Is there an analytical solution to $\int_1^\infty \frac {dx}{\prod_{i=0}^n (x+i)}$

Solution 1:

It is relatively straightforward to show that for positive integers $n$, $$f_n(x) = B(x,n+1) = \frac{\Gamma(x)\Gamma(n+1)}{\Gamma(x+n+1)} = \frac{1}{x\binom{x+n}{n}} = \sum_{k=0}^n \frac{(-1)^k \binom{n}{k}}{x+k}.$$ Consequently, $$\int_{x=1}^\infty f_n(x) \, dx = \lim_{x \to \infty} \sum_{k=0}^n (-1)^k \binom{n}{k} \left(\log (x+k) - \log(1+k)\right).$$ The limit for the upper endpoint terms of course is zero; this leaves us with $$\int_{x=1}^\infty f_n(x) \, dx = \sum_{k=0}^n (-1)^{k+1} \binom{n}{k} \log (1+k).$$ From here we could start to do things with even and odd cases of $n$. I haven't the time to look at large $n$ asymptotics but it should not be difficult.