Computing $\int_0^\infty\frac1{(x+1)(x+2)\cdots(x+n)}\mathrm dx $ I would like to compute:
$$\int_0^\infty\frac1{(x+1)(x+2)\cdots(x+n)}\mathrm dx $$ $$ n\geq 2$$
So my question is how can I find the partial fraction expansion of
$$\frac1{(x+1)(x+2)\cdots(x+n)}\ ?$$
 A: 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.
A: 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
$$\frac{1}{(n-1)!}\sum_{k=0}^{n-1} (-1)^{k-1} {n-1 \choose k} \ln(1+k) \; .$$
A: 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 $P(x)$.
