MHB A generalization of triple and higher power polylog integrals

[alyafey22]

Inspired by this http://mathhelpboards.com/calculus-10/powers-polylogarithms-7998.html we look at the generalization

$$L^m_n(p,q)=\int^1_0 \frac{\mathrm{Li}_p(x)^m\, \mathrm{Li}_q(x)^n}{x} \, dx $$​

This is NOT a tutorial. Any comments, attempts or suggestions are always welcomed.

 

[alyafey22]

We explore some properties

\begin{align}
L^1_1(p,q) = \mathscr{H}(p,q)&= \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}

Let $$n=1\,\,\, q=p-1$$

$$L^m_1(p,p-1) = \frac{\zeta(p)^{m+1}}{m+1}$$

Similarly

Let $$m=1\,\,\, p=q-1$$

$$L^1_n(q-1,q) = \frac{\zeta(q)^{n+1}}{n+1}$$

Hence we have

$$\tag{1} \, L^m_1(p,p-1)+L^1_n(q-1,q) = \frac{\zeta(p)^{m+1}}{m+1}+\frac{\zeta(q)^{n+1}}{n+1}$$

We showed that

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

Which can be rewritten as

$$\tag{2} L^2_1(q,q)-L^2_1(q+1,q-2)= \zeta(q+1)\zeta(q)^2-\zeta(q-1)\zeta^2(q+1)$$

Let $m=2\,\,\, n=1$ and $p=q$ in (1) to obtain

$$ \tag{3} L^2_1(q,q-1)+ \mathscr{H} (q-1,q) = \frac{\zeta(q)^3}{3}+\frac{\zeta(q)^2}{2}$$

By adding (3) and (2) we get

\begin{align}
L^2_1(q,q)+L^2_1(q,q-1)-L^2_1(q+1,q-2)&= \frac{\zeta(q)^3}{3}+ \zeta(q+1)\zeta(q)^2-\zeta(q-1)\zeta^2(q+1)+\frac{\zeta(q)^2}{2}\\ & \,\, \, \, - \mathscr{H} (q-1,q) \end{align}

 

[alyafey22]

$$\tag{1}\int^1_0 x^{n-1} \mathrm{Li}_k(x)\, dx = (-1)^{k-1}\frac{H_{n}}{n^k}+\sum_{m\geq 0}^{k-2}(-1)^m \frac{\zeta(k-m)}{n^{m+1}}$$

Also we define the following

$$\tag{2} S_{p^m \, , \, q}\sum_{n\geq 1} \frac{(H^{(p)})^m}{n^q}$$

Using that formula we attempt to find the solution for the following

$$L^1_2(p,1)=\int^1_0 \frac{\mathrm{Li}_p(x) \log^2(1-x)}{x} \, dx$$

First we use the generating function

$$\sum_{n\geq 1}H_n x^n = -\frac{\log(1-x)}{1-x}$$

$$\sum_{n\geq 1}\frac{H_n}{n} x^n = \mathrm{Li}_2(x)+\frac{\log^2(1-x)}{2}$$

Hence we have

$$\log^2(1-x)= 2\left(\sum_{n\geq 1}\frac{H_n}{n} x^n -\mathrm{Li}_2(x)\right)$$

Consequently we have

\begin{align}
L^1_2(p,1)&=2\int^1_0 \frac{\mathrm{Li}_p(x)}{x}\left(\sum_{n\geq 1}\frac{H_n}{n} x^n - \mathrm{Li}_2(x)\right) \, dx\\
&=2\int^1_0 \mathrm{Li}_p(x) \sum_{n\geq 1}\frac{H_n}{n} x^{n-1} \, dx-2\int^1_0 \frac{\mathrm{Li}_p(x)\mathrm{Li}_2(x)}{x} \, dx\\
&=2 \sum_{n\geq 1}\frac{H_n}{n} \int^1_0 x^{n-1}\mathrm{Li}_p(x)\, dx-2 \mathscr{H} (p,2)\\
&=2 \sum_{n\geq 1}\frac{H_n}{n}\left((-1)^{p-1}\frac{H_{n}}{n^p}+2\sum_{m\geq 0}^{p-2}(-1)^m \frac{\zeta(p-m)}{n^{m+1}} \right)-2 \mathscr{H} (p,2)\\
&= 2 (-1)^{p-1}\sum_{n\geq 1}\frac{H^2_n}{n^{p+1}}+2\sum_{m\geq 0}^{p-2}(-1)^m \zeta(p-m)\sum_{n\geq 1}\frac{H_n}{n^{m+2}}-2 \mathscr{H} (p,2)\\

