Prove that: $\int_{0}^{1}\frac{(\ln x)^k}{1-x}dx=(-1)^kk!\sum_{n=1}^{+\infty}\frac{1}{n^{k+1}}, \ \ k\in\mathbb{N}^*$ Prove that: $\int_{0}^{1}\frac{(\ln x)^k}{1-x}dx=(-1)^kk!\sum_{n=1}^{+\infty}\frac{1}{n^{k+1}}, \ \ k\in\mathbb{N}^*$
I have $\forall x\in (0;1), \frac{(\ln x)^k}{1-x}=\sum_{n=0}^{+\infty}x^n(\ln x)^k$
Examining a mapping $f_n:[0;1]\rightarrow \mathbb{R}$, $f_n(x)=\left\{\begin{matrix}
x^n(\ln x)^k\ \ if \ \ x\neq 0\\ 
0 \ \ if \ \ x = 0
\end{matrix}\right.$
$R_n(x)=\sum_{p=n+1}^{+\infty}f_p(x)$
I want to prove that $\int_{0}^{1}R_n(x)dx \rightarrow 0$ (*) as $n \rightarrow +\infty$ to use $\int_{0}^{1}\frac{(\ln x)^k}{1-x}dx=\sum_{n=0}^{+\infty}\int_{0}^{1}x^n(\ln x)^kdx$
Could you help me to prove that (*)? Thank for helping.
 A: In order to have
\begin{equation*}
\int_{0}^{1}\frac{(\ln x)^k}{1-x}d x=\int_{0}^{1}(\ln x)^k\sum_{n=0}^{\infty}x^ndx=\sum_{n=0}^{\infty}\int_{0}^{1}(\ln x)^kx^ndx=\sum_{n=0}^{\infty}\frac{(-1)^kk!}{(n+1)^{k+1}}
\end{equation*}
we have to prove some things.
First, we proof $\int_{0}^{1}(\ln x)^kx^ndx=\frac{(-1)^kk!}{(n+1)^{k+1}}$:
Consider
\begin{equation*}
\int_{0}^{1}x^ndx=\frac{1}{n+1}.
\end{equation*}
Take $n$ as a parameter and differentiate both sides with respect to $n$, you get
\begin{equation*}
\int_{0}^{1}x^n\ln(x)dx=-\frac{1}{(n+1)^2}=\frac{(-1)^11!}{(n+1)^{1+1}}.
\end{equation*}
That's the case $k=1$, applying induction you get the result.
Now, to have $\int_{0}^{1}(\ln x)^k\sum_{n=0}^{\infty}x^ndx=\sum_{n=0}^{\infty}\int_{0}^{1}(\ln x)^kx^ndx$, we need to prove that $\sum_{n=0}^{\infty}x^n(\ln x)^k$ is uniformly convergent. 
I could prove uniform convergence only in case $k >1$. In that case, we use Weiestrass M-Test: If there is a sequence of positive real numbers $M_n$ such that $|f_n|_{A} \leq M_n$ for all $n \in \mathbb{N}$ and $\sum M_n < \infty$, then the series $\sum f_n$ converges uniformly in $A$. Here $||_{A}$ denothes the supremum norm.  
With $f_n=x^n(\ln x)^k$, we have
\begin{equation*}
|f_n|_{(0,1]}=|x^n(\ln x)^k|_{(0,1]}\leq (\frac{k}{ne})^k=M_n,
\end{equation*}
because over $(0,1]$, $|x^n (\ln x)^k|$ reach its maximum value in $x = e^{-\frac{k}{n}}$ (you get that with the help of calculus, or Wolfram), and that maximum value is $(\frac{k}{ne})^k$.
Furthermore, 
\begin{equation*}
\sum_{n=0}^{\infty} M_n=\sum_{n=0}^{\infty} (\frac{k}{ne})^k=(\frac{k}{e})^k\sum_{n=0}^{\infty} \frac{1}{n^k} < \infty
\end{equation*}
in case $k > 1$.
So $\sum_{n=0}^{\infty}x^n(\ln x)^k$ is uniformly convergent if $k > 1$.
Case $k=1$ should be considered apart. Personally couldn't do it.
