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  1. زيد اليافعي
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    #1
    Inspired by this 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.

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  3. زيد اليافعي
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    #2 Thread Author
    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}

  4. زيد اليافعي
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    #3 Thread Author
    $ \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 by ZaidAlyafey; December 10th, 2013 at 16:24.

  5. زيد اليافعي
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    #4 Thread Author
    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)$

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    #5
    Congratulations, friend!!! It's a beauty!


    Have you considered writing paper about this and the other Polylog integrals/series you evaluated on the other thread...? Seems a mighty worthy topic.

  7. زيد اليافعي
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    #6 Thread Author
    Quote Originally Posted by DreamWeaver View Post
    Congratulations, friend!!! It's a beauty!


    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 !

  8. زيد اليافعي
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    #7 Thread Author
    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)$

    We already know that

    $ \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.

  9. زيد اليافعي
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    #8 Thread Author
    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.

  10. زيد اليافعي
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    #9 Thread Author
    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 by ZaidAlyafey; December 13th, 2013 at 17:56.

  11. زيد اليافعي
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    #10 Thread Author
    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 by ZaidAlyafey; December 13th, 2013 at 18:34.

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