&=2 (-1)^{p-1}S_{1^2,\,p+1}+2\sum_{m\geq 0}^{p-2}(-1)^m \zeta(p-m)S_{1,\,m+2}-2 \mathscr{H} (p,2)
\end{align}

For remainder we know that

\begin{align}\mathscr{H}(p,q) &= \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) \,\,\, (3) \end{align}

$$\tag{4} S_{1,\,q}=\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

And the final generalization is

$$\tag{5}\int^1_0 \frac{\mathrm{Li}_p(x) \log^2(1-x)}{x} \, dx=2 (-1)^{p-1}S_{1^2,\,p+1}+2\sum_{m\geq 0}^{p-2}(-1)^m \zeta(p-m)S_{1,\,m+2}-2 \mathscr{H} (p,2)$$

That was the most interesting formula I have ever,ever obtained. I hope it is new.

 

[alyafey22]

Well, I was trying to test the integral and it works fine

$$\tag{1} L^1_2(1,1) = -\int^1_0 \frac{\log^3(1-x)}{x}\, dx=\frac{\pi^4}{15}$$

$$\tag{2} L^1_2(2,1) = \int^1_0 \frac{\mathrm{Li}_2(x)\log^2(1-x)}{x}\, dx=2\zeta(2)\zeta(3)-\zeta(5)$$

$$\tag{3} L^1_2(3,1) = \int^1_0 \frac{\mathrm{Li}_3(x)\log^2(1-x)}{x}\, dx=\frac{97}{12} \zeta(6)-\zeta^2(3)-\zeta^3(2)$$

 

[DreamWeaver]

Congratulations, friend! It's a beauty! (Yes)(Yes)(Yes)Have you considered writing paper about this and the other Polylog integrals/series you evaluated on the other thread...? Seems a mighty worthy topic.

 

[alyafey22]

Congratulations, friend! It's a beauty! (Yes)(Yes)(Yes)Have you considered writing paper about this and the other Polylog integrals/series you evaluated on the other thread...? Seems a mighty worthy topic.

Hey Dw , thanks for the comment. I am working on a more generalized version. I hate publishing papers because the process takes a long time. I don't know whether it is worth it !

 

[alyafey22]

I derived an interesting formula

$$ \sum_{k\geq 1}(-1)^k H_k^{(p)}\, x^k = \frac{\mathrm{Li}_p(-x)}{1+x}$$

$$ \sum_{k\geq 1}(-1)^k H_k^{(p)}\, x^{k-1} = \frac{\mathrm{Li}_p(-x)}{x(1+x)}= \frac{\mathrm{Li}_p(-x)}{x}-\frac{\mathrm{Li}_p(-x)}{1+x} $$

Multiply through by $$\mathrm{Li}_q(x)$$

$$ \sum_{k\geq 1}(-1)^{k-1} H_k^{(p)}\, x^k\mathrm{Li}_q(x) = \frac{\mathrm{Li}_p(-x)\mathrm{Li}_q(x)}{1+x}$$

Now take the integral $$\int^1_0$$

$$ \sum_{k\geq 1}(-1)^{k-1} H_k^{(p)}\, \int^1_0 x^k\mathrm{Li}_q(x)\, dx = \int^1_0\frac{\mathrm{Li}_p(-x)\mathrm{Li}_q(x)}{x}\, dx -\int^1_0 \frac{\mathrm{Li}_p(-x)\mathrm{Li}_q(x)}{1+x}\, dx$$

