I need to find the general nth degree polynomial expansion of $x(x+1)(x+2)\dots(x+n)$ in order to solve an integral. Expanding $x(x+1)(x+2)\dots(x+n)$ has been challenging for me.
Say $n=5$, you will notice many patterns like $x^n+(n-1)x^{n-1}+\dotsb$ et cetera, but does there exist a general polynomial expansion for any $n$? This has been driving me nuts. I need a whiteboard.
The integral is $$\frac{1}{x(x+1)(x+2)\dots(x+n)}$$ and will be easy once I expand the polynomial. I am trying to avoid partial fraction decomposition.
 A: The roots of the polynomial $x(x+1)(x+2)\dots(x+n)$ are the numbers $0,-1,-2\dots,-n$. Therefore by Vieta's theorem you can expand
$x(x+1)(x+2)\dots(x+n)=x^{n+1}+a_1x^n+a_2x^{n-1}+\dots+a_nx+a_{n+1}$, 
where $a_k=(-1)^k\sigma_k(0,-1,-2,\dots,-n)$, where $\sigma_k(x_1,\dots,x_{n+1})$ is the $k$-th elementary symmetric polynomial:
$\sigma_k(x_1,\dots,x_{n+1})=\sum_{1\leq i_1<i_2<\dots<i_k\leq n+1}x_{i_1}x_{i_2}\dots x_{i_k}$
A: $f(x) = \frac{1}{(x-a)(x-b)\cdots(x-k)} = \frac {A}{x-a} + \frac {B}{x-b} \cdots \frac {K}{(x-k)}$
$A = \lim_\limits {x\to a} (x-a)f(x) = \frac {1}{(a-b)(a-c)\cdots(a-k)}$
$f(x) = {1}{(x)(x+1)\cdots(x+n)} = \frac 1{n!}\frac {1}{x} - \frac {1}{(n-1)!}\frac {1}{x+1} + \frac {1}{2(n-2)!}\frac {1}{x+2} \cdots\\
f(x) = \sum_{i=0}^n\frac {(-1)^n}{i!(n-i)!} \frac{1}{x+i}$
By the way, supposing you had a nice way to expand the polynomial in the denominator, then what?  It still does not integrate without the partial fraction decomposition.
A: Even if you were to find coefficients $\{a_m\}_{m = 1}^{n+1}$ such that $$f_n(x) = \prod_{k=0}^n (x+k) = \sum_{m=1}^{n+1} a_m x^m$$ this doesn't help you to evaluate $$\int \frac{1}{f_n(x)} \, dx.$$  The reason is because even a "simple" polynomial in the denominator such as $$\int \frac{dx}{x^3 + 5x + 1}$$ does not admit a straightforward antiderivative.  You have also not proposed a mechanism by which the reciprocal of general polynomials with real coefficients could be integrated.
The reason why $1/f_n(x)$ can be integrated in a relatively straightforward manner is precisely because it is the product of factors of linear degree over the integers; thus it is amenable to partial fraction decomposition in a way that my earlier example is not (although it is expressible in terms of the roots of $x^3 + 5x + 1$).
My recommendation is to find coefficients $\{b_j\}_{j=1}^{n+1}$ such that $$\frac{1}{f_n(x)} = \sum_{j=1}^{n+1} \frac{b_j}{x+j} = \frac{1}{f_n(x)} \sum_{j=1}^{n+1} b_j \prod_{k\ne j} (x+k).$$  This is not difficult:  for example, when $x = -j$, we find $$b_j = \left(\prod_{k\ne j} (k-j) \right)^{-1}.$$  This immediately leads to the desired decomposition and the evaluation of the integral is trivial.
A: Let us find the partial fraction decomposition of the rational function
$$\frac{1}{x(x + 1)(x + 2) \cdots (x + n)}, \quad n = 0,1,2,\ldots$$
using the Heaviside cover-up method.
As we have $n + 1$ distinct linear factors we can write the rational function as
$$\frac{1}{x(x + 1)(x + 2) \cdots (x + n)} = \frac{A_0}{x} + \frac{A_1}{x + 1} + \frac{A_2}{x + 2} + \cdots + \frac{A_n}{x + n}.$$
To find each of the unknown constants $A_0$ through to $A_n$ we "cover-up" the first linear factor in the denominator on the left hand side then substitute in the value of $x$ that makes the factor covered up equal to zero which then gives the value for $A_0$. In the first case $x = 0$. We then repeat this process for each linear factor in the denominator. Doing so we find
\begin{align*}
A_0 &= \frac{1}{1.2.\ldots n} = \frac{1}{0! n!}\\
A_1 &= \frac{1}{(-1).1.2\ldots (n - 1)} = -\frac{1}{1!(n - 1)!}\\
A_2 &= \frac{1}{(-2)(-1).1.2\ldots (n - 2)} = \frac{1}{2!(n - 2)!}\\
& \vdots\\
A_n &= \frac{1}{(-1)(-n + 1)(-n + 2) \cdots (-1)} = \frac{(-1)^n}{n! 0!}
\end{align*}
So we see the $k$th constant $A_k$ where $0 \leqslant k \leqslant n$ will be given by
$$A_k = \frac{(-1)^k}{(n - k)! k!}, \quad 0 \leqslant k \leqslant n.$$
The partial fraction decomposition for the rational function will therefore be
$$\frac{1}{x(x + 1)(x + 2) \cdots (x + n)} = \sum_{k = 0}^n \frac{(-1)^k}{(n - k)! k!} \frac{1}{x + k}.$$
Now the integral you are after can be readily found. Here we have
$$\int \frac{dx}{x(x + 1)(x + 2) \cdots (x + n)} = \sum_{k = 0}^n \frac{(-1)^k}{(n - k)! k!} \ln |x + k| + C.$$
