Another "closed form expression" for a generating function involving harmonic numbers.

Question

This question is closely related to Calculating alternating Euler sums of odd powers.

Let $p\ge 1$ and $q\ge 1$ be integers and $1> x\ge 0$ be real. We use the following definition: \begin{equation} {\bf H}^{(p)}_q(x):=\sum\limits_{n=1}^\infty \frac{H_n^{(p)}}{n^q} \cdot x^n \end{equation} Now, by dividing the closed form expression for ${\bf H}^{(1)}_3(x)$ given in the answer to Generating function of squares of generalized harmonic numbers by $x$ and integrating we derived the following expression below: \begin{eqnarray} &&{\bf H}^{(1)}_4(x)=\\ &&-\text{Li}_5(1-x)+\text{Li}_5\left(\frac{x-1}{x}\right)+2 \text{Li}_5(x)-\frac{1}{2} \text{Li}_3(x) \log ^2(1-x)+\\ &&-\frac{1}{3} \text{Li}_3(1-x) \left(3 \log ^2(1-x)-3 \log (x) \log (1-x)+\pi ^2\right)-\frac{1}{4} \text{Li}_2(x){}^2 \log (1-x)+\\ &&+\frac{1}{12} \text{Li}_2(x) \left(\pi ^2-9 \log (1-x) \log (x)\right) \log (1-x)+\\ &&\text{Li}_4(1-x) \log (1-x)-\text{Li}_4\left(\frac{x-1}{x}\right) \log (1-x)-\text{Li}_4(1-x) \log (x)+\frac{1}{2} \zeta (3) \log ^2(1-x)+\\ &&\frac{1}{20} \log ^5(1-x)-\frac{\log ^5(x)}{120}-\frac{1}{4} \log (x) \log ^4(1-x)+\frac{1}{18} \pi ^2 \log ^3(1-x)-\frac{1}{36} \pi ^2 \log ^3(x)+\\ &&\frac{1}{24} \pi ^2 \log (x) \log ^2(1-x)-\frac{1}{4} \log ^2(x) \log ^3(1-x)+\frac{1}{12} \log ^3(x) \log ^2(1-x)+\\ &&\frac{1}{48} \pi ^4 \log (1-x)-\frac{1}{120} \pi ^4 \log (x)+\\ &&\zeta (5)+\frac{\pi ^2 \zeta (3)}{3}+\\ &&-{\mathcal S}_2^{(2)}(x)+{\mathcal S}_2^{(2)}(\frac{x}{x-1})-\frac{5}{4} ({\mathcal S}_2^{(2)}(1)-{\mathcal S}_2^{(2)}(1-x))+\\ &&\frac{1}{4}({\mathcal D}^{(2)}_2(1)-{\mathcal D}^{(2)}_2(1-x)) \end{eqnarray} where \begin{eqnarray} {\mathcal S}^{(2)}_2(x)&:=& \int\limits_0^x \frac{Li_2(t)^2}{t} dt \\ &=&2 \cdot \left( {\bf H}^{(2)}_3(x) - Li_5(x) \right) + 4 \left( {\bf H}^{(1)}_4(x) - Li_5(x) \right) \end{eqnarray} and \begin{eqnarray} {\mathcal D}^{(2)}_2(x)&:=&\int\limits_0^x \log(t)^2 \cdot \frac{Li_2(t)}{1-t}dt\\ &=& 2 \left( {\bf H}^{(2)}_3(x)- Li_5(x) \right) - 2 \log(x) \cdot \left({\bf H}^{(2)}_2(x)-Li_4(x) \right) + [\log(x)]^2 \cdot \left( {\bf H}^{(2)}_1(x) - Li_3(x)\right) \end{eqnarray} where \begin{eqnarray} {\bf H}^{(2)}_1(x) &=& 2 Li_3(1-x) - 2 \zeta(3) - \frac{\pi^2}{3} \log(1-x) + \log(1-x)^2 \log(x) + \log(1-x) Li_2(x) + Li_3(x) \\ {\bf H}^{(2)}_2(x) &=& -2 \zeta(4) + 2 Li_4(1-x) - 2 \log(1-x) \zeta(3) - \frac{\pi^2}{6} \log(1-x)^2 - \frac{1}{12} \log(1-x)^4 +\\ && \frac{1}{3} \log(1-x)^3 \log(x) + \frac{1}{2} Li_2(x)^2 + 2 \log(1-x) Li_3(x) - Li_4(x) - 2 Li_4(\frac{x}{x-1}) \end{eqnarray} Now I have two questions. The first one is a rhetoric one and it reads can the residual terms,i.e. the integrals on the very bottom, be expressed via poly-logarithms and elementary functions only? The second one can we derive closed form expressions for other values of $p$ and $q$ especially when $p+q \ge 5$.