Ok, we know that

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

We already know that

$$\mathscr{C}(q , k) = \sum_{m=1}^{q-1}(-1)^{m-1}\frac{\zeta(q-m+1)}{k^m}+(-1)^{q-1}\frac{H_k}{k^q}$$

$$ \sum_{k\geq 1}(-1)^{k-1} H_k^{(p)}\left( \sum_{m=1}^{q-1}(-1)^{m-1}\frac{\zeta(q-m+1)}{k^m}+(-1)^{q-1}\frac{H_k}{k^q} \right) = \int^1_0\frac{\mathrm{Li}_p(-x)\mathrm{Li}_q(x)}{x}\, dx -\int^1_0 \frac{\mathrm{Li}_p(-x)\mathrm{Li}_q(x)}{1+x}\, dx$$

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

\begin{align}
\int^1_0 \frac{\mathrm{Li}_p(-x)\mathrm{Li}_q(x)}{x(1+x)}\, dx &= \sum_{m=1}^{q-1}(-1)^{m-1} \zeta(q-m+1) \sum_{k\geq 1} (-1)^{k-1} \frac{H_k^{(p)}} {k^m} + (-1)^{q-1} \sum_{k\geq 1} (-1)^{k-1} \frac{H_k^{(p)} H_k}{k^q}
\end{align}

It would be interesting if we can find a similar formula for positive arguments of the polylogarithm. The evaluation of alternating Euler sums is a little more challenging than the regular sums.

 

[alyafey22]

If we consider the following

$$\sum_k H_k x^k = \frac{\mathrm{Li}_p(x)}{1-x}$$

$$\sum_k H_k x^{k-1} = \frac{\mathrm{Li}_p(x)}{x}+\frac{\mathrm{Li}_p(x)}{1-x}$$

Then multiply by $$\mathrm{Li}_q(x)$$

$$\sum_k H_k \, x^{k-1} \mathrm{Li}_q(x)= \frac{\mathrm{Li}_p(x) \mathrm{Li}_q(x)}{x}+\frac{\mathrm{Li}_p(x) \mathrm{Li}_q(x)}{1-x}$$

Then if we integrate both sides $$\int^1_0 $$ we will have an obvious problem of divergence. I think there is a way to remove the singularity from both sides.

We may consider the general integral

$$\int^x_0 \frac{\mathrm{Li}_p(t) \mathrm{Li}_q(t)}{1-t} \, dt$$

Then take the limit $$x \to 1$$ which will cancel with some terms in RHS . Not sure whether that will work. Anyways , I will continue tomorrow. (Headbang)

 

[alyafey22]

Here is a rough sketch of an evaluation

$$\sum_k H_k \, x^{k-1} \mathrm{Li}_q(x)= \frac{\mathrm{Li}_p(x) \mathrm{Li}_q(x)}{x}+\frac{\mathrm{Li}_p(x) \mathrm{Li}_q(x)}{1-x}$$

$$\sum_k H_k \, \int^1_0 x^{k-1} \mathrm{Li}_q(x)\, dx= \mathscr{H}(p,q)+\int^1_0\frac{\mathrm{Li}_p(x) \mathrm{Li}_q(x)}{1-x}\, dx$$

$$\sum_k H_k \, \int^1_0 x^{k-1} \mathrm{Li}_q(x)\, dx= \mathscr{H}(p,q)-\lim_{s\to 1}\,\log(1-s)\mathrm{Li}_p(s) \mathrm{Li}_q(s)+\int^1_0\frac{\mathrm{Li}_{p-1}(x) \mathrm{Li}_q(x) \log(1-x)}{x}\, dx+\int^1_0\frac{\mathrm{Li}_p(x) \mathrm{Li}_{q-1}(x) \log(1-x)}{x}\, dx$$

$$\sum_{k\geq 1} H_k^{(p)}\left( \sum_{m=1}^{q-1}(-1)^{m-1}\frac{\zeta(q-m+1)}{k^m}+(-1)^{q-1}\frac{H_k}{k^q} \right) = \mathscr{H}(p,q)-\lim_{s\to 1}\,\log(1-s)\mathrm{Li}_p(s) \mathrm{Li}_q(s)+\int^1_0\frac{\mathrm{Li}_{p-1}(x) \mathrm{Li}_q(x) \log(1-x)}{x}\, dx+\int^1_0\frac{\mathrm{Li}_p(x) \mathrm{Li}_{q-1}(x) \log(1-x)}{x}\, dx$$

