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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.

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2 Answers 2

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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.


Note added:

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

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

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  • $\begingroup$ So according to @Jack answer since it's related to zeta function maybe that series can be simplified even more. Thank you! $\endgroup$
    – Zacky
    Apr 12, 2018 at 12:13
  • $\begingroup$ @Zacky I have completely rewritten the answer to give an explicit expression for the integral. Surprisingly it differs from the expression given in WP, but is correct (and is in my view much simpler and nicer). I did not check the WP formula. $\endgroup$
    – user
    Apr 13, 2018 at 13:13
  • $\begingroup$ Nice! For the record: Define $p_m(x):=\prod_{j=1}^m(x+j-1)$, which encapsulates the (absolute) Stirling numbers. Then in umbral notation, i.e. replacing $\zeta^r$ by $\zeta(r)$ after expanding, this becomes simply $$\int_0^1\frac{\log^n(1-u)}{u^{m+1}}du=\frac{(-1)^n n!}{m!}\zeta^{n}p_m(\frac 1\zeta).$$ $\endgroup$
    – Wolfgang
    May 7 at 13:31
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$$\int_{0}^{1} x^s (1-x)^{-m}\,dx =B(s+1,1-m)=\frac{\Gamma(s+1)\Gamma(1-m)}{\Gamma(s+2-m)}$$ 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 $s\in\{2,3,4,\ldots\}$.

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