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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 $\mathcal S.$

Ali Shadhar
  • 25,498

1 Answers1

2

Different approach by Cornel:

By master theorem we have

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

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

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

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.

Ali Shadhar
  • 25,498