$$\sum_{k\geq 1} H_k^{(p)}\left( \sum_{m=2}^{q-1}(-1)^{m-1}\frac{\zeta(q-m+1)}{k^m}+(-1)^{q-1}\frac{H_k}{k^q} \right)+\zeta(q) \lim_{s \to 1}\sum_{k\geq 1}\frac{H^{(p)}_k}{k}s^k= \mathscr{H}(p,q)-\zeta(p)\zeta(q)\lim_{s\to 1}\,\log(1-s) +\int^1_0\frac{\mathrm{Li}_{p-1}(x) \mathrm{Li}_q(x) \log(1-x)}{x}\, dx+\int^1_0\frac{\mathrm{Li}_p(x) \mathrm{Li}_{q-1}(x) \log(1-x)}{x}\, dx$$

\begin{align} \int^1_0\frac{\mathrm{Li}_{p-1}(x) \mathrm{Li}_q(x) \log(1-x)}{x}\, dx+\int^1_0\frac{\mathrm{Li}_p(x) \mathrm{Li}_{q-1}(x) \log(1-x)}{x}\, dx &= \sum_{m=2}^{q-1}(-1)^{m-1} \zeta(q-m+1) S_{p,m} \\&+ (-1)^{q-1} \sum_{k\geq 1} \frac{H_k^{(p)} H_k}{k^q}-\mathscr{H}(p,q)+\\&\zeta(q) \lim_{s \to 1}\left( \sum_{k\geq 1}\frac{H^{(p)}_k}{k}s^k +\zeta(p)\log(1-s) \right)
\end{align}

I haven't checked whether the evaluations are correct. I still have to evaluate the limit which I hope vanishes.

 

[alyafey22]

Now we will evaluate the limit. Consider the following generating function

$$\sum_{k\geq 1} H_k^{(p)} x^k = \frac{\mathrm{Li}_p(x)}{1-x}$$

Dividing by $x$ we have

$$\sum_{k\geq 1} H_k^{(p)} x^{k-1} = \frac{\mathrm{Li}_p(x)}{x}+\frac{\mathrm{Li}_p(x)}{1-x}$$

Now integrate with respect to $x$ to get

$$\sum_{k\geq 1} \frac{H_k^{(p)}}{k} x^{k}= \mathrm{Li}_{p+1}(x)+\int^x_0 \frac{\mathrm{Li}_p(x)}{1-x} \, dx$$

Now use integration by parts to obtain

$$\sum_{k\geq 1} \frac{H_k^{(p)}}{k} x^{k}= \mathrm{Li}_{p+1}(x)-\log(1-x) \mathrm{Li}_p(x) +\int^x_0 \frac{\mathrm{Li}_{p-1}(t) \log(1-t)}{t} \, dx$$

$$\sum_{k\geq 1} \frac{H_k^{(p)}}{k} x^{k}+\log(1-x) \mathrm{Li}_p(x) = \mathrm{Li}_{p+1}(x)+\int^x_0 \frac{\mathrm{Li}_{p-1}(t) \log(1-t)}{t} \, dx$$

Taking the limit $$x \to 1$$ we have

$$\lim_{x \to 1}\left( \sum_{k\geq 1}\frac{H^{(p)}_k}{k}x^k+\mathrm{Li}_p(x)\log(1-x) \right)= \zeta(p+1)-\mathscr{H}(p-1,1)$$

To conclude we have the general formula

\begin{align} \int^1_0\frac{\mathrm{Li}_{p-1}(x) \mathrm{Li}_q(x) \log(1-x)}{x}\, dx+\int^1_0\frac{\mathrm{Li}_p(x) \mathrm{Li}_{q-1}(x) \log(1-x)}{x}\, dx &= \sum_{m=2}^{q-1}(-1)^{m-1} \zeta(q-m+1) S_{p,m} \\&+ (-1)^{q-1} \sum_{k\geq 1} \frac{H_k^{(p)} H_k}{k^q}-\mathscr{H}(p,q)\\& +\zeta(q) \zeta(p+1)-\zeta(q)\mathscr{H}(p-1,1)
\end{align}

For the special case $$p=q $$ we get

\begin{align}2\int^1_0\frac{\mathrm{Li}_q(x) \mathrm{Li}_{q-1}(x) \log(1-x)}{x}\, dx &= \sum_{m=2}^{q-1}(-1)^{m-1} \zeta(q-m+1) S_{q,m} + (-1)^{q-1} \sum_{k\geq 1} \frac{H_k^{(q)} H_k}{k^q}-\mathscr{H}(q,q)\\& +\zeta(q) \zeta(q+1)-\zeta(q)\mathscr{H}(q-1,1)
\end{align}

 

[DreamWeaver]

Great stuff, Zaid! :D Keep 'em coming...

