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[SOLVED] Powers of polylogarithms

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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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DreamWeaver

Well-known member
Sep 16, 2013
337
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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ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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.
 

DreamWeaver

Well-known member
Sep 16, 2013
337
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



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... ;) )
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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.
 

DreamWeaver

Well-known member
Sep 16, 2013
337
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)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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 \(\displaystyle q=2\)

\(\displaystyle \mathscr{H}(2,2)=\int^1_0 \frac{ \mathrm{Li}_2(x) ^2 \, }{x}\, dx=2\zeta(3)\zeta(2)-3\zeta(5)\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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

\(\displaystyle \int^1_0 \frac{\mathrm{Li}_{1}(x)\mathrm{Li}_{2}(x)\mathrm{Li}_{3}(x)}{x}\, dx \)​
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Excellent work, Zaid! (Yes)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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

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

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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.
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Thanks for the link, Zaid! (Hug)

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

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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 .