Evaluate the integral $\int \frac{dx}{x(x+1)(x+2) \cdot \cdot \cdot (x+n)}$

Question

I found the integral in this pdf. $$\int \frac{dx}{x(x+1)(x+2) \cdot \cdot \cdot (x+n)}$$

My first thought was to use partial fraction decomposition. Namely, $$ \begin{align*}\prod_{k=0}^n \frac{1}{(x+k)} &= \sum_{k=0}^n \frac{A_k}{(x+k)} \\ 1&=\prod_{k=0}^n(x+k) \sum_{k=0}^n \frac{A_k}{(x+k)} \end{align*} $$ Where $A_k$ is a constant. However, I'm not even sure if my progress is correct, or if there is a simpler way to approach this. Any help would be appreciated.

Answer 1

In general, if $P(x)$ is a polynomial whose roots $\lambda_1, \ldots, \lambda_n$ are all simple and $Q(x)$ is another polynomial with degree lower than that of $P(x)$, we have $$\frac{Q(x)}{P(x)} = \sum_{k=1}^n \frac{Q(\lambda_k)}{(x-\lambda_k)P'(\lambda_k)}$$ In particular, if $\displaystyle\;P(x) = \prod_{k=1}^n (x-\lambda_k)\;$, then $\displaystyle\;P'(\lambda_k) = \prod_{j=1,\ne k}^n (\lambda_k - \lambda_j)\;$.

For your case, this becomes $$\prod_{k=0}^n\frac{1}{x+k} = \sum_{k=0}^n\frac{1}{x+k}\prod_{j=0,\ne k}^n\frac{1}{j-k} = \sum_{k=0}^n\frac{(-1)^k}{(x+k)(n-k)!k!} $$ The partial fraction decomposition for the integrand isn't that complicated at all.

Answer 2

$$ \prod_{k=0}^n \frac{1}{x+k} = \sum_{k=0}^n \frac{A_k}{x+k}, $$ $$ \prod_{m=0,\,m\not= k}^n \frac{1}{x+m} = A_k + \sum_{m=0,\,m\not= k}^n \frac{x+k}{x+m}A_m, $$ $$ x\to -k: \quad\prod_{m=0,\,m\not= k}^n \frac{1}{m-k} = A_k.$$