# Computing $\int_{0}^{\infty} \frac{1}{(x+1)(x+2)…(x+n)} \mathrm dx$

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

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

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## 3 Answers

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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Somehow I've never seen this trick before - very clever! (This doesn't work so well unless all terms are inverse-linear, but it's going straight into my toolbox...) – Steven Stadnicki Feb 28 '12 at 0:28
Thank you very much for this answer! – Chon Feb 28 '12 at 10:36

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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From my answer, you can see that the coefficient is $\frac{1}{(1-k)(2-k)\dots((k-1)-k)(k+1-k)\dots(n-k)} = \frac{(-1)^k}{(k-1)!(n-k)!}$ – Aryabhata Feb 28 '12 at 1:22
@Aryabhata: Many thanks! The first coefficient is positive. Shouldn't it be $\dfrac{(-1)^{k-1}}{(k-1)!(n-k)!}$, for $k=1,2,\dots n$? – Américo Tavares Feb 28 '12 at 1:35

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Yes, it is $(-1)^{k-1}$. – Aryabhata Feb 28 '12 at 1:38
So now we have to find $$\lim_{a\rightarrow\infty} \sum_{k=1}^n \frac{(-1)^{k-1}}{(k-1)!(n-k)!}\ln(1+\frac{a}{k})$$ – Chon Feb 28 '12 at 8:59
@Chon: Actually, the part with the limit to infinity is $0$. It's the lower bound that gives something interesting. I've computed the first few values for $n=2,3$ and $4$ and they are resp. $\ln 2$, $\ln(2/\sqrt{3})$ and $\ln(2/\sqrt{3})$. – Raskolnikov Feb 28 '12 at 9:25
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