Improper integral: $\int_1^{+\infty}\frac{\mathrm dx}{x(x+1)(x+2)\cdots(x+n)}$ I've tried many ways, but it seems that it didn't work:
$$
   \int_1^{+\infty}\frac{\mathrm dx}{x(x+1)(x+2)\cdots(x+n)} 
    = \int_0^1\frac{x^{n-1}}{(x+1)(2x+1)\cdots(nx+1)}\mathrm dx 
    = \cdots
$$
Any help would be appreciated!

We have the partial fraction decomposition
$$\frac{1}{x(x+1)(x+2)\cdots(x+n)}=\frac{a_0}{x}+\frac{a_1}{x+1}+\frac{a_2}{x+2}+\ldots+\frac{a_n}{x+n}$$
Now as $x\to 0$, the part with $a_0$ dominates, and we have
$$a_0=\lim_{x\to0}\frac{x}{x(x+1)(x+2)\cdots(x+n)}=\frac1{n!}$$
Similarily, as $x\to -1$, the part with $a_1$ dominates, and we have
$$a_1=\lim_{x\to-1}\frac{x+1}{x(x+1)(x+2)\cdots(x+n)}=-\frac1{(n-1)!}$$
and if you were to continue this idea, you'd find that
$$a_k=\lim_{x\to -k}\frac{x+k}{x(x+1)(x+2)\cdots(x+n)}= \frac{(-1)^k}{k!(n-k)!}=\frac{(-1)^k {n\choose k}}{n!}$$
so that
$$\frac{1}{x(x+1)(x+2)\cdots(x+n)}=\frac1{n!}\sum_{k=0}^n\frac{(-1)^k {n\choose k}}{(x+k)} $$
and so 
$$\int_1^{\infty}\frac{\mathrm dx}{x(x+1)(x+2)\cdots(x+n)}
=\frac{1}{n!}\int_1^{\infty}\sum_{k=0}^n\frac{(-1)^k {n\choose k}}{(x+k)}\\
=\frac1{n!}\left(\sum_{k=0}^n(-1)^k {n\choose k}\ln(x+k)\right)\left. \right|_1^\infty$$
For $x=\infty$, consider the binomial expansion of $(1-1)^k$. It implies that the sum of the positive binomial terms is equal to the sum of the negative binomial terms, and so if we combine all the logs into one, we get the logarithm of a polynomial divided by a polynomial of the same degree, so the limit to infinity is hence $\ln(1) = 0$. So then the integral is thus equal to the negative of the antiderivative above at $x=1$, so
$$\frac{1}{x(x+1)(x+2)\cdots(x+n)}=\frac1{n!}\sum_{k=0}^n(-1)^{k+1} {n\choose k}\ln(k+1)\\=\frac1{n!}\sum_{k=2}^{n+1}(-1)^{k} {n\choose k-1}\ln(k)$$
