5

First let me introduce the two recursive relations:

$$\int_0^1 x^{n-1}\ln^a(1-x)dx=f(a,n),$$

where

$$f(a,n)=(a-1)!\sum_{j=0}^{a-1}\frac{(-1)^{a-j}}{j!}f(j,n) H_n^{(a-j)},\quad f(0,n)=\frac1n.\tag{1}$$

Cases using Mathematica:

graph1

$$(-1)^a\frac{\ln^a(1-x)}{1-x}=\sum_{n=0}^\infty g(a,n) x^n,$$

where

$$g(a,n)=-(a-1)!\sum_{j=0}^{a-1}\frac{(-1)^{a-j}}{j!}g(j,n)H_n^{(a-j)},\quad g(0,n)=1.\tag{2}$$

Cases using Mathematica:

graph2

Question: Are $(1)$ and $(2)$ known in the literature? If so, any reference?

Proof of $(1)$:

Take the logarithm of both sides of

$$\operatorname{B}(m,n)=\Gamma(n)\prod_{k=0}^{n-1} \frac{1}{k+m},$$

we have \begin{gather*} \ln\operatorname{B}(m,n)=\ln\Gamma(n)+\ln\prod_{k=0}^{n-1}\frac{1}{k+m}\\ \left\{\text{ use $\ln\prod a_n=\sum \ln(a_n)$}\right\}\\ =\ln\Gamma(n)-\sum_{k=0}^{n-1} \ln(k+m). \end{gather*} Differentiate both sides with respect to $m$, $$\frac{\frac{\partial}{\partial m}\operatorname{B}(m,n)}{\operatorname{B}(m,n)}=-\sum_{k=0}^{n-1} \frac{1}{k+m}$$ or \begin{equation} \frac{\partial}{\partial m}\operatorname{B}(m,n)=-\operatorname{B}(m,n)\sum_{k=0}^{n-1} \frac{1}{k+m}. \end{equation}

Let's keep differentiating w.r.t $m$:

\begin{equation} \frac{\partial^2}{\partial m^2}\operatorname{B}(m,n)=\operatorname{B}(m,n)\sum_{k=0}^{n-1} \frac{1}{(k+m)^2}-\frac{\partial}{\partial m}\operatorname{B}(m,n)\sum_{k=0}^{n-1} \frac{1}{k+m}, \end{equation}

\begin{equation} \frac{\partial^3}{\partial m^3}\operatorname{B}(m,n)=-2\operatorname{B}(m,n)\sum_{k=0}^{n-1} \frac{1}{(k+m)^3}+2\frac{\partial}{\partial m}\operatorname{B}(m,n)\sum_{k=0}^{n-1} \frac{1}{(k+m)^2}\\ -\frac{\partial^2}{\partial m^2}\operatorname{B}(m,n)\sum_{k=0}^{n-1} \frac{1}{k+m}, \end{equation}

\begin{equation} \frac{\partial^4}{\partial m^4}\operatorname{B}(m,n)=6\operatorname{B}(m,n)\sum_{k=0}^{n-1} \frac{1}{(k+m)^4}-6\frac{\partial}{\partial m}\operatorname{B}(m,n)\sum_{k=0}^{n-1} \frac{1}{(k+m)^3}\\ +3\frac{\partial^2}{\partial m^2}\operatorname{B}(m,n)\sum_{k=0}^{n-1} \frac{1}{(k+m)^2} -\frac{\partial^3}{\partial m^3}\operatorname{B}(m,n)\sum_{k=0}^{n-1} \frac{1}{k+m}, \end{equation}

so in general $$\frac{\partial^a}{\partial m^a}\operatorname{B}(m,n)=(a-1)!\sum_{j=0}^{a-1}\frac{(-1)^{a-j}}{j!}\frac{\partial^j}{\partial m^j}\operatorname{B}(m,n)\sum_{k=0}^{n-1} \frac{1}{(k+m)^{a-j}}.$$

