I've been trying to solve explicitly the following indefinite integral:
$$\int\frac{dx}{x(x+1)(x+2)\cdot\space...\space\cdot(x+n)}$$
I tried to perform partial fraction decomposition, and after substituting some natural n's, I figured the Binomial Theorem might help here, but I couldn't figure out how to use it.
Thank you and have a good day/night!
Let
$\begin{array}\\ I_n &=\int\dfrac{dx}{x(x+1)(x+2)\cdot\space...\space\cdot(x+n)}\\ &=\int\dfrac{dx}{\prod_{k=0}^n (x+k)}\\ \end{array} $
Let's try partial fractions.
If $\dfrac1{\prod_{k=0}^n (x+k)} =\sum_{k=0}^n \dfrac{a_k}{x+k} $, then $1 =\sum_{k=0}^n \dfrac{a_k\prod_{j=0}^n (x+j)}{x+k} =\sum_{k=0}^n a_k\prod_{j=0,j \ne k}^n (x+j) $.
Setting $x = -m, 0 \le m \le n$,
$\begin{array}\\ 1 &=\sum_{k=0}^n a_k\prod_{j=0,j \ne k}^n (-m+j)\\ &= a_m\prod_{j=0,j \ne m}^n (-m+j)\\ &= a_m\prod_{j=0}^{m-1} (-m+j)\prod_{j=m+1}^n (-m+j)\\ &= a_m(-1)^m\prod_{j=0}^{m-1} (m-j)\prod_{j=m+1}^n (j-m)\\ &= a_m(-1)^m\prod_{j=1}^{m} j\prod_{j=1}^{n-m} j\\ &= a_m(-1)^mm!(n-m)!\\ \end{array} $
so $a_m =\dfrac{(-1)^m}{m!(n-m)!} $.
Therefore
$\begin{array}\\ I_n &=\int\dfrac{dx}{\prod_{k=0}^n (x+k)}\\ &=\int \sum_{k=0}^n \dfrac{a_k}{x+k}dx\\ &=\sum_{k=0}^n a_k\int \dfrac1{x+k}dx\\ &=\sum_{k=0}^n a_k(\ln(x+k)+c_k)\\ &=\sum_{k=0}^n a_k\ln(x+k)+C\\ \end{array} $
This is undoubtedly well-known, but I did work it out independently.
