MHB What is the product of polylogarithms in a series?

[alyafey22]

As a continuation on the work done on this http://mathhelpboards.com/calculus-10/generalization-triple-higher-power-polylog-integrals-8044.html. In this thread we are looking at the product

$$\int^1_ 0 \mathrm{Li}_p(x) \mathrm{Li}_q(x) \mathrm{Li}_r(x) \,dx$$​

This is not a tutorial as I have no idea how to solve for a general formula. I'll keep posting my attempts on it. As always all other attempts and suggestions are welcomed.

 

[alyafey22]

Start by the easiest case

$$\int^1_0 \mathrm{Li}_p(x)\,dx $$

Integrate by parts

$$\int^1_0 \mathrm{Li}_p(x)\,dx = \zeta(p)-\int^1_0 \mathrm{Li}_{p-1}(x) \,dx $$

Continuing this way we get the following

$$\int^1_0 \mathrm{Li}_p(x)\,dx = \zeta(p)-\cdots +(-1)^k\zeta(p-k)-(-1)^k\int^1_0 \mathrm{Li}_{p-k-1}(x)\,dx $$

Now let $p-k = 2$$$\int^1_0 \mathrm{Li}_p(x)\,dx = \zeta(p)-\cdots (-1)^p \zeta(2)-(-1)^p\int^1_0 \mathrm{Li}_{1}(x) \, dx $$$$\int^1_0 \mathrm{Li}_p(x)\,dx = \sum_{k=2}^p (-1)^{p+k}\zeta(k)+(-1)^p\int^1_0 \log(1-x)\,dx $$

$$\int^1_0 \mathrm{Li}_p(x)\,dx = \sum_{k=2}^p (-1)^{k+p}\zeta(k)-(-1)^p $$

 

[alyafey22]

Now let us look at the integral

$$\int^1_0 x^{n-1}\, \mathrm{Li}_p(x)\,dx$$

This can be written as

$$\int^1_0 x^{n-1}\, \mathrm{Li}_p(x)\,dx = \sum _{k=0}^\infty \frac{1}{k^p} \int^1_0x^{k+n}\,dx = \sum _{k=1}^\infty \frac{1}{k^p(n+k)}\,dx $$

We already proved that

$$\sum _{k=1}^\infty \frac{1}{k^p(n+k)}\,dx=\sum_{k=1}^{p-1}(-1)^{n-1}\frac{\zeta(p-k+1)}{n^k}+(-1)^{p-1}\frac{H_n}{k^p}$$

Hence we have

$$\int^1_0 x^{n-1}\, \mathrm{Li}_p(x)\,dx=\sum_{k=1}^{p-1}(-1)^{k-1}\frac{\zeta(p-k+1)}{n^k}+(-1)^{p-1}\frac{H_n}{n^p}$$

Setting $n=1$ we get our integral $$\int^1_0 \mathrm{Li}_p(x)\,dx=\sum_{k=1}^{p-1}(-1)^{k-1}{\zeta(p-k+1)}+(-1)^{p-1}$$

The sum can be converted to

$$\int^1_0 \mathrm{Li}_p(x)\,dx=\sum_{k=2}^{p}(-1)^{k+p}{\zeta(k)}+(-1)^{p-1}$$

 

[alyafey22]

We proved that

$$\int^1_0 x^{n}\, \mathrm{Li}_p(x)\,dx=\sum_{k=1}^{p-1}(-1)^{k-1}\frac{\zeta(p-k+1)}{(n+1)^k}+(-1)^{p-1}\frac{H_n}{(n+1)^p}$$

Now divide by $n^q$$$\int^1_0 \frac{x^n}{n^q}\, \mathrm{Li}_p(x)\,dx=\sum_{k=1}^{p-1}(-1)^{k-1}\frac{\zeta(p-k+1)}{n^q(n+1)^k}+(-1)^{p-1}\frac{H_n}{n^q(n+1)^p}$$

Sum with respect to $n$

$$\int^1_0 \mathrm{Li}_q(x) \mathrm{Li}_p(x)\,dx= \sum_{n=1}^\infty\sum_{k=1}^{p-1}(-1)^{k-1}\frac{\zeta(p-k+1)}{n^q(n+1)^k}+(-1)^{p-1}\sum_{n=1}^\infty\frac{H_n}{n^q(n+1)^p}$$

Next we look at the sum

$$\sum_{n=1}^\infty\frac{1}{n^q(n+1)^k}$$

 

[alyafey22]

Now we look at the series

$$\sum_{n=1}^\infty\frac{1}{n^q(n+1)^k}$$

First notice that

$$\int^1_0 x^n \,dx= \frac{1}{n+1} $$

By induction we get

$$\frac{(-1)^{k-1}}{(k-1)!}\int^1_0 x^n \log^{k-1}(x)\,dx= \frac{1}{(n+1)^k} $$

Hence we get the series

$$\sum_{n=1}^\infty\frac{1}{n^q(n+1)^k} = \frac{(-1)^{k-1}}{(k-1)!}\int^1_0\sum_{n=1}^\infty\frac{1}{n^q} x^n \log^{k-1}(x)\,dx $$

$$\sum_{n=1}^\infty\frac{1}{n^q(n+1)^k} = \frac{(-1)^{k-1}}{(k-1)!}\int^1_0\mathrm{Li}_q(x) \log^{k-1}(x)\,dx $$

Now let us look at the following

$$\sum_{n=1}H_n^{(q)} x^n = \frac{\mathrm{Li}_q(x)}{1-x}$$

Or we have

$$\sum_{n=1}H_n^{(q)} (1-x)x^n =\mathrm{Li}_q(x)$$

$$\sum_{n=1}H_n^{(q)} \int^1_0 x^n(1-x)\log^{k-1}(x)\,dx =\int^1_0\mathrm{Li}_q(x)\log^{k-1}(x) \,dx$$

Eventually we get that

$$\frac{(-1)^{k-1}}{(k-1)!}\int^1_0 x^n(1-x)\log^{k-1}(x)\,dx = \frac{1}{(n+1)^k}-\frac{1}{(n+2)^k} $$

Finally we get

$$\sum_{n=1}^\infty\frac{1}{n^q(n+1)^k} = \sum_{n=1}^\infty\frac{H_n^{(q)}}{(n+1)^k}-\frac{H_n^{(q)}}{(n+2)^k} $$