Evaluating $\int_{0}^{1} \frac{\ln^{n} x}{(1-x)^{m}} \, \mathrm dx$

Question

On another site, someone asked about proving that

$$ \int_{0}^{1} \frac{\ln^{n}x}{(1-x)^{m}} \, dx = (-1)^{n+m-1} \frac{n!}{(m-1)!} \sum_{j=1}^{m-1} (-1)^{j} s (m-1,j) \zeta(n+1-j), \tag{1} $$

where $n, m \in \mathbb{N}$, $n \ge m$, $m \ge 2$, and $s(m-1,j)$ are the Stirling numbers of the first kind.

My attempt at proving $(1)$:

$$ \begin{align}\int_{0}^{1} \frac{\ln^{n}x}{(1-x)^{m}} \, dx &= \frac{1}{(m-1)!} \int_{0}^{1} \ln^{n} x \sum_{k=m-1}^{\infty} k(k-1) \cdots (k-m+2) \ x^{k-m+1} \ dx \\ &= \frac{1}{(m-1)!} \sum_{k=m-1}^{\infty} k(k-1) \cdots (k-m+2) \int_{0}^{1} x^{k-m+1} \ln^{n} x \, dx \\ &= \frac{1}{(m-1)!} \sum_{k=m-1}^{\infty} k(k-1) \cdots (k-m+2) \frac{(-1)^{n} n!}{(k-m+2)^{n+1}}\\ &= \frac{1}{(m-1)!} \sum_{k=m-1}^{\infty} \sum_{j=0}^{m-1} s(m-1,j) \ k^{j} \ \frac{(-1)^{n} n!}{(k-m+2)^{n+1}} \\ &= (-1)^{n} \frac{n!}{(m-1)!} \sum_{j=0}^{m-1} s(m-1,j) \sum_{k=m-1}^{\infty} \frac{k^{j}}{(k-m+2)^{n+1}} \end{align} $$

But I don't quite see how the last line is equivalent to the right side of $(1)$. (Wolfram Alpha does say they are equivalent for at least a few particular values of $m$ and $n$.)

Answer 1

Substitute $x=e^{-t}$ and get that the integral is equal to

$$(-1)^n \int_0^{\infty} dt \, e^{-t} \frac{t^n}{(1-e^{-t})^m} $$

Now use the expansion

$$(1-y)^{-m} = \sum_{k=0}^{\infty} \binom{m+k-1}{k} y^k$$

and reverse the order of summation and integration to get

$$\sum_{k=0}^{\infty} \binom{m+k-1}{k} \int_0^{\infty} dt \, t^n \, e^{-(k+1) t}$$

I then get as the value of the integral:

$$\int_0^1 dx \, \frac{\ln^n{x}}{(1-x)^m} = (-1)^n\, n!\, \sum_{k=0}^{\infty} \binom{m+k-1}{k} \frac{1}{(k+1)^{n+1}}$$

Note that when $m=0$, the sum reduces to $1$; every term in the sum save that at $k=0$ is zero.

Note also that this sum gives you the ability to express the integral in terms of a Riemann zeta function for various values of $m$, which will provide the Stirling coefficients.

Answer 2

Here is another approach based on the beta function

$$ F = \int_{0}^{1}x^{\alpha}(1-x)^{-m} dx = \beta( \alpha+1,1-m ),\quad \alpha+1>0,\, 1-m>0. $$

Now, our integral $I$ can be evaluated as

$$ I = \lim_{\alpha \to 0} \frac{d^n}{d \alpha^n} \beta( \alpha+1,1-m ),\quad n>m-1,\,m\geq 2. $$

Here is a useful identity

$$ {\partial \over \partial x} \mathrm{B}(x, y) = \mathrm{B}(x, y) \left( {\Gamma'(x) \over \Gamma(x)} - {\Gamma'(x + y) \over \Gamma(x + y)} \right) = \mathrm{B}(x, y) (\psi(x) - \psi(x + y)), $$

where $\psi(x)$ is the digamma function.