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
$$ \frac{1}{(x+1)(x+2)...(x+n)} \; ?$$
Tell me more
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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 $$\frac{1}{(n-1)!}\sum_{k=0}^{n-1} (-1)^{k-1} {n-1 \choose k} \ln(1+k) \; .$$ |
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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 $P(x)$. |
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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. |
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