Answer 1

This is not an answer but just your expression for ${\bf H}^{(1)}_4(x)$ typed in in Mathematica in order to check it.

If it contains typing errors then please correct it.

EDIT 4.12.17

It did indeed contain a typing error (spurious line break). The correct expression was provided by Przemo in his answer of today.

Answer 2

This is an answer to Dr. Wolfgang Hintze's comment above. First of all thank you for your interest in this question:). Secondly the way I derive the various expressions is error proof since, before I make any change, I always test whether the right hand side equals the left hand side for some randomly chosen parameters (in this case for the parameter $x$). Therefore I am usually pretty confident that my expressions are correct. Yet, having said that, I admit that I haven't done the series expansion which you did before. I did it now and the results are below:

x =.; Clear[h4]; Clear[S22]; Clear[D22];
S22[x_] := Integrate[PolyLog[2, t]^2/t, {t, 0, x}];
D22[x_] := Integrate[Log[t]^2 PolyLog[2, t]/(1 - t), {t, 0, x}];
h4[x_] := -PolyLog[5, 1 - x] + PolyLog[5, (-1 + x)/x] + 
   2 PolyLog[5, x] +
   -(1/2) Log[1 - x]^2 PolyLog[3, x] - 
   1/3 (\[Pi]^2 + 3 Log[1 - x]^2 - 3 Log[1 - x] Log[x]) PolyLog[3, 
     1 - x] +
   -(1/4) Log[1 - x] PolyLog[2, x]^2 + 
   1/12 Log[1 - x] (\[Pi]^2 - 9 Log[1 - x] Log[x]) PolyLog[2, x] +
   Log[1 - x] PolyLog[4, 1 - x] - Log[1 - x] PolyLog[4, (-1 + x)/x] - 
   Log[x] PolyLog[4, 1 - x] + 1/2 Log[1 - x]^2 Zeta[3] +
   1/20 Log[1 - x]^5 - Log[x]^5/120 - 1/4 Log[1 - x]^4 Log[x] + 
   1/18 \[Pi]^2 Log[1 - x]^3 - 1/36 \[Pi]^2 Log[x]^3 + 
   1/24 \[Pi]^2 Log[1 - x]^2 Log[x] - 1/4 Log[1 - x]^3 Log[x]^2 + 
   1/12 Log[1 - x]^2 Log[x]^3 + 1/48 \[Pi]^4 Log[1 - x] - 
   1/120 \[Pi]^4 Log[x] + 1/3 \[Pi]^2 Zeta[3] + Zeta[5] -
   S22[x] + S22[x/(x - 1)] +
   -5/4 (S22[1] - S22[1 - x]) +
   1/4 (D22[1] - D22[1 - x]);
Simplify[Normal[Series[h4[x], {x, 0, 3}]], Assumptions -> 0 < x < 1]

This gives us after two minutes of Mathematica's "thinking the following output:

x + (3 x^2)/32 + (11 x^3)/486

Now, from my experience with Mathematica it follows that the code is sometimes prone to some redundant trailing characters. It happens at times that if you break a line some special character is inserted which may scramble the result. Therefore I always make sure that there are no unnecessary line breaks in a long code.