 

[alyafey22]

In the previous post we proved that

\begin{align}2\int^1_0\frac{\mathrm{Li}_q(x) \mathrm{Li}_{q-1}(x) \log(1-x)}{x}\, dx &= \sum_{m=2}^{q-1}(-1)^{m-1} \zeta(q-m+1) S_{q,m} + (-1)^{q-1} \sum_{k\geq 1} \frac{H_k^{(q)} H_k}{k^q}-\mathscr{H}(q,q)\\& +\zeta(q) \zeta(q+1)-\zeta(q)\mathscr{H}(q-1,1)
\end{align}

Letting $q=3$ we have

\begin{align}-2\int^1_0\frac{\mathrm{Li}_3(x) \mathrm{Li}_{2}(x) \mathrm{Li}_1(x)}{x}\, dx &= - \zeta(2) S_{3,2} +\sum_{k\geq 1} \frac{H_k^{(3)} H_k}{k^3}-\mathscr{H}(3,3)\\& +\zeta(3) \zeta(4)-\zeta(3)\mathscr{H}(2,1)
\end{align}

We already now that

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

Hence we have

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

$$L^2_1(2,2) = \int^1_0\frac{\mathrm{Li}_{2}(x)^3}{x}\, dx = \zeta(3)\zeta(2)^2- \zeta(2) S_{3,2} +\sum_{k\geq 1} \frac{H_k^{(3)} H_k}{k^3}-\mathscr{H}(3,3)+\zeta(3) \zeta(4)-\zeta(3)\mathscr{H}(2,1)$$

 

[alyafey22]

Consider the following general from

\begin{align}2\int^1_0\frac{\mathrm{Li}_q(x) \mathrm{Li}_{q-1}(x) \log(1-x)}{x}\, dx &= \sum_{m=2}^{q-1}(-1)^{m-1} \zeta(q-m+1) S_{q,m} + (-1)^{q-1} \sum_{k\geq 1} \frac{H_k^{(q)} H_k}{k^q}-\mathscr{H}(q,q)\\& +\zeta(q) \zeta(q+1)-\zeta(q)\mathscr{H}(q-1,1)
\end{align}

Now, let $q=2$ then

\begin{align}-2\int^1_0\frac{\mathrm{Li}_2(x) \log^2(1-x)}{x}\, dx &= - \sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^2}-\mathscr{H}(2,2)+\zeta(2) \zeta(3)-\zeta(2)\mathscr{H}(1,1)
\end{align}

\begin{align} \sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^2}=2L^1_2(1,2)-\mathscr{H}(2,2)+\zeta(2) \zeta(3)-\zeta(2)\mathscr{H}(1,1)
\end{align}

Eventually we have the result

$$\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$$​

We already know that

$$\sum_{k\geq 1}\frac{H_k^{(3)}}{k^2}+\sum_{k\geq 1}\frac{H_k^{(2)}}{k^3}=\zeta(2)\zeta(3)+\zeta(5)$$

Hence we have

$$\sum_{k\geq 1}\frac{H_k^{(3)}}{k^2}+\sum_{k\geq 1}\frac{H_k^{(2)}}{k^3}=\sum_{k\geq 1}\frac{H_k^{(2)}H_k^{(1)} }{k^2}$$​

 

[alyafey22]

Consider the following general case

$$\int^1_0\frac{\mathrm{Li}_q(x) \mathrm{Li}_{q-1}(x) \log(1-x)}{x}\, dx$$

Integrate by parts

$$\int^1_0\frac{\mathrm{Li}_q(x) \mathrm{Li}_{q-1}(x) \log(1-x)}{x}\, dx=-\zeta(2)\zeta(q-1)\zeta(q)+\int^1_0\frac{ \mathrm{Li}_{q-1}(x)^2 \mathrm{Li}_2(x)}{x}\, dx+\int^1_0\frac{\mathrm{Li}_q(x) \mathrm{Li}_{q-2}(x) \mathrm{Li}_2(x)}{x}\, dx$$

