A generalized form of an integral

Continuing the post from here we give the generalization which is due to Ovidiu Furdui. 

Let $k$ be a positive integer. Prove that:

$$\int_{0}^{1}\ln^k (1-x)\ln x\, {\rm d}x =(-1)^{k+1}k! \left ( k+1 - \zeta(2)-\zeta(3)- \cdots -\zeta(k+1) \right )$$

where $\zeta$ is the Riemann zeta function.

Solution



We have that:

$$\begin{align*}
\int_{0}^{1}\ln^k (1-x) \ln x \, {\rm d}x& \overset{u=1-x}{=\! =\! =\! =\!} \int_{0}^{1}\ln^k x \ln (1-x)\, {\rm d} x\\
 &=-\sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{1}x^n \ln^k x  \, {\rm d}x \\
 &= -(-1)^k k! \sum_{n=1}^{\infty}\frac{1}{n \left ( n+1 \right )^{k+1}}
\end{align*}$$

Let $\displaystyle S_k = \sum_{n=1}^{\infty}\frac{1}{n \left ( n+1 \right )^{k+1}}$ . Then:

$$\begin{align*}
S_k &=\sum_{n=1}^{\infty}\frac{1}{(n+1)^k}\left [ \frac{1}{n}- \frac{1}{n+1} \right ] \\
 &= S_{k-1} - \sum_{n=2}^{\infty}\frac{1}{n^{k+1}}\\
 &= S_{k-1} +1 - \zeta(k+1)\\
 &= \left ( S_{k-2} +1 -\zeta(k)  \right )+1 -\zeta(k+1) \\
 &=S_{k-2} +2 -\zeta(k)-\zeta(k+1) \\
 &= \cdots \\
 &=S_0 +k -\zeta(2)- \zeta(3) - \cdots - \zeta(k+1)\\
 &=\sum_{n=1}^{\infty}\left [ \frac{1}{n} - \frac{1}{n+1} \right ] +k - \zeta(2) - \zeta(3)-\cdots -\zeta(k+1) \\
 &=1+k -\zeta(2)-\zeta(3)-\cdots -\zeta(k+1)
\end{align*}$$

and the result follows.

The exercise can also be found in mathematica.gr