Let $m$ approach $1$ and call $\displaystyle\left.\frac{\partial^a}{\partial m^a}\operatorname{B}(m,n)\right|_{m\to1}=f(a,n)$,

$$f(a,n)=(a-1)!\sum_{j=0}^{a-1}\frac{(-1)^{a-j}}{j!}f(j,n)H_n^{(a-j)}$$

and the proof finishes on observing \begin{gather*} \left.\frac{\partial^a}{\partial m^a}\operatorname{B}(m,n)\right|_{m=1}=\left.\int_0^1\frac{\partial^a}{\partial m^a} x^{n-1}(1-x)^{m-1}\mathrm{d}x\right|_{m=1}\\ =\left.\int_0^1 x^{n-1}\ln^a(1-x)(1-x)^{m-1}\mathrm{d}x\right|_{m=1}\\ =\int_0^1 x^{n-1}\ln^a(1-x)\mathrm{d}x=f(a,n) \end{gather*}

and $f(0,n)=\int_0^1 x^{n-1}\mathrm{d}x=\frac1n.$

Proof of $(2)$:

We have

$$\frac{1}{(1-x)^m}=(1-x)^{-m}=\sum_{n=0}^\infty \binom{m+n-1}{n}x^n$$

Take the $a$-th derivative of both sides w.r.t $m$

$$(-1)^a\frac{\ln^a(1-x)}{(1-x)^m}=\sum_{n=0}^\infty \frac{\partial^a}{\partial^am}\binom{m+n-1}{n} x^n$$

Let $m\to 1$ and call $\displaystyle\left.\frac{\partial^a}{\partial m^a}\binom{m+n-1}{n}\right|_{m\to 1}=g(a,n)$

$$(-1)^a\frac{\ln^a(1-x)}{1-x}=\sum_{n=0}^\infty g(a,n) x^n$$

Note that

$$\frac{\partial}{\partial m}\binom{m+n-1}{n}=\binom{m+n-1}{n}\left(\psi(m+n)-\psi(m)\right)=\binom{m+n-1}{n}\sum_{k=0}^{n-1}\frac{1}{k+m}$$

which can be generalized to

$$\frac{\partial^a}{\partial m^a}\binom{m+n-1}{n}=-(a-1)!\sum_{j=0}^{a-1}\frac{(-1)^{a-j}}{j!}\frac{\partial^j}{\partial m^j}\binom{m+n-1}{n}\sum_{k=0}^{n-1} \frac{1}{(k+m)^{a-j}}.$$

and so

$$\left.\frac{\partial^a}{\partial m^a}\binom{m+n-1}{n}\right|_{m\to 1}$$ $$=-(a-1)!\sum_{j=0}^{a-1}\frac{(-1)^{a-j}}{j!}\left.\frac{\partial^j}{\partial m^j}\binom{m+n-1}{n}\right|_{m\to 1}\,\left.\sum_{k=0}^{n-1} \frac{1}{(k+m)^{a-j}}\right|_{m\to 1}$$

or

$$g(a,n)=-(a-1)!\sum_{j=0}^{a-1}\frac{(-1)^{a-j}}{j!}g(j,n)H_n^{(a-j)}$$

and the proof completes on observing

$$g(0,n)=\left.\binom{m+n-1}{n}\right|_{m\to 1}=\binom{n}{n}=1$$

Note: A recursive relation, similar to $(2)$, was introduced by @Marko Riedel here.

Ali Shadhar
  • 25,498

1 Answers1

2

As I mentioned in a comment the integral can be expressed by a finite sum. Here is the derivation.

For an integer $m\ge 0$ let

$$f(m,n) = \int_{0}^{1} x^{n-1}\log^m(1-x)\;dx$$

We can rewrite this as

$$\begin{align} &f(m,n) \\ &\overset{x\to 1-e^{-t}}=\int_{0}^{\infty} e^{-t} (1-e^{-t})^{n-1}(-1)^m t^m\;dt\\ &\overset{\text{binom exp}}=\int_{0}^{\infty} e^{-t} \sum_{k=0}^{n-1}\binom{n-1}{k}(-1)^k e^{-k t}(-1)^m t^m\;dt\\ &\overset{\text{sum <-> int}}=(-1)^m\sum_{k=0}^{n-1}\binom{n-1}{k}(-1)^k \int_{0}^{\infty}e^{-(k+1) t} t^m\;dt\\ &\overset{\text{int ->}\Gamma}= (-1)^m\sum_{k=0}^{n-1}\binom{n-1}{k}(-1)^k \Gamma(m+1) \frac{1}{(1+k)^{m+1}}\\ &\overset{\text{k->k-1}}=(-1)^m m!\sum_{k=1}^{n}\binom{n-1}{k-1}(-1)^{k-1} \frac{1}{k^{m+1}} \end {align}$$

which is the announced finite sum.

Dr. Wolfgang Hintze
  • 12,465