Put $q=4$ to get

$$\int^1_0\frac{\mathrm{Li}_4(x) \mathrm{Li}_{3}(x) \log(1-x)}{x}\, dx=-\zeta(2)\zeta(3)\zeta(4)+\frac{\zeta(3)^3}{3}+\int^1_0\frac{\mathrm{Li}_4(x) \mathrm{Li}_2(x)^2}{x}\, dx$$

So we get

$$\int^1_0\frac{\mathrm{Li}_4(x) \mathrm{Li}_2(x)^2}{x}\, dx=\frac{\zeta(3)^3}{3}-\zeta(2)\zeta(3)\zeta(4)-\int^1_0\frac{\mathrm{Li}_4(x) \mathrm{Li}_{3}(x) \log(1-x)}{x}\, dx$$

$$\int^1_0\frac{\mathrm{Li}_4(x) \mathrm{Li}_2(x)^2}{x}\, dx=\frac{\zeta(3)^3}{3}-\zeta(2)\zeta(3)\zeta(4)-\frac{1}{2} \left( \sum_{m=2}^{3}(-1)^{m-1} \zeta(5-m) S_{4,m}- \sum_{k\geq 1} \frac{H_k^{(4)} H_k}{k^4}-\mathscr{H}(4,4) +\zeta(4) \zeta(5)-\zeta(4)\mathscr{H}(3,1) \right)$$

 

[alyafey22]

To finish the last result we need the following results found easily by our previous generalizations

$$\mathscr{H}(3,1)=-\zeta(2)\zeta(3)+3\zeta(5)$$

$$\mathscr{H}(4,4)=2\zeta(4)\zeta(5) +2\zeta(2)\zeta(7) − 5\zeta(9)$$

For the Euler sums we will refer to the following paper which gives a general formula for $$p+q$$ is odd due to Browein .

Finish up later when I have time.

 

[alyafey22]

If we consider again the formula

\begin{align}\int^1_0\frac{\mathrm{Li}_{p-1}(x) \mathrm{Li}_q(x) \log(1-x)}{x}\, dx+\int^1_0\frac{\mathrm{Li}_p(x) \mathrm{Li}_{q-1}(x) \log(1-x)}{x}\, dx &= \sum_{m=2}^{q-1}(-1)^{m-1} \zeta(q-m+1) S_{p,m} \\&+ (-1)^{q-1} \sum_{k\geq 1} \frac{H_k^{(p)} H_k}{k^q}-\mathscr{H}(p,q)\\& +\zeta(q) \zeta(p+1)-\zeta(q)\mathscr{H}(p-1,1) \end{align}

Then letting $$p=2,q=3$$ yields

$$-\int^1_0\frac{ \mathrm{Li}_3(x) \log^2(1-x)}{x}\, dx+\int^1_0\frac{\mathrm{Li}_2(x)^2 \log(1-x)}{x}\, dx = -\zeta(2) S_{2,2} + \sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^3}-\mathscr{H}(2,3)+ \zeta^2(3)-\zeta(3) \mathscr{H}(1,1)$$

Hence we have

$$\sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^3} = -\int^1_0\frac{ \mathrm{Li}_3(x) \log^2(1-x)}{x}\, dx+\zeta(2) S_{2,2} + \mathscr{H}(2,3)-\zeta^2(3)+\zeta(3) \mathscr{H}(1,1)+\frac{\zeta(2)^3}{3}$$

Now we use that

$$\int^1_0\frac{ \mathrm{Li}_3(x) \log^2(1-x)}{x}\, dx=\frac{97}{12} \zeta(6)-\zeta^2(3)-\zeta^3(2)$$

$$S_{2,2}=\frac{7}{4}\zeta(4)$$

$$\mathscr{H}(2,3)=\frac{\zeta(3)^2}{2}$$

$$\mathscr{H}(1,1)=2\zeta(3)$$

So we have the interesting result

$$\sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^3} =- \frac{97}{12} \zeta(6)+\frac{7}{4}\zeta(4)\zeta(2) + \frac{5}{2}\zeta(3)^2+\frac{2}{3}\zeta(2)^3$$​

