Coefficients of polynomial $(x+1)(x+2)...(x+n)$

Question

I was trying to calculate the integral $$ I(m,n)=\int_0^\infty\frac{x^me^x}{(1-e^x)^n}\mathrm{d}x $$ It's possible to evaluate this in closed form by using the zeta function, for example: $$ I(m,4)=m!(\zeta(m-2)+3\zeta(m-1)+2\zeta(m)) $$ This follows from the fact that $(x+1)(x+2)=x^2+3x+2$. In general, we have that $$ I(m,n)=\sum_{k=1}^\infty\frac{(k+1)(k+2)...(k+n-2)}{k^m} $$ Clearly this will always give a "polynomial" in the sense of sending $k^j\mapsto\zeta(m-j)$.

This got me wondering about the coefficients of the polynomial $$P_n(x)= \prod_{i=1}^n(x+i)$$ Some coefficients are obvious: the first one is $1$, the last one is $n!$, the second one is $n(n+1)/2$. The third one I also managed to evaluate using a recursion relation, however, the answers here and here provide a simpler derivation for the third and fourth coefficients using Vieta's formulas.

It's pretty clear that all the coefficients will be given by polynomials in $n$. Does anyone know how to find an explicit form for these coefficients, perhaps involving finite sums? (For example, robjohn's answers in the links hint at a possible closed-form expression using finite sums of binomial coefficients.)

Answer 1

The coefficient of $x^k$ in $(x+1)\ldots(x+n)$ is $|S_1(n+1,k+1)|$ where $S_1(\cdot,\cdot)$ are the Stirling numbers of the first kind.

Answer 2

$\newcommand{\usn}[2]{\begin{bmatrix}#1 \\ #2 \end{bmatrix}}$

It is well known that the unsigned Stirling numbers of the first kind can be written as

$$\begin{align} \usn{x}{x-1}&=\binom{x}{2} \\ \usn{x}{x-2}&=2\binom{x}{3}+3\binom{x}{4} \\ \usn{x}{x-3}&=6\binom{x}{4}+20\binom{x}{5} + 15\binom{x}{6} \\ \usn{x}{x-4}&=24\binom{x}{5}+130\binom{x}{6} + 210\binom{x}{7} + 105\binom{x}{8} \\ &\vdots \\ \usn{x}{x-n}&=\sum_{k=n+1}^{2n}p_{n,k}\binom{x}{n+k} \end{align}$$ where the $p_{n,k}$ satisfy the recurrence $$p_{n+1,k}=(n+k)(p_{n,k}+p_{n,k-1})\text{.}$$

See OEIS A259456.

Answer 3

If you have the polinomial $$ p(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_{n-s}x^{n-s}+\cdots+a_1x+a_0 $$ with roots $r_1,r_2,\cdots,r_n$, then:

the sum of the products of the roots taken in groups of $m$ without repetition is equal to $(-1)^{m}\dfrac{a_{n-m}}{a_n}$

So, the coefficient you want is:

$$ a_n\cdot(-1)^{m}\cdot\left(\displaystyle\sum_{\begin{smallmatrix} 1<i<j<\cdots<k<n \\ m \small\mbox{ elements}\end{smallmatrix} } r_ir_j\cdots r_k \right)= a_{n-m} $$

For example:

$p(x)=4x^5-2x^4+{\color{red} 1}x^3-x^2+2x-7$ with roots $r_1,\cdots,r_5$, then:

$\dfrac{1}{4}=(-1)^{2}(r_1r_2+r_1r_3+r_1r_4+r_1r_5+r_2r_3+r_2r_4+r_2r_5+r_3r_4+r_3r_5+r_4r_5)$