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

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.

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.

• Gorgeous. Out of curiosity, what led you to see the Gamma function? – Brevan Ellefsen Jun 25 '16 at 15:43
• @BrevanEllefsen not sure I understand your question...would you elaborate? – Pierpaolo Vivo Jun 25 '16 at 15:47
• of course! I was working on solving this problem when I saw your answer, and I was instantly intrigued, as I failed to see the integral substitutions to a form where the Gamma function could be extracted. I am simply wondering what your thought process was so that I could improve in the future in spotting such transforms. – Brevan Ellefsen Jun 25 '16 at 15:50
• @BrevanEllefsen glad you liked the answer. I think when you have integer exponents as in this case, it is always good to try to exploit the binomial theorem first, to get a finite (and not an infinite) summation, so the first substitution was quite natural. As for the second, I think with a little bit of experience you realize that logarithms in the integrand are usually quite nasty, so the combination power x logarithm can be often traded for exp x power, and that led immediately to the solution (I had not seen the Gamma coming when I started off, though...) – Pierpaolo Vivo Jun 25 '16 at 15:57
• wonderful! Thank you for taking the time to explain all that. – Brevan Ellefsen Jun 25 '16 at 16:06

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.

Another closed form follows by differentiating the beta function multiple times and applying the Faà di Bruno's formula.

Claim. For positive integers $m$ and $n$, $$\mathcal{J}_{n,m} := \int_0^1 x^{n-1}\log^m (1-x) \, \mathrm{d}x = (-1)^m \frac{m!}{n} \sum_{\alpha\in I_m} \prod_{k=1}^m \frac{1}{\alpha_k!} \bigg(\frac{H_n^{(k)}}{k}\bigg)^{\alpha_k} \tag{1}$$ where $\alpha$ runs over the set of indices $$I_m = \{(\alpha_1,\cdots,\alpha_m)\in\Bbb{N}_0^m : 1\cdot\alpha_1+\cdots+m\cdot\alpha_m=m\}.$$

This formula gives an almost explicit formula for $\mathcal{J}_{n,m}$ in terms of polynomial of $H_n^{(1)}, \cdots, H_n^{(n)}$ at the expense of introducing certain combinatorial object, namely $I_m$.

Proof. Notice that

$$\int_0^1 x^{n-1}(1-x)^s \, \mathrm{d}x = \frac{(n-1)!}{(s+1)\cdots(s+n)} = (n-1)!\exp\left(-\sum_{j=1}^n \log(s+j) \right).$$

Letting $f(s) = -\sum_{j=1}^n \log(s+j)$ and applying the Faà di Bruno's formula, we have

$$\mathcal{J}_{n,m} = (n-1)!e^{f(0)} \sum_{\alpha \in I_m} m! \prod_{k=1}^{m} \frac{1}{\alpha_k !} \bigg( \frac{f^{(k)}(0)}{k!} \bigg)^{\alpha_k}. \tag{2}$$

Plugging $f(0) = -\log n!$ and

$$f^{(k)}(0) = \sum_{j=1}^n (-1)^k (k-1)! (s+j)^{-k} \bigg|_{s=0} = (-1)^k (k-1)! H_n^{(k)}$$

into $\text{(2)}$ and simplifying the resulting expression yields $\text{(1)}$.

• how do you estimate the $m$th derivative of $f(z) = B(n,z+1)$ at $z=1$ with di Bruno's formula ? – reuns Sep 9 '16 at 23:07
• @user1952009, My idea is rather straightforward, and it is to write $B(n, s+1) = (n-1)!\exp(f(s))$ with $f(s)=\log\Gamma(1+s)-\log\Gamma(n+1+s)$ and apply the Faà di Bruno's formula. – Sangchul Lee Sep 9 '16 at 23:11
• ok tks, I missed that relation between $\log B(n,.)$ and $\log \Gamma(.)$ (and for the reader : $B(n,s+1) = \int_0^1 x^{n-1} (1-x)^s dx = \sum_{k=0}^\infty \frac{\int_0^1 x^{n-1} \log^k(1-x) dx}{k!} s^k$ is the generating function of $\int_0^1 x^{n-1} \log^k(1-x) dx$) – reuns Sep 9 '16 at 23:14