 

[alyafey22]

Consider the following general case

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

Letting $$p=1$$

$$\sum_{m=1}^{q-1}(-1)^{m-1} \zeta(q-m+1) \sum_{k\geq 1} (-1)^{k-1} \frac{H_k} {k^m}+ (-1)^{q-1} \sum_{k\geq 1} (-1)^{k-1} \frac{H^2_k}{k^q} = -\int^1_0\frac{\log(1+x)\mathrm{Li}_q(x)}{x}\, dx +\int^1_0 \frac{\log(1+x)\mathrm{Li}_q(x)}{1+x}\, dx$$

Integrating the first integral by parts we got

\begin{align}\int^1_0 \frac{\log(1+x)\mathrm{Li}_q(x)+\mathrm{Li}_{q+1}(x)}{1+x}\, dx &= \sum_{m=1}^{q-1}(-1)^{m-1} \zeta(q-m+1) \sum_{k\geq 1} (-1)^{k-1} \frac{H_k} {k^m}+ (-1)^{q-1} \sum_{k\geq 1} (-1)^{k-1} \frac{H^2_k}{k^q}\\&+ \log(2)\zeta(q+1)
\end{align}

For $q=2$ we have

\begin{align}\int^1_0 \frac{\log(1+x)\mathrm{Li}_2(x)+\mathrm{Li}_{3}(x)}{1+x}\, dx &= \zeta(2) \sum_{k\geq 1} (-1)^{k-1} \frac{H_k} {k}- \sum_{k\geq 1} (-1)^{k-1} \frac{H^2_k}{k^2}+ \log(2)\zeta(3)
\end{align}

 

[alyafey22]

Our next hope is finding a general formula for

$$\sum_{n\geq 1} H_n H^{(p)}_n x^n $$

Consider the following

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

If we sum from $$k=1 \to n $$ we have

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

Well, that might be hopeless but will try it later!

 

[alyafey22]

I've got what is seems a systematic way of solving Higher Euler sums

According to Nielsen we have the following :

If $$f(x)= \sum_{n\geq 0}a_n x^n $$

Then we have the following $$\tag{1}\int^1_0 f(xt)\, \mathrm{Li}_2(t)\, dx=\frac{\pi^2}{6x}\int^x_0 f(t)\, dt
-\frac{1}{x}\sum_{n\geq1}\frac{a_{n-1} H_{n}}{n^2}x^n$$

Now let $a_n = H_n$ then we have the following

$$f(x)=\sum_{n\geq 1}H_n x^n=-\frac{\log(1-x)}{1-x}$$

$$-\int^1_0 \frac{\log(1-xt)}{1-xt} \mathrm{Li}_2(t)\, dt=-\frac{\pi^2}{6x}\int^x_0 \frac{\log(1-t)}{1-t} dt-\sum_{n\geq1}\frac{H_{n-1} H_{n}}{n^2}x^{n-1}$$

Hence we have the following by gathering the integrals and $x\to 1$

$$\sum_{n\geq1}\frac{H_{n-1} H_{n}}{n^2}=\int^1_0\frac{\log(1-x)\left(\mathrm{Li}_2(x)-\zeta(2)\right)}{1-x} dx$$

Integrating by parts we have

$$\sum_{n\geq1}\frac{H_{n-1} H_{n}}{n^2}=-\frac{1}{2}\int^1_0\frac{\log(1-x)^3}{x} dx$$

Hence we have

$$\sum_{n\geq1}\frac{ H^2_{n}}{n^2}=\sum_{n\geq1}\frac{ H_{n}}{n^3}-\frac{1}{2}\int^1_0\frac{\log(1-x)^3}{x} dx=\frac{17 \pi^4}{360}$$

I'll try to generalize the approach in the next thread.

