Log integrals I

In this example the value of the integral \begin{align} I_{3} = \int_{0}^{1} \frac{\ln^{3}(1+x)}{x} \, dx \end{align} was derived. The purpose of this question is to determine the value of the more general form of the integral, namely, \begin{align} I_{n} = \int_{0}^{1} \frac{\ln^{n}(1+x)}{x} \, dx . \end{align} Also is there a corresponding value for the integral \begin{align} J_{n} = \int_{0}^{1} \frac{\ln^{n}(1+x)}{1+x} \, dx \hspace{3mm} ? \end{align}

Solution 1:

Result: $$\mathcal{I}_n=n!\zeta(n+1)-\frac{n\ln^{n+1}(2)}{n+1}-\sum^{n-1}_{j=0}\frac{n!\ln^{j}(2)}{j!}{\rm Li}_{n-j+1}\left(\frac{1}{2}\right)$$

Derivation: \begin{align} \mathcal{I}_n &=\int^2_1\frac{\ln^n{x}}{x-1}{\rm d}x\tag1\\ &=(-1)^n\int^1_{\frac{1}{2}}\frac{\ln^n{x}}{x-x^2}{\rm d}x\tag2\\ &=\frac{\ln^{n+1}(2)}{n+1}+(-1)^n\sum^\infty_{k=1}\int^1_{\frac{1}{2}}x^{k-1}\ln^n{x}\ {\rm d}x\tag3\\ &=\frac{\ln^{n+1}(2)}{n+1}+(-1)^n\sum^\infty_{k=1}\frac{\partial^n}{\partial k^n}\left[\frac{1}{k}-\frac{1}{k2^k}\right]\\ &=\frac{\ln^{n+1}(2)}{n+1}+(-1)^n\sum^\infty_{k=1}\left[\frac{(-1)^nn!}{k^{n+1}}-\sum^n_{j=0}\binom{n}{j}\frac{(-1)^jj!}{k^{j+1}}\frac{(-1)^{n-j}\ln^{n-j}(2)}{2^k}\right]\tag4\\ &=\frac{\ln^{n+1}(2)}{n+1}+n!\sum^\infty_{k=1}\left[\frac{1}{k^{n+1}}-\sum^n_{j=0}\frac{\ln^{n-j}(2)}{k^{j+1}2^k(n-j)!}\right]\\ &=\frac{\ln^{n+1}(2)}{n+1}+n!\left[\zeta(n+1)-\sum^n_{j=0}\frac{\ln^{n-j}(2)}{(n-j)!}{\rm Li}_{j+1}\left(\frac{1}{2}\right)\right]\tag5\\ &=n!\zeta(n+1)-\frac{n\ln^{n+1}(2)}{n+1}-\sum^{n-1}_{j=0}\frac{n!\ln^{j}(2)}{j!}{\rm Li}_{n-j+1}\left(\frac{1}{2}\right)\tag6 \end{align}

Explanation:
$(1): \text{Let $x \mapsto x-1$}$.
$(2): \text{Let $x \mapsto x^{-1}$}$.
$(3): \text{Partial fractions, expand $(1-x)^{-1}$}$.
$(4): \text{Apply Leibniz's Generalised Product Rule}$.
$(5): \text{Recognise the zeta and polylog terms}$.
$(6): \text{Let $j\mapsto n-j$, rearrange terms}$.

Solution 2:

As M.N.C.E. has provided a nice answer for $I_{n}$ the only change that could be made it to cast the result into the form \begin{align} I_{n} &= \int_{0}^{1} \frac{\ln^{n}(1+x)}{x} \, dx = - \frac{n}{n+1} \, \ln^{n+1}(2) + n! \zeta(n+1) - \sum_{k=1}^{n} \frac{n! \, \ln^{n-k}(2)}{(n-k)!} \, Li_{k+1}\left(\frac{1}{2}\right). \end{align}

The second integral is \begin{align} J_{n} = \int_{0}^{1} \frac{\ln^{n}(1+x)}{1+x} \, dx = \left[ \frac{\ln^{n+1}(1+x)}{n+1} \right]_{0}^{1} = \frac{\ln^{n+1}(2)}{n+1} \end{align}