Answer 3

You probably know this already, but in case you don't, here we go. ${\bf H}^{(3)}_2 (x)$ satisfies your requirement of $p + q \geqslant 5$ and is very easy to find.

From $${\bf H}^{(3)}_1 (x) = \sum^\infty_{n = 1} \frac{H^{(3)}_n}{n} x^n = \text{Li}_4 (x) - \text{Li}_3 (x) \ln (1 - x) - \frac{1}{2} \text{Li}^2_2 (x),$$ dividing throughout by $x$ before integrating from $0$ to $x$ gives \begin{align*} {\bf H}^{(3)}_2 (x) &= \text{Li}_5 (x) - \int^x_0 \frac{\text{Li}_3 (u) \ln (1 - u)}{u} \, du - \frac{1}{2} \int^x_0 \frac{\text{Li}_2^2 (u)}{u} \, du. \end{align*} As $$\int^x_0 \frac{\text{Li}_3 (u) \ln (1 - u)}{u} \, du = - \text{Li}_2 (x) \text{Li}_3 (x) + \int^x_0 \frac{\text{Li}^2_2 (u)}{u} \, du,$$ a result which can be readily found by parts, we have $${\bf H}^{(3)}_2 (x) = \text{Li}_5 (x) + \text{Li}_2 (x) \text{Li}_3 (x) - \frac{3}{2} \mathcal{S}^{(2)}_2 (x),$$ where, in your notation, $$\mathcal{S}^{(2)}_2 (x) = \int^x_0 \frac{\text{Li}^2_2 (u)}{u} \, du.$$ Of course, as you point out, since $${\mathcal S}^{(2)}_2(x) = \int\limits_0^x \frac{\text{Li}^2_2(u)}{u} du = 2 \left( {\bf H}^{(2)}_3(x) - \text{Li}_5(x) \right) + 4 \left( {\bf H}^{(1)}_4(x) - \text{Li}_5(x) \right),$$ this makes $\mathcal{S}^{(2)}_2 (x)$ a really troublesome term to have to deal with.

Answer 4

This is not an answer to this question but, as I believe, it is a step towards it. Again, let $p\ge 1$ and $q\ge 1$ be integers and $x\in (-1,1)$. Then the following identity holds: \begin{eqnarray} &&{\bf H}^{(p)}_q(x)=\\ &&Li_p(x) Li_q(x) + Li_{p+q}(x) + \\ &&\sum\limits_{j=1}^{p-1} (-1)^j \binom{q+j-1}{j} Li_{q+j}(x) Li_{p-j}(x)+\\ &&(-1)^p\sum\limits_{j=p}^{q+p-1} \binom{j-1}{p-1} Li_{q-j+p}(x) Li_j(x)+\\ &&\sum\limits_{j_2=0}^{q-1} [\log(x)]^{j_2} \cdot \sum\limits_{j=p}^{(q+p-1-j_2)} \sum\limits_{j_1=1}^{(q+p-j-j_2)} \frac{(-1)^{q+1-j-j_1-j_2}}{(j_2)!(q+p-j-j_1-j_2)!} \binom{j-1}{p-1} \cdot \\ &&\int\limits_0^x [\log(t)]^{q+p-j-j_1-j_2} \cdot \frac{Li_{j_1}(t) Li_{j-1}(t)}{t} dt \end{eqnarray} I have derived this identity using integration by parts. Below is a piece of Mathematica's code that verifies it:

