# [SOLVED]Powers of polylogarithms

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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.

Last edited:

#### DreamWeaver

##### Well-known member
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

$$\displaystyle (06) \quad \int_0^z \frac{ \text{Li}_1(x) \text{Li}_6(x) }{x}\,dx=$$

$$\displaystyle \text{Li}_2(z) \text{Li}_6(z)- \text{Li}_3(z) \text{Li}_5(z) + \frac{1}{2} \text{Li}_4(z)^2$$

Last edited:

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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
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.

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
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
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.

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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
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
Excellent work, Zaid!

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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
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
Thanks for the link, Zaid!

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

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Thanks for the link, Zaid!

I read that paper years ago, and have been trying to locate it again more recently.
Yup , very interesting . He treats a generalization of my results. I am glad I found it .