 

[alyafey22]

Here is a little bit of a generalization

Now let $a_n = H^{(p)}_n$ then we have the following

$$f(x)=\sum_{n\geq 1}H_n x^n=\frac{\mathrm{Li}_p(x)}{1-x}$$

$$\int^1_0 \frac{\mathrm{Li}_p(xt)}{1-xt} \mathrm{Li}_2(t)\, dx=\frac{\pi^2}{6x}\int^x_0 \frac{\mathrm{Li}_p(t)}{1-t} dt-\sum_{n\geq1}\frac{H^{(p)}_{n-1} H_{n}}{n^2}x^{n-1}$$

$$\int^1_0 \frac{\mathrm{Li}_p(xt)(\zeta(2)-\mathrm{Li}_2(t))}{1-xt} dt=\sum_{n\geq1}\frac{H^{(p)}_{n-1} H_{n}}{n^2}x^{n-1}$$

or

$$\int^1_0 \frac{\mathrm{Li}_p(t)(\zeta(2)-\mathrm{Li}_2(t))}{1-t} dt=\sum_{n\geq1}\frac{H^{(p)}_{n-1} H_{n}}{n^2}$$

Eventually we have

$$\sum_{n\geq1}\frac{H^{(p)}_{n} H_{n}}{n^2}=\int^1_0 \frac{\mathrm{Li}_p(t)(\zeta(2)-\mathrm{Li}_2(t))}{1-t} dt-\sum_{n\geq 1}\frac{H_n}{n^{p+2}}$$

The Euler sum seems reducible when $p$ is ODD.

 

[alyafey22]

In our on-going journey we have to find a general formula for

$$\int^x_0 \frac{\mathrm{Li}_p(t)}{1-t}\, dt $$

This integral will have an anti-derivative if $p$ is odd or equal to $2$.

The idea is integrating by parts

\begin{align}
\int^x_0 \frac{\mathrm{Li}_p(t)}{1-t}\, dt & = \mathrm{Li}_p(x)\mathrm{Li}_1(x)-\int^x_0 \frac{\mathrm{Li}_{p-1}(t)\mathrm{Li}_1(t)}{t}\, dt\\ &=\mathrm{Li}_p(x)\mathrm{Li}_1(x)-\mathrm{Li}_{p-1}(x)\mathrm{Li}_2(x)+\int^x_0 \frac{\mathrm{Li}_{p-2}(t)\mathrm{Li}_2(t)}{t}\,dt\\&=\mathrm{Li}_p(x) \mathrm{Li}_1(x) -\mathrm{Li}_{p-1}(x)\mathrm{Li}_2(x)+\mathrm{Li}_{p-2}(x)\mathrm{Li}_3(x)-\int^x_0 \frac{\mathrm{Li}_{p-3}(t)\mathrm{Li}_3(t)}{t}\,dt\\ & = \,\,.\\ & = \,\,. \\ & = \,\,. \\&=\sum_{n=1}^k(-1)^{n-1}\mathrm{Li}_n(x)\mathrm{Li}_{p-n+1}(x)+(-1)^{k+1} \int^x_0 \frac{\mathrm{Li}_{p-k}(t)\mathrm{Li}_{k}(t)}{t}\,dt
\end{align}

Now let $p=2l-1$ hence we have

$$\sum_{n=1}^k(-1)^{n-1}\mathrm{Li}_n(x)\mathrm{Li}_{2l-n}(x)+(-1)^{k+1} \int^x_0 \frac{\mathrm{Li}_{2l-k-1}(t)\mathrm{Li}_{k}(t)}{t}\,dt$$

by letting $k=l-1$ we have

$$\sum_{n=1}^{l-1}(-1)^{n-1}\mathrm{Li}_n(x)\mathrm{Li}_{2l-n}(x)+(-1)^{l} \int^x_0 \frac{\mathrm{Li}_{l}(t)\mathrm{Li}_{l-1}(t)}{t}\,dt$$

Eventually we have

$$\sum_{n=1}^{l-1}(-1)^{n-1}\mathrm{Li}_n(x)\mathrm{Li}_{2l-n}(x)+(-1)^{l}\frac{\mathrm{Li}_l(x)^2}{2}$$

$$\int^x_0 \frac{\mathrm{Li}_{2k-1}(t)}{1-t}\, dt =\sum_{n=1}^{k-1}(-1)^{n-1}\mathrm{Li}_n(x)\mathrm{Li}_{2k-n}(x)+(-1)^{k}\frac{\mathrm{Li}_k(x)^2}{2}$$