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.

  • $\begingroup$ your approach is correct and standard $\endgroup$ – SiXUlm Dec 30 '15 at 22:24
  • $\begingroup$ @SiXUlm so I see the pattern $1 = A_1 \left[(x+1)(x+2)(x+3)\cdot\cdot\cdot (x+n)\right] + A_2 \left[x(x+2)(x+3)\cdot\cdot\cdot (x+n) \right] + A_3 \left[x(x+1)(x+2)\cdot\cdot\cdot (x+n)\right]$ but I'm not sure how to continue. $\endgroup$ – rgarci0959 Dec 30 '15 at 22:31
  • $\begingroup$ Now, substitute $x=0$, you will have is $A_1 n! = 1$ . So $A_1 = 1/n!$. Next, substitute $x = -1$, you have $-A_2 (n-1)! = 1$, and so on for $x = -2, ..., -n$ $\endgroup$ – SiXUlm Dec 30 '15 at 22:38

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.


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


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