M = 1000;
x = RandomReal[{-1, 1}, WorkingPrecision -> 50];
AA = Map[N[Take[Accumulate[#], -5], 30] &, 
   Table[HarmonicNumber[n, p] x^n/n^q, {p, 1, 5}, {q, 1, 5}, {n, 1, 
     M}], {2}];
AA1 = Table[
   PolyLog[q, x] PolyLog[p, x] +  PolyLog[q + p, x] +  
    Sum[(-1)^j Binomial[q + j - 1, j] PolyLog[p - j, x] PolyLog[q + j,
        x], {j, 1, p - 1}] + (-1)^p Sum[  
      Binomial[j - 1, p - 1] PolyLog[q - j + p, x] PolyLog[j, x], {j, 
       p, q + p - 1}] + 
    Sum[ Log[x]^j2 ((-1)^(
      q + 1 - j - j2 + j1)) /((j2)! (q + p - j - j2 - j1)!)
       Binomial[j - 1, p - 1] NIntegrate[
       Log[t]^(q + p - j - j1 - j2) ( 
        PolyLog[j1, t] PolyLog[j - 1, t])/t , {t, 0, x}, 
       WorkingPrecision :> 30], {j2, 0, q - 1}, {j, p, 
      q + p - 1 - j2}, {j1, 1, q + p - j - j2}], {p, 1, 5}, {q, 1, 
    5}];
MatrixForm[#] & /@ {AA1, AA[[All, All, -1]] - AA1}

After running it we are getting a following output:

Now let us take $p=1$ and $q=4$. Then we have: \begin{equation} {\bf H}^{(1)}_4(x)= -2 Li_2(x) Li_3(x)+ \log(1-x) Li_4(x) + Li_5(x) + \sum\limits_{j_2=0}^3 [\log(x)]^{j_2} {\mathcal s}_{4-j_2} \end{equation} where \begin{eqnarray} {\mathcal s}_1&=&\frac{1}{12} \log(1-x)^2\\ {\mathcal s}_2&=&\frac{1}{12} \left(-6 \text{Li}_3(1-x)-12 \text{Li}_2(x) \log (1-x)-6 \log (x) \log ^2(1-x)+\pi ^2 \log (1-x)+6 \zeta (3)\right)\\ {\mathcal s}_3&=&\text{Li}_2(x){}^2-\text{Li}_4(1-x)+\text{Li}_4(x)+\text{Li}_4\left(\frac{x}{x-1}\right)+2 \text{Li}_2(x) \log (x) \log (1-x)-2 \text{Li}_3(x) \log (1-x)+\text{Li}_3(1-x) \log (x)+\zeta (3) \log (1-x)-\zeta (3) \log (x)+\frac{1}{24} \log ^4(1-x)-\frac{1}{6} \log (x) \log ^3(1-x)+\frac{3}{4} \log ^2(x) \log ^2(1-x)+\frac{1}{12} \pi ^2 \log ^2(1-x)-\frac{1}{6} \pi ^2 \log (x) \log (1-x)+\frac{\pi ^4}{90}\\ {\mathcal s}_4&=&-\frac{1}{12} \log (1-x) \left(12 \text{Li}_4(x)-12 \text{Li}_3(x) \log (x)+\log (1-x) \log ^3(x)\right)+\text{Li}_2(x) \left(2 \text{Li}_3(x)-\frac{3}{4} \log (1-x) \log ^2(x)\right)-\frac{5}{4} \text{Li}_2(x){}^2 \log (x)+\frac{1}{4}\left( {\mathcal S}^{(2)}_2(x) - {\mathcal D}^{(2)}_2(x)\right) \end{eqnarray} Below is the Mathematica code that verifies the expression above:

