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I would like to compute:

$ \int_{0}^{\infty} \frac{1}{(x+1)(x+2)...(x+n)} \mathrm dx $ $ n\geq 2$

So my question is how can I find the partial fraction expansion of

$ \frac{1}{(x+1)(x+2)...(x+n)} \; ?$

3 Answers 3

16

HINT: Here's a trick to find partial fraction expansions. Compute

$\lim_{x\to -k} \frac{(x+k)}{(x+1)(x+2)...(x+n)} \; .$

This should give you the coefficient of the term $1/(x+k)$ in the expansion.

EDIT: As Américo points out, the partial fraction expansion is

$\frac{1}{\left( x+1\right) \left( x+2\right) \cdots \left( x+n\right) } =\sum_{k=1}^{n}\frac{(-1)^{k-1}}{(k-1)!\left( n-k\right) !}\cdot\frac{1}{x+k} \; . $

The indefinite integral of that expansion is

$\ln\left( \prod_{k=1}^{n}(x+k)^{\frac{(-1)^{k-1}}{(k-1)!\left( n-k\right) !}} \right) \; .$

When you fill in the upper bound, you can see that the result must be zero as the leading power in $x$ for the product is $0$ because

$0 = (1-1)^{n-1} = \sum_{k=0}^{n-1} \frac{(-1)^{k} (n-1)!}{(k)!\left( (n-1)-k\right) !} = (n-1)! \sum_{k=1}^{n} \frac{(-1)^{k-1}}{(k-1)!\left( n-k\right) !} \; .$

Therefore, we are left with the lower bound

$-\ln\left( \prod_{k=1}^{n}(k)^{\frac{(-1)^{k-1}}{(k-1)!\left( n-k\right) !}} \right) \; .$

For $n=2,3$ and $4$ you get resp. $\ln 2$, $\ln(2/\sqrt{3})$ and $\ln(2^5/3^3)/6$.

The lower bound can also be written as

$\frac{1}{(n-1)!}\sum_{k=0}^{n-1} (-1)^{k-1} {n-1 \choose k} \ln(1+k) \; .$

  • 0
    Thank you very much for this answer!2012-02-28
13

If $\frac{1}{(x+1)(x+2)\dots(x+n)} = \sum_{i=1}^{n} \frac{A_i}{x+i}$

To compute $A_k$, multiply by $(x+k)$ and set $x = -k$.

In fact, this can be used to show, that for any polynomial $P(x)$ with distinct roots $\alpha_1, \alpha_2, \dots \alpha_n$, that

\frac{1}{P(x)} = \sum_{j=1}^{n} \frac{1}{P'(\alpha_j)(x-\alpha_j)}

where P'(x) is the derivative of $P(x)$.

8

Based on my computations in SWP for $2\leq n\leq 8$ I conjecture the following expansion

$\begin{equation*} \frac{1}{\left( x+1\right) \left( x+2\right) \cdots \left( x+n\right) } =\sum_{k=1}^{n}\frac{(-1)^{k-1}}{(k-1)!\left( n-k\right) !}\cdot\frac{1}{x+k}. \end{equation*}$

Added. How to prove or disprove? Induction doesn't seem easy.

Added 2. It follows from Aryabhata's answer. See comment below.

  • 0
    I made a mistake, for $n=4$ the result should be $\ln(2^5/3^3)/6$.2012-02-28