Evaluating $\int_0^1 \frac{\ln^m (1+x)\ln^n x}{x}\; dx$ for $m,n\in\mathbb{N}$

Evaluate $\displaystyle \int\limits_0^1 \dfrac{\ln^m (1+x)\ln^n x}{x}\; dx$ for $m,n\in\mathbb{N}$

I was wondering if the above had some kind of a closed form, here some of the special cases have been discussed but this one is really a fascinating one.

I guess there's no general taylor expansion for $\ln^m (1+x)$ and so transforming into a series wouldn't be that easy.

Solution 1:

Stirling numbers of the first kind might be useful here, Consider

$$m! \sum_{k=m}^\infty (-1)^{k-m} \left[k\atop m\right] \frac{x^k}{k!} = \log^m(1+x)$$

$$\int\limits_0^1 \dfrac{\log^m (1+x)\log^n x}{x}\; dx = m! \sum_{k=m}^\infty (-1)^{k-m} \left[k\atop m\right] \frac{1}{k!} \int^1_0 x^{k-1} \log^n(x)\,dx$$

Now it is easy to see that

$$\int^1_0 x^{k-1} \,dx = \frac{1}{k}$$

By differentiation $n$ times with respect to $k$

$$\int^1_0 x^{k-1} \log^n(x)\,dx = (-1)^n\frac{n!}{k^{n+1}}$$

Substituting back we have

$$\int\limits_0^1 \dfrac{\log^m (1+x)\log^n x}{x}\; dx =(m!)(n!) \sum_{k=m}^\infty (-1)^{k-m+n} \left[k\atop m\right] \frac{1}{k!\, k^{n+1}}$$

Now the Striling numbers could related to Euler sums through equations like

$$\frac{\left[k\atop 3\right]}{k!} =\frac{ (H_{k-1})^2-H^{(2)}_{k-1}}{2k}$$

and

$$\frac{\left[k\atop 4\right]}{k!} =\frac{ (H_{k-1})^3-3H^{(2)}_{k-1}H_{k-1}+2H^{(3)}_{k-1}}{6k}$$

I don't think there exist a simple formula but this procedure should work.

Case $m=2 , n=2$

$$\int\limits_0^1 \dfrac{\log^2 (1+x)\log^2 x}{x}\; dx =4 \sum_{k=2}^\infty (-1)^{k} \left[k\atop 2\right] \frac{1}{k!\, k^{3}}$$

Note that

$$\frac{\left[k\atop 2\right]}{k!} = \frac{H_{k-1}}{k}$$

Hence we deduce that

$$\int\limits_0^1 \dfrac{\log^2 (1+x)\log^2 x}{x}\; dx =4 \sum_{k=2}^\infty (-1)^{k} \frac{H_{k-1}}{\, k^{4}}$$

Note that

$$\begin{align} \sum_{k=2}^\infty (-1)^{k} \frac{H_{k-1}}{\, k^{4}} &=\sum_{k=2}^\infty (-1)^{k} \frac{H_{k}}{ k^{4}} -\sum_{k=2}^\infty (-1)^{k} \frac{1}{ k^{5}} \\ &=\sum_{k=1}^\infty (-1)^{k} \frac{H_{k}}{ k^{4}} -\sum_{k=1}^\infty (-1)^{k} \frac{1}{ k^{5}}\\ &= \frac{\zeta(2) \zeta(3)}{2} - \frac{ 29\zeta(5)}{32} \end{align}$$

We deduce that

$$\boxed{\int\limits_0^1 \dfrac{\log^2 (1+x)\log^2 x}{x}\; dx = 2\zeta(2) \zeta(3)- \frac{ 29}{8}\zeta(5)}$$

This implies we can represent the special case $m=2$

$$\int\limits_0^1 \dfrac{\log^2 (1+x)\log^n x}{x}\; dx =2 (-1)^n(n!) \left[ \sum_{k=1}^\infty (-1)^{k} \frac{H_k}{ k^{n+2}} + \left(1-2^{-n-2} \right) \zeta(n+3) \right]$$

General formula in terms of nonlinear Euler sums

Define $\{ m\}$ as the $l$ partitions of $m$ where $m = i_1r_1+\cdots i_l r_l$

$$ \frac{1}{(m+1)!} \log^{m+1}(1+x) =\sum_{\{m\}} \sum_{k=1}^\infty \prod^l_{j=1}\frac{(-1)^{i_j+1}}{(i_j)!} \left( \frac{H_{k-1}^{(r_j)}}{r_j}\right)^{i_j} \frac{(-x)^k}{k} $$

Substitute back in the integral

$$\int\limits_0^1 \dfrac{\log^{m} (1+x)\log^n x}{x}\; dx = (-1)^{n+1}(n!) (m)! \sum_{\{m-1\}} \sum_{k=1}^\infty \frac{(-1)^k}{k^{n+2}} \prod^{l'}_{j=1}\frac{(-1)^{i_j}}{(i_j)!} \left( \frac{H_{k-1}^{(r_j)}}{r_j}\right)^{i_j}$$

Reference: https://arxiv.org/pdf/math/0607514.pdf

Solution 2:

Given the integral, $$I(n,p) = \int\limits_0^1 \dfrac{ \ln^{n-1}(x)\ln^p (1+x)}{x}\; dx$$

which is the notation consistent with Nielsen polylogs. Then closed-forms in terms of ordinary polylogarithms are known only for the following cases,

$$I(1,p) \\ I(n,1) \\ I(2k+1,2) \\ I(2,2) \\ I(2,3)$$

and no more. See this more general post on Nielsen polylogs.