x =.; t =.;
S22[x_] := Integrate[ PolyLog[2, t]^2/t, {t, 0, x}];
D22[x_] := Integrate[(Log[t]^2 PolyLog[2, t])/ (1 - t), {t, 0, x}];
ss := {1/12 Log[1 - x]^2,
   1/12 (\[Pi]^2 Log[1 - x] - 6 Log[1 - x]^2 Log[x] - 
      12 Log[1 - x] PolyLog[2, x] - 6 PolyLog[3, 1 - x] + 
      6 Zeta[3]), \[Pi]^4/90 + 1/12 \[Pi]^2 Log[1 - x]^2 + 
    1/24 Log[1 - x]^4 - 1/6 \[Pi]^2 Log[1 - x] Log[x] - 
    1/6 Log[1 - x]^3 Log[x] + 3/4 Log[1 - x]^2 Log[x]^2 + 
    2 Log[1 - x] Log[x] PolyLog[2, x] + PolyLog[2, x]^2 + 
    Log[x] PolyLog[3, 1 - x] - 2 Log[1 - x] PolyLog[3, x] - 
    PolyLog[4, 1 - x] + PolyLog[4, x] + PolyLog[4, x/(-1 + x)] + 
    Log[1 - x] Zeta[3] - Log[x] Zeta[3],
   -(5/4) Log[x] PolyLog[2, x]^2 + 
    PolyLog[2, x] (-(3/4) Log[1 - x] Log[x]^2 + 2 PolyLog[3, x]) - 
    1/12 Log[
      1 - x] (Log[1 - x] Log[x]^3 - 12 Log[x] PolyLog[3, x] + 
       12 PolyLog[4, x]) + 1/4 (-D22[x] + S22[x])};

{p, q} = {1, 4};
{t0, dummy} = Timing[Simplify[Normal[Series[
     -2 PolyLog[2, x] PolyLog[3, x] + Log[1 - x] PolyLog[4, x] + 
      PolyLog[5, x] + 
      Total[Table[Log[x]^j2 ss[[4 - j2]], {j2, 0, q - 1}]], {x, 0, 
      3}]], Assumptions -> -1 < x < 1]]

If you run that code after about 105 seconds Mathematica 9.0 produces the following result:

as it should be. Likewise let us take $p=2$ and $q=3$ then we have: \begin{equation} {\bf H}^{(2)}_3(x) = 4 Li_2(x) Li_3(x) +Li_5(x) + \sum\limits_{j_2=0}^2 [\log(x)]^{j_2} {\mathcal s}_{3-j_2} \end{equation} where \begin{eqnarray} {\mathcal s}_1 &=& \text{Li}_3(1-x)-\text{Li}_2(1-x) \log (1-x)-\frac{1}{2} \log (x) \log ^2(1-x)-\zeta (3)\\ {\mathcal s}_2 &=& -\frac{3 \text{Li}_2(x){}^2}{2}+2 \text{Li}_4(1-x)-2 \text{Li}_4(x)-2 \text{Li}_4\left(\frac{x}{x-1}\right)-2 \text{Li}_2(x) \log (x) \log (1-x)+2 \text{Li}_3(x) \log (1-x)-2 \text{Li}_3(1-x) \log (x)-2 \zeta (3) \log (1-x)+2 \zeta (3) \log (x)-\frac{1}{12} \log ^4(1-x)+\frac{1}{3} \log (x) \log ^3(1-x)-\log ^2(x) \log ^2(1-x)-\frac{1}{6} \pi ^2 \log ^2(1-x)+\frac{1}{3} \pi ^2 \log (x) \log (1-x)-\frac{\pi ^4}{45}\\ {\mathcal s}_3 &=& \frac{1}{2} \text{Li}_2(x) \left(-8 \text{Li}_3(x)+4 \text{Li}_2(x) \log (x)+\log (1-x) \log ^2(x)\right)+ \frac{1}{2} {\mathcal D}^{(2)}_2(x) \end{eqnarray} Again , below is the Mathematica code that verifies that:

