that I wasn't able to solve is the following

\(\displaystyle \int^1_0 \frac{\mathrm{Li}_{q-1}(x)\mathrm{Li}_{q}(x)\mathrm{Li}_{q+1}(x)}{x}\, dx \,\,\,\, q>1\)​

For simplicity we can consider

\(\displaystyle \int^1_0 \frac{\mathrm{Li}_{1}(x)\mathrm{Li}_{2}(x)\mathrm{Li}_{3}(x)}{x}\, dx \)​
 
  • #9
Excellent work, Zaid! (Yes)
 
  • #10
Consider the following integral

\(\displaystyle \int^1_0 \frac{\mathrm{Li}_{q-1}(x)\mathrm{Li}_{q}(x)\mathrm{Li}_{q+1}(x)}{x}\, dx \)

Integrating by parts we obtain

\(\displaystyle \int^1_0 \frac{\mathrm{Li}_{q-1}(x)\mathrm{Li}_{q}(x)\mathrm{Li}_{q+1}(x)}{x}\, dx= \frac{\zeta(q+1)\zeta(q)^2}{2}-\frac{1}{2}\int^1_0 \frac{\mathrm{Li}_{q}(x)^3}{x}\, dx\)

Hence we have

\(\displaystyle \int^1_0\frac{\mathrm{Li}_{q}(x)^3}{x}\, dx = \zeta(q+1)\zeta(q)^2-2\int^1_0 \frac{\mathrm{Li}_{q-1}(x)\mathrm{Li}_{q}(x)\mathrm{Li}_{q+1}(x)}{x}\, dx\)

Integrating by parts again we obtain

\(\displaystyle \int^1_0 \frac{\mathrm{Li}_{q-1}(x)\mathrm{Li}_{q}(x)\mathrm{Li}_{q+1}(x)}{x}\, dx=\frac{\zeta(q-1)\zeta^2(q+1)}{2}-\frac{1}{2}\int^1_0 \frac{\mathrm{Li}_{q+1}(x)^2\mathrm{Li}_{q-2}(x)}{x}\, dx\)

So we have the following formula

\(\displaystyle \int^1_0\frac{\mathrm{Li}_{q}(x)^3-\mathrm{Li}_{q+1}(x)^2\mathrm{Li}_{q-2}(x)}{x}\, dx = \zeta(q+1)\zeta(q)^2-\zeta(q-1)\zeta^2(q+1)\)
 
  • #11
Hey , I found a paper by PEDRO FREITAS that exactly dealt with the form I had in mind namely the author discusses

\(\displaystyle J(m,p,q)= \int^1_0 x^m \mathrm{Li}_p(x) \mathrm{Li}_q(x) \, dx\)

The author gives a general formula for \(\displaystyle J(-1,p,q) =\mathscr{H}(p,q)\) which is just like the formula I gave proof for but he uses a different method. He discusses many special cases for \(\displaystyle m=-2\).

Here is a link for the paper.
 
  • #12
Thanks for the link, Zaid! (Hug)

I read that paper years ago, and have been trying to locate it again more recently. :D
 
  • #13
DreamWeaver said:
Thanks for the link, Zaid! (Hug)

I read that paper years ago, and have been trying to locate it again more recently. :D

Yup , very interesting . He treats a generalization of my results. I am glad I found it .
 

FAQ: What is the general form of Powers of Polylogarithms?

What are polylogarithms?

Polylogarithms are mathematical functions that are defined as the sum of terms involving powers of logarithms raised to a given exponent. They are typically denoted by Lis(x), where s is the exponent and x is the argument.

What are the properties of polylogarithms?

Polylogarithms have several important properties, including:

  • They are analytic functions, meaning they are differentiable and have a power series representation.
  • They have branch cuts along the negative real axis, and are multivalued functions.
  • They have functional equations, such as the duplication formula and the reflection formula.

How are polylogarithms used in mathematics?

Polylogarithms have a wide range of applications in mathematics, including:

  • Number theory, particularly in the study of special values of zeta functions.
  • Complex analysis, where they are used to study the behavior of analytic functions.
  • Combinatorics, where they have connections to the theory of partitions and combinatorial sums.
  • Physics, specifically in the study of quantum field theory and conformal field theory.

What are the main properties of the powers of polylogarithms?

The powers of polylogarithms have several important properties, including:

  • They satisfy a generalization of the Cauchy product formula, known as the generalized Cauchy product formula.
  • They have integral representations, which can be useful for evaluating special values.
  • They exhibit duality properties, such as the duality between the Riemann zeta function and the polylogarithm Li1(x).

How do polylogarithms relate to other special functions?

Polylogarithms have connections to several other special functions, including:

  • The Riemann zeta function, where the polylogarithm Li1(x) is related to the Riemann zeta function by a duality property.
  • The Lerch transcendent, which is a generalization of the polylogarithm.
  • The Clausen function, which is related to the polylogarithm by a special integral representation.

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