Closed Form for $\int_{0}^1\frac{\ln(x)\ln^a(1-x)}{x^3}dx$

Question

I'm looking for a closed form for $a\in\mathbb{N}$ with $a\geq 3$: $$I(a)=\int_{0}^1\frac{\ln(x)\ln^a(1-x)}{x^3}dx$$ $I(a)$ can be rewritten as: $$I(a)=\int_0^1\frac{\ln(1-x)\ln^a(x)}{(1-x)^3}dx$$ $$I(a)=\frac{1}{4}\int_{0}^1\ln\left(\frac{2x}{1+x}\right)\ln^a\left(\frac{1-x}{1+x}\right)\left(\frac{1+x}{x^3}\right)dx$$ The problem can also be boiled down by using the binomial theorem, but that makes the integrals arguably more complicated to solve. The first couple values have solutions in terms of combinations of $\zeta$ functions: $$I(3)=\int_{0}^1 \frac{\ln(x)\ln^3(1-x)}{x^3}dx=-\frac{9}{2}\zeta(2)+\frac{9}{2}\zeta(3)+\frac{15}{4}\zeta(4)$$ $$I(4)=\int_{0}^1 \frac{\ln(x)\ln^4(1-x)}{x^3}dx=-9\zeta\left(4\right)+18\zeta\left(3\right)+12\zeta\left(2\right)\zeta\left(3\right)-36\zeta\left(5\right)$$ $$I(5)=\int_{0}^1 \frac{\ln(x)\ln^5(1-x)}{x^3}dx=-90\zeta(4)+105\zeta(6)-60\zeta(2)\zeta(3)-30\zeta^2(3)+150\zeta(5)$$

This is really an extension from my last question which begs the more general question: Is there a closed form for $$\int_{0}^1\frac{\ln(x)\ln^a(1-x)}{x^n}dx$$ Where $a,n\in\mathbb{N}$ and $a\geq n$.

If I am not mistaken, there does not seem to be a closed form for $n=1$ (but I could very well be wrong). So maybe imposing the restriction that $n\geq 2$ may be necessary.

Answer 1

Following the same approach here: $$I(a)=\int_0^1\frac{\ln x\ln^a(1-x)}{x^3}dx=\int_0^1\frac{\ln(1-x)\ln^ax}{(1-x)^3}dx\\\overset{IBP}{=}\frac12\int_0^1\frac{\ln^ax}{(1-x)^3}dx-\frac12a\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{(1-x)^2x}dx$$.

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The first integral:

$$\int_0^1\frac{\ln^ax}{(1-x)^3}dx=\frac12\sum_{n=2}^\infty n(n-1)\int_0^1 x^{n-2}\ln^ax\ dx\\=\frac12(-1)^aa!\sum_{n=2}^\infty\frac{n(n-1)}{(n-1)^{a+1}}=\frac12(-1)^aa!\sum_{n=1}^\infty\frac{n+1}{n^a}\\=\boxed{\frac12(-1)^aa!(\zeta(a-1)+\zeta(a))}$$

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The second integral:

Use the partial fraction decomposition: $\frac1{(1-x)^2x}=\frac1x+\frac1{1-x}+\frac1{(1-x)^2}$

$$\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{(1-x)^2x}dx\\=\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{x}dx+\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{1-x}dx+\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{(1-x)^2}dx$$

where the first integral:

$$\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{x}dx=-\sum_{n=1}^\infty\frac1n\int_0^1 x^{n-1}\ln^{a-1}x\ dx\\=(-1)^a(a-1)!\sum_{n=1}^\infty\frac1{n^{a+1}}=(-1)^a(a-1)!\zeta(a+1)$$

and the second integral

$$\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{1-x}dx=-\sum_{n=1}^\infty H_n\int_0^1x^n\ln^{a-1}x\ dx\\=(-1)^a(a-1)!\sum_{n=1}^\infty\frac{H_n}{(n+1)^a}\\=(-1)^a(a-1)!\left(\sum_{n=1}^\infty\frac{H_n}{n^a}-\zeta(a+1)\right)$$

As for the third integral, its already calculated in the link I mentioned above which shows

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

then we can write

$$\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{(1-x)^2}dx=(-1)^a (a-1)!\left(\sum_{n=1}^\infty\frac{H_n}{n^{a-1}}-\zeta(a-1)\right)$$

Collect the three sub-integrals to get

$$\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{(1-x)^2x}dx=\boxed{(-1)^a(a-1)!\left(\sum_{n=1}^\infty\frac{H_n}{n^a}+\sum_{n=1}^\infty\frac{H_n}{n^{a-1}}-\zeta(a-1)\right)}$$

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Combine the boxed results we get

$$I(a)=\frac14(-1)^aa!\left(3\zeta(a-1)+\zeta(a)-2\sum_{n=1}^\infty\frac{H_n}{n^a}-2\sum_{n=1}^\infty\frac{H_n}{n^{a-1}}\right)$$

where $$\sum_{n=1}^\infty \frac{H_n}{n^a}=\left(1+\frac a2\right)\zeta(a+1)-\frac12\sum_{k=1}^{a-2}\zeta(k+1)\zeta(a-k)$$ is the generalized Euler identity.