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I've been trying to solve explicitly the following indefinite integral:


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!


marked as duplicate by José Carlos Santos calculus Mar 8 at 0:26

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  • 1
    $\begingroup$ Use the partial fraction expansion given in math.stackexchange.com/q/715706 (I have been able to retrieve it by using "approach0" formula finder). $\endgroup$ – Jean Marie Mar 7 at 21:17


$\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)!} $.


$\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.

  • $\begingroup$ Wonderful!!! Thank you very much! $\endgroup$ – Amit Zach Mar 7 at 21:44

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