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How to prove without using the derivative of Beta function that

$$\int_0^1\frac{5\ln^4(1-x)\ln^4x-8\ln^3(1-x)\ln^5x+4\ln^2(1-x)\ln^6x}{x(1-x)}\ dx\\=5760\left\{4\zeta(9)-\zeta(2)\zeta(7)-\zeta(4)\zeta(5)-\zeta(3)\zeta(6)\right\}$$

Clearly its not a good idea to use the derivative of Beta function $\text{B}(a,b)$ as we need the $8^{th}$ derivative for each integral and that's too much calculations and what makes the calculations even harder is the case of $a$ and $b$ approach zero here. So any other way?

Ali Shadhar
  • 25,498

1 Answers1

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First, lets establish some identities:

From here we have

$$\sum_{n=1}^\infty (H_n^2-H_n^{(2)})x^{n}=\frac{\ln^2(1-x)}{1-x}\tag1$$

multiply both sides by $\frac{\ln^6x}{6!x}$ then integrate from $x=0$ to $1$ we have

$$\sum_{n=1}^\infty \frac{H_n^2-H_n^{(2)}}{n^7}=\frac1{6!}\int_0^1\frac{\ln^2(1-x)\ln^6x}{x(1-x)}\ dx\tag2$$

Integrate both sides of $(1)$ from $x=0$ to $x$

$$\sum_{n=1}^\infty (H_n^2-H_n^{(2)})\frac{x^{n+1}}{n+1}=-\frac13\ln^3(1-x)$$

By reindexing we get

$$\ln^3(1-x)=-3\sum_{n=1}^\infty \left(H_n^2-H_n^{(2)}-\frac{2H_n}{n}+\frac2{n^2}\right)\frac{x^{n}}{n}\tag3$$

$$\int_0^1\frac{\ln^4(1-x)\ln^4x}{x(1-x)}\ dx=\int_0^1\frac{\ln^4(1-x)\ln^4x}{x}\ dx+\underbrace{\int_0^1\frac{\ln^4(1-x)\ln^4x}{1-x}\ dx}_{1-x\mapsto x}$$

$$=2\int_0^1\frac{\ln^4(1-x)\ln^4x}{x}\ dx\overset{IBP}{=}\frac85\int_0^1\frac{\ln^3(1-x)\ln^5x}{1-x}\ dx$$

use the identity $\frac1{1-x}=\frac1{x(1-x)}-\frac1x$ for the last integral,

$$\int_0^1\frac{\ln^4(1-x)\ln^4x}{x(1-x)}\ dx=\frac85\int_0^1\frac{\ln^3(1-x)\ln^5x}{x(1-x)}\ dx-\frac85\int_0^1\frac{\ln^3(1-x)\ln^5x}{x}\ dx\tag4$$

Lets use the identity $(3)$ for the last integral in$(4)$,

$$\int_0^1\frac{\ln^3(1-x)\ln^5x}{x}\ dx=-3\sum_{n=1}^\infty \left(H_n^2-H_n^{(2)}-\frac{2H_n}{n}+\frac2{n^2}\right)\frac{1}{n}\int_0^1 x^{n-1}\ln^5x\ dx$$

$$=3\cdot 5!\sum_{n=1}^\infty \left(H_n^2-H_n^{(2)}-\frac{2H_n}{n}+\frac2{n^2}\right)\frac{1}{n^7}$$

$$=3\cdot 5!\sum_{n=1}^\infty \frac{H_n^2-H_n^{(2)}}{n^7}-6\cdot 5!\left(\sum_{n=1}^\infty\frac{H_n}{n^8}-\zeta(9)\right)$$

$$\overset{(2)}{=}\frac12\int_0^1\frac{\ln^2(1-x)\ln^6x}{x(1-x)}\ dx-6\cdot 5!\left(\sum_{n=1}^\infty\frac{H_n}{n^8}-\zeta(9)\right)\tag5$$

Plug $(5)$ in $(4)$

$$\int_0^1\frac{\ln^4(1-x)\ln^4x}{x(1-x)}\ dx=\frac85\int_0^1\frac{\ln^3(1-x)\ln^5x}{x(1-x)}\ dx-\frac45\int_0^1\frac{\ln^3(1-x)\ln^5x}{x}\ dx+1152\left(\sum_{n=1}^\infty\frac{H_n}{n^8}-\zeta(9)\right)$$

Multiply by $5$ and rearrange

$$\int_0^1\frac{5\ln^4(1-x)\ln^4x-8\ln^3(1-x)\ln^5x+4\ln^2(1-x)\ln^6x}{x(1-x)}\ dx=5760\left(\sum_{n=1}^\infty\frac{H_n}{n^8}-\zeta(9)\right)$$

And finally by substituting

$$\sum_{n=1}^\infty\frac{H_n}{n^8}=5\zeta(9)-\zeta(2)\zeta(7)-\zeta(4)\zeta(5)-\zeta(3)\zeta(6)$$

we get the desired answer.

Note that the last sum follows from using the Euler identity.

I hope we can use this equality for calculating some challenging integrals/ Euler sums.

Ali Shadhar
  • 25,498