Closed form for $\int_0^1 \frac {\log^n(x)}{(1-x)^m} dx$

Question

How can one find a general form for $\int_0^1 \frac {\log(x)}{(1-x)} dx=-\zeta(2) \,?$ Namely $\int_0^1 \frac {\log^n(x)}{(1-x)^m} dx\,$ where $n,m\ge1$ Similar to the original integral I let $1-x=u\,$ which gives $$\int_{-1}^0 \frac {\log^n(1+x)}{x^m} dx$$ and expanding into series we have: $\int_{-1}^0x^{-m}(\sum_{k=1}^{\infty}\frac{(-1)^{k+1}x^k}{k})^n\,dx$ Now this might be doable with a computer using Cauchy product's but otherwise it's a madness.

Another try is to let $I(k)=\int_0^1 \frac {x^k}{(1-x)^m}\,dx$ And take derivate n times while assuming $k\ge n$ so: $$\frac{d^n}{dx^n}I(k)=\int_0^1\frac{x^k\log^n(x)}{(1-x)^m}dx$$ Plugging $(1-x)^{-m}=\sum_{j=0}^{\infty} \binom{-m}{j}(-1)^jx^j $ in integral and make use of Tonelli s theorem we get: $$\frac{d^n}{dx^n}I(k)=\sum_{j=0}^{\infty} \binom{-m}{j}(-1)^j\int_0^1 x^{(k+j)}\log^n(x)dx=\sum_{j=0}^{\infty} \binom{-m}{j}(-1)^{(n+j)} n! (k+j+1)^{-(n+1)}$$ But I don't know how to evaluate the latter series.

Answer 1

$$\int_{0}^{1} x^s (1-x)^{-m}\,dx =B(s+1,1-m)=\frac{\Gamma(s+1)\Gamma(1-m)}{\Gamma(s+2-m)}$$ and both sides can be differentiated with respect to $s$ multiple times, then evaluated at $s\to 0^+$.
For differentiating the RHS it is practical to exploit $f'(z)=f(x)\cdot\frac{d}{dz}\log f(z)$ and the fact that $\psi(x)=\frac{d}{dx}\log\Gamma(x)$ fulfills $$ \psi'(a)=\sum_{n\geq 0}\frac{1}{(n+a)^2} $$ hence $\int_{0}^{1}\frac{\log(x)^n}{(1-x)^m}\,dx$ is naturally related to the values of $\zeta(s)$ for $s\in\{2,3,4,\ldots\}$.

Answer 2

Be careful: for convenience of the following derivation I have changed $m$ to $m+1$.

We are going to prove that for all integer $n>m\ge0$: $$ S(n,m):=\int_0^1\frac{\log^n(1-u)}{u^{m+1}}du=\frac{(-1)^n n!}{m!}\sum_{i=0}^{m}{m \brack i}\zeta(n+1-i).\tag{1} $$ where ${m \brack i}$ are the Stirling numbers of the first kind and $\zeta(n)$ are the Riemann functions.

First we check that the expression is valid for $m=0$ and arbitrary $n>0$: $$\begin{align} (-1)^nS(n,0)&=(-1)^n\int_0^1\frac{\log^n(1-u)}{u}du\\ &\stackrel{1-u\mapsto e^{-t}}{=}\int_0^{\infty}\frac{t^n e^{-t}}{1-e^{-t}}dt\\ &=\int_0^{\infty} t^n\sum_{k=1}^\infty e^{-kt}\; dt\\ &=\sum_{k=1}^\infty\int_0^{\infty} t^n e^{-kt}\; dt\\ &\stackrel{t\mapsto z/k}{=} \sum_{k=1}^\infty\frac{1}{k^{n+1}} \int_0^{\infty}z^n e^{-z}\; dz\\ &=n!\zeta(n+1). \end{align}$$

Assume now that (1) is valid for some $m\ge0$ and arbitrary $n> m$. We will show that this implies that the expression is valid for $m+1$ and arbitrary $n> m+1$.

$$\begin{align} S(n,m)&=\int_0^1\frac{\log^{n}(1-u)}{u^{m+1}}du\\ &=-\frac{1}{n+1}\underbrace{\left[\frac{(1-u)\log^{n+1}(1-u)}{u^{m+1}}\right]_0^1}_{=0}\\ &\quad\quad+\frac{1}{n+1}\int_0^1\left(\frac{m}{u^{m+1}}-\frac{m+1}{u^{m+2}}\right)\log^{n+1}(1-u)du\\ &=\frac{m}{n+1}S(n+1,m)-\frac{m+1}{n+1}S(n+1,m+1) \end{align}$$ or $$\begin{align} S(n+1,m+1)&=\frac{m}{m+1}S(n+1,m)-\frac{n+1}{m+1}S(n,m)\\ &\stackrel{I.H.}{=}\frac{m}{m+1}\frac{(-1)^{n+1}(n+1)!}{m!}\sum_{i=0}^{m}{m \brack i}\zeta(n+2-i)\\ &\quad\quad-\frac{n+1}{m+1}\frac{(-1)^n n!}{m!}\sum_{i=0}^{m}{m \brack i}\zeta(n+1-i)\\ &=\frac{(-1)^{n+1}(n+1)!}{(m+1)!}\left[\sum_{i=0}^{m}m{m \brack i}\zeta(n+2-i)+\sum_{i=1}^{m+1}{m \brack i-1}\zeta(n+2-i)\right]\\ &\stackrel{*}{=}\frac{(-1)^{n+1}(n+1)!}{(m+1)!}\sum_{i=0}^{m+1}{m+1 \brack i}\zeta(n+2-i), \end{align}$$ where in ($\stackrel{*}{=}$) the well-known recurrence identity: $$ m{m \brack i}+{m \brack i-1}={m+1 \brack i} $$ was used.

Thus, by induction the claim $(1)$ is proved.

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Note added:

If one considers formally the case of "negative" $m$ an interesting kind of symmetry can be observed:

$$ \int_0^1u^m\log^n(1-u)\;du=(-1)^n n!\sum_{i=0}^{m}\binom{m}{i}\frac{(-1)^i}{(i+1)^{n+1}}. $$