{p, q} = {2, 3}; x =.; t =.;
D22[x_] := Integrate[(Log[t]^2 PolyLog[2, t])/ (1 - t), {t, 0, x}];
ss := {-(1/2) Log[1 - x]^2 Log[x] - Log[1 - x] PolyLog[2, 1 - x] + 
    PolyLog[3, 1 - x] - Zeta[3], -(\[Pi]^4/45) - 
    1/6 \[Pi]^2 Log[1 - x]^2 - 1/12 Log[1 - x]^4 + 
    1/3 \[Pi]^2 Log[1 - x] Log[x] + 1/3 Log[1 - x]^3 Log[x] - 
    Log[1 - x]^2 Log[x]^2 - 2 Log[1 - x] Log[x] PolyLog[2, x] - 
    3/2 PolyLog[2, x]^2 - 2 Log[x] PolyLog[3, 1 - x] + 
    2 Log[1 - x] PolyLog[3, x] + 2 PolyLog[4, 1 - x] - 
    2 PolyLog[4, x] - 2 PolyLog[4, x/(-1 + x)] - 
    2 Log[1 - x] Zeta[3] + 2 Log[x] Zeta[3], 
   1/2 PolyLog[2, 
      x] (Log[1 - x] Log[x]^2 + 4 Log[x] PolyLog[2, x] - 
       8 PolyLog[3, x]) + 1/2 D22[x]};
{t0, dummy} = 
 Timing[Simplify[
   Normal[Series[
     4 PolyLog[2, x] PolyLog[3, x] + PolyLog[5, x] + 
      Total[Table[Log[x]^j2 ss[[3 - j2]], {j2, 0, q - 1}]], {x, 0, 
      10}]], Assumptions -> 0 < x < 1]]

The output being:

Now let us take $p=3$ and $q=2$. Then we have: \begin{equation} {\bf H}^{(3)}_2(x) = -2 Li_2(x) Li_3(x)+Li_5(x) + \sum\limits_{j_2=0}^1 [\log(x)]^{j_2} {\mathcal s}_{2-j_2} \end{equation} where \begin{eqnarray} {\mathcal s}_1 &=& \frac{1}{2} [Li_2(x)]^2 \\ {\mathcal s}_2 &=& -\frac{1}{2} \log(x) Li_2(x)^2 + 3 Li_2(x) Li_3(x) - \frac{3}{2} {\mathcal S}^{(2)}_2(x) \end{eqnarray} The code that verifies this along with its output are below:

{p, q} = {3, 2}; x =.; t =.;
S22[x_] := Integrate[ PolyLog[2, t]^2/t, {t, 0, x}];
ss := {1/2 PolyLog[2, x]^2, -1/2 Log[x] PolyLog[2, x]^2  + 
    3 PolyLog[2, x] PolyLog[3, x] - 3/2 S22[x]};
{t0, dummy} = 
 Timing[Simplify[
   Normal[Series[-2 PolyLog[2, x] PolyLog[3, x] + PolyLog[5, x] + 
      Total[Table[Log[x]^j2 ss[[2 - j2]], {j2, 0, q - 1}]], {x, 0, 
      10}]], Assumptions -> 0 < x < 1]]

Finally let us take $p=4$ and $q=1$. Then we have: \begin{equation} {\bf H}^{(4)}_1(x) = -\log(1-x) Li_4(x) + Li_5(x) + \sum\limits_{j_2=0}^0 [\log(x)]^{j_2} {\mathcal s}_{1-j_2} \end{equation} where \begin{eqnarray} {\mathcal s}_1 &=& -Li_2(x) Li_3(x) + {\mathcal S}^{(2)}_2(x) \end{eqnarray} with the code and its output given below:

{p, q} = {4, 1}; x =.; t =.;
S22[x_] := Integrate[ PolyLog[2, t]^2/t, {t, 0, x}];
ss := {-PolyLog[2, x] PolyLog[3, x] + S22[x]}

{t0, dummy} = 
 Timing[Simplify[
   Normal[Series[-Log[1 - x] PolyLog[4, x] + PolyLog[5, x] + 
      Total[Table[Log[x]^j2 ss[[1 - j2]], {j2, 0, q - 1}]], {x, 0, 
      10}]], Assumptions -> 0 < x < 1]]

The overall conclusion from this exercise is that whenever $p+q\ge 5$ the generating functions in question cannot be reduced to poly-logarithms only. If $p+q=5$ then the additional quantities that enter are ${\mathcal S}^{(2)}_2(x)$ and ${\mathcal D}^{(2)}_2(x)$ which themselves are independent from each other and also cannot be reduced to other generating functions.