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How to prove the following equality

$$\mathcal S=\sum_{n=1}^\infty\frac{H_n^4-6H_n^2H_n^{(2)}+8H_nH_n^{(3)}+3\left(H_n^{(2)}\right)^2-6H_n^{(4)}}{n^5}\\=672\zeta(9)-240\zeta(2)\zeta(7)-105\zeta(3)\zeta(6)-168\zeta(4)\zeta(5)+24\zeta^3(3)$$

Where $H_n^{(r)}=\sum_{k=1}^n\frac1{k^r}$ is the harmonic number and $\zeta$ is The Riemann zeta function.


Here is my approach and would like to see different ways.

From here we have

$$\frac{\ln^4(1-x)}{1-x}=\sum_{n=1}^\infty\left(H_n^4-6H_n^2H_n^{(2)}+8H_nH_n^{(3)}+3\left(H_n^{(2)}\right)^2-6H_n^{(4)}\right)x^n$$

Multiply both sides by $\frac{\ln^4x}{4!x}$ then integrate from $x=0$ to $1$

and use the fact that $\frac1{4!}\int_0^1 x^{n-1}\ln^4x\ dx=\frac1{n^5}$ to have

\begin{align} \mathcal S&=\frac1{4!}\int_0^1\frac{\ln^4(1-x)\ln^4x}{x(1-x)}\ dx\\ &=\frac1{4!}\int_0^1\frac{\ln^4(1-x)\ln^4x}{x} dx+\frac1{4!}\underbrace{\int_0^1\frac{\ln^4(1-x)\ln^4x}{1-x}dx}_{1-x\mapsto x}\\ &=\frac2{4!}\int_0^1\frac{\ln^4(1-x)\ln^4x}{x}dx\overset{IBP}{=}\frac1{15}\int_0^1\frac{\ln^3(1-x)\ln^5x}{1-x}dx\tag1 \end{align}

The interesting part in this solution is that we can calculate the last integral without using the derivative of beta function:

We proved here

$$\int_0^1\frac{x^n\ln^m(x)\ln^3(1-x)}{1-x}dx=\frac1{4}\frac{\partial^m}{\partial n^m}\left(H_n^4+6H_n^2H_n^{(2)}+8H_nH_n^{(3)}+3\left(H_n^{(2)}\right)^2+6H_n^{(4)}\right)$$

Set $m=5$ then let $n$ approach $0$ we get

$$\int_0^1\frac{\ln^3(1-x)\ln^5x}{1-x}\ dx\\=10080\zeta(9)-3600\zeta(2)\zeta(7)-1575\zeta(3)\zeta(6)-2520\zeta(4)\zeta(5)+360\zeta^3(3)$$

Substitute this result in $(1)$ we get the closed form of $\mathcal S.$

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    $\begingroup$ So what do you need? $\endgroup$ – user712576 Oct 24 '19 at 18:04
  • $\begingroup$ This is very impressive. Why not submit this to the knowledge base, as it would be highly unlikely for someone who did not have your extensive knowledge would have any idea how to even start on this. $\endgroup$ – marty cohen Oct 24 '19 at 18:48
  • $\begingroup$ @martycohen thank you for the kind words but sorry i didnt get what you mean by " knowledge base ". would you mind explaining? $\endgroup$ – Ali Shather Oct 24 '19 at 21:07
  • $\begingroup$ I think there is a wiki of contributed results somewhere here. $\endgroup$ – marty cohen Oct 24 '19 at 23:47
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Different approach by Cornel:

By master theorem we have

$$\sum_{k=1}^\infty\frac{H_n^4-6H_n^2H_n^{(2)}+8H_nH_n^{(3)}+3\left(H_n^{(2)}\right)^2-6H_n^{(4)}}{(k+1)(k+n+1)}\\=\frac{H_n^5+10H_n^3H_n^{(2)}+15H_n\left(H_n^{(2)}\right)^2+20H_n^2H_n^{(3)}+20H_n^{(2)}H_n^{(3)}+30H_nH_n^{(4)}+24H_n^{(5)}}{5n}$$

Multiply both sides by $n$ then differentiate with respect to $n$ we have

$$\sum_{k=1}^\infty\frac{H_n^4-6H_n^2H_n^{(2)}+8H_nH_n^{(3)}+3\left(H_n^{(2)}\right)^2-6H_n^{(4)}}{(k+n+1)^2}\\=\frac{\partial}{\partial n}\frac{H_n^5+10H_n^3H_n^{(2)}+15H_n\left(H_n^{(2)}\right)^2+20H_n^2H_n^{(3)}+20H_n^{(2)}H_n^{(3)}+30H_nH_n^{(4)}+24H_n^{(5)}}{5}$$

By differentiating both sides with respect to $n$ three times then let $n\mapsto -1$, the result of $\mathcal S$ follows.


The identity from the master theorem I used above can be found in the book Almost Integrals, Sums and Series page 291.

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