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Is there a closed form for $b\in\mathbb{N}$ and $b\geq 2$: $$I(b)=\int_{0}^1 \frac{\ln(x)\ln(1-x)^b}{x^2}dx $$

I've calculated the first couple of values using Mathematica: $$I(2)=\frac{\pi^2}{3}-4\zeta(3)$$ $$I(3)=\frac{\pi^4}{12}-6\zeta(3)$$ $$I(4)=\frac{4\pi^4}{15}+4\pi^2\zeta(3)-72\zeta(5)$$ $$I(5)=\frac{2\pi^6}{9}-60\zeta(3)^2-120\zeta(5)$$

I've managed to show that: $$I(a)=\left(-1\right)^{\left(a+1\right)}a!\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{m}{n\left(n+m\right)^{\left(a+1\right)}} $$ But I'm not too good with double infinite sums.

My question: Can $I$ be expressed as a combination of $\zeta$ functions?

Ali Shadhar
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Tom Himler
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1 Answers1

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Here is an approach without using the derivative of beta function :

$$I(b)=\int_0^1\frac{\ln x\ln^b(1-x)}{x^2}dx\overset{IBP}{=}\int_0^1\frac{\ln^b(1-x)}{x^2}dx-b\int_0^1\frac{\ln x\ln^{b-1}(1-x)}{x(1-x)}dx$$

where

$$\int_0^1\frac{\ln^b(1-x)}{x^2}dx=\int_0^1\frac{\ln^bx}{(1-x)^2}dx=\sum_{n=1}^\infty n\int_0^1 x^{n-1}\ln^bx\ dx\\=(-1)^b b!\sum_{n=1}^\infty\frac{n}{n^{b+1}}=\boxed{(-1)^b b!\zeta(b)}$$

and

$$\int_0^1\frac{\ln x\ln^{b-1}(1-x)}{x(1-x)}dx=\int_0^1\frac{\ln(1-x)\ln^{b-1}x}{x(1-x)}dx\\=-\sum_{n=1}^\infty H_n\int_0^1 x^{n-1}\ln^{b-1}x \ dx=\boxed{-(-1)^{b-1}(b-1)!\sum_{n=1}^\infty\frac{H_n}{n^b}}$$

Combine the two results we have

$$I(b)=(-1)^b b!\left(\zeta(b)-\sum_{n=1}^\infty\frac{H_n}{n^b}\right)$$

where the generalized Euler sum:

$$\sum_{n=1}^\infty \frac{H_n}{n^b}=\left(1+\frac b2\right)\zeta(b+1)-\frac12\sum_{k=1}^{b-2}\zeta(k+1)\zeta(b-k)$$

Ali Shadhar
  • 25,498