• MHB
• alyafey22
In summary, the conversation discusses a generalized formula for the polylogarithm function and explores some of its properties. It also mentions a potential paper topic on evaluating polylogarithm integrals and series. The conversation also includes a newly derived formula for evaluating alternating Euler sums.
alyafey22
Gold Member
MHB
Inspired by this http://mathhelpboards.com/calculus-10/powers-polylogarithms-7998.html we look at the generalization

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

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

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

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

Similarly

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

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

Hence we have

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

We showed that

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

Which can be rewritten as

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

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

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

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}

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

Also we define the following

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

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

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

First we use the generating function

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

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

Hence we have

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

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}

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

And the final generalization is

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

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

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

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

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

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

DreamWeaver said:

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 !

I derived an interesting formula

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

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

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

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

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

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

Ok, we know that

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

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

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

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

\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.

If we consider the following

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

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

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

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

Then if we integrate both sides $$\displaystyle \int^1_0$$ 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

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

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

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.

Last edited:
Now we will evaluate the limit. Consider the following generating function

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

Dividing by $x$ we have

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

Now integrate with respect to $x$ to get

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

Now use integration by parts to obtain

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

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

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

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

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 $$\displaystyle p=q$$ 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}

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

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}

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

Hence we have

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

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

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

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

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

Hence we have

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

Consider the following general case

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

Integrate by parts

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

Put $q=4$ to get

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

So we get

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

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

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

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

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

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

Finish up later when I have time.

Last edited:
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 $$\displaystyle p=2,q=3$$ yields

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

Hence we have

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

Now we use that

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

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

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

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

So we have the interesting result

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

Consider the following general case

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

Letting $$\displaystyle p=1$$

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

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}

Last edited:
Our next hope is finding a general formula for

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

Consider the following

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

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

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

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

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.

Last edited:
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.

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

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

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

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

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

1. What is a generalization of triple and higher power polylog integrals?

A generalization of triple and higher power polylog integrals is a mathematical concept that extends the traditional polylogarithm function to include integrals with powers higher than three. This generalization allows for the evaluation of more complex integrals and has applications in various areas of mathematics, including number theory and quantum field theory.

The traditional polylogarithm function, also known as the Lerch transcendent, is defined as the sum of powers of natural logarithms. The generalization of triple and higher power polylog integrals extends this concept by including integrals with powers higher than three, making it a more versatile and powerful tool for solving mathematical problems.

3. What are some practical applications of this generalization?

This generalization has practical applications in various areas of mathematics, including number theory, where it is used to study the distribution of prime numbers. It also has applications in quantum field theory, where it is used to calculate Feynman integrals and other complex mathematical expressions.

4. Are there any limitations to this generalization?

Like any mathematical concept, there are limitations to the generalization of triple and higher power polylog integrals. It may not be applicable to every mathematical problem and may have restrictions on the values of the powers used in the integrals.

5. How is this generalization relevant to current research in mathematics?

This generalization is a topic of ongoing research in mathematics, with various mathematicians and scientists exploring its applications and properties. It has the potential to provide new insights and solutions to complex mathematical problems, making it a relevant and exciting area of study.

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