Is there a closed form for the integral $\int_0^1 x^n \log^m (1-x) \, {\rm d}x$?

Question

Let $n \in \mathbb{N}$. We know that:

$$\int_0^1 x^n \log(1-x) \, {\rm d}x = - \frac{\mathcal{H}_{n+1}}{n+1}$$

Now, let $m , n \in \mathbb{N}$. What can we say about the integral

$$\int_0^1 x^n \log^m (1-x) \, {\rm d}x$$

For starters we know that $\displaystyle \log^m (1-x)=m! \sum_{k=m}^{\infty} (-1)^k \frac{s(k, m)}{k!} x^k$ where $s(k, m)$ are the Stirling numbers of first kind.

Thus

\begin{align*} \int_{0}^{1} x^n \log^m (1-x) \, {\rm d}x &=m! \int_{0}^{1}x^n \sum_{k=m}^{\infty} (-1)^k \frac{s(k, m)}{k!} x^k \\ &= m! \sum_{k=m}^{\infty} (-1)^k \frac{s(k, m)}{m!} \int_{0}^{1}x^{n+m} \, {\rm d}x\\ &= m! \sum_{k=m}^{\infty} (-1)^k \frac{s(k, m)}{m!} \frac{1}{m+n+1} \end{align*}

Can we simplify? I know that Striling numbers are related to the Harmonic number but I don't remember all identities.

Answer 1

Change variables $x=1-t$: $$ I=\int_0^1 x^n \log^m (1-x) \, {\rm d}x=\int_0^1dt (1-t)^n\log^m t=\sum_{k=0}^n {n\choose k}(-1)^{n-k}\int_0^1 dt \ t^{n-k}\log^m t\ . $$ Now change variable $t=\exp(z)$ and get $$ I=\sum_{k=0}^n {n\choose k}(-1)^{n-k}\int_{-\infty}^0 dz\ e^{(n-k+1)z}z^m= \boxed{\Gamma (m+1)\sum_{k=0}^n {n\choose k}\frac{(-1)^{n-k+m} }{ (n+1-k)^{m+1}}}\ , $$ which is a finite sum.

Answer 2

We have $$(-1)^m n \int_0^1 x^{n-1} \ln^m(1-x)dx=\lim_{z\to 0} \frac{d^m}{dz^m} \,n \int_0^1 x^{n-1} (1-x)^{-z}dx\\ =\lim_{z\to 0} \frac{d^m}{dz^m} \, \frac{n!\, \Gamma(1-z)}{\Gamma(n-z+1)}=\lim_{z\to 0} \frac{d^m}{dz^m} \, \prod_{k=1}^n \frac{1}{1-z/k}. \tag{1}$$

But, in view of the generating function of the Complete homogeneous symmetric polynomials, we have $$\prod_{k=1}^n \frac{1}{1-z/k} = \sum_{k=0}^{\infty} h_k\,z^k\tag{2}$$

Where $$h_k \equiv h_k(1,1/2,\dots ,1/n)=\sum_{1 \leq a_1\leq a_2\leq\dots\leq a_k\leq n} \,\frac1{a_1\,a_2\cdots a_k}\tag{3}$$

Hence $$(-1)^m n \int_0^1 x^{n-1} \ln^m(1-x)dx= m!\,h_m. \tag{4}$$

The Newton–Girard formulae connect the symmetric polynomials to their corresponding power sums, which in our case are the generalized harmonic numbers.

The first few cases are: $$-\int_0^1 x^{n-1} \ln(1-x) dx=\frac{H_n}{n}\tag{5}$$ $$\int_0^1 x^{n-1} \ln^2(1-x) dx=\frac{H_n^2+H_n^{(2)}}{n}\tag{6}$$

$$-\int_0^1 x^{n-1} \ln^3(1-x) dx=\frac1{n}(H_n^3+3H_n\,H_n^{(2)}+2H_n^{(3)})\tag{7}$$ $$\int_0^1 x^{n-1} \ln^4(1-x) dx=\frac1{n}(H_n^4+6H_n^2\,H_n^{(2)}+3H_n^{(2)2}+8H_n\,H_n^{(3)}+6H_n^{(4)})\tag{8}$$

With some more algebraic manipulations, we can get the form $$f(m)=\int_0^1 x^{n-1} \ln^m(1-x) dx=\frac{m!}{n\,(2m)!} \frac{d^{2m}}{dx^{2m}} \exp\left(\sum_{k=1}^{\infty} \frac{(-1)^k x^{2k}}{k} H_n^{(k)}\right)\Bigg{|}_{x=0}\tag{9}$$

Mathematica code:

f[m_] := m!/(n (2 m)!) D[ Exp[Sum[(-1)^k x^(2 k)/k H[k], {k, 1, 2 m}]], {x, 2 m}] /. x -> 0 /. H[int_] :> HarmonicNumber[n, int] // Simplify

This code becomes less efficient for large $m$'s.