TITLE: Is there an analytical solution to $\int_1^\infty \frac {dx}{\prod_{i=0}^n (x+i)}$
QUESTION [11 upvotes]: In a problem from physics, I have to deal with this apparently simple function $$I_n=n! \int_1^\infty \frac {dx}{\prod_{i=0}^n (x+i)}$$ $(n\geq 1)$, the result of which being, for sure, something like $\log\left(\frac {p_n}{q_n}\right)$ where $p_n,q_n$ are whole numbers which become to be very large even for small values of $n$ (e.g : $p_5=67108864$, $q_5=61509375$).
For any particular $n$, the value of $I_n$ can be exactly computed (partial fraction decomposition) but I need accurate (or better, exact) results for large values of $n$.
In the past, someone proposed as a first approximation   $$I_n\approx\frac{0.694}{n^{1.285}}$$ which I have been able to reproduce almost exactly curve fitting the values for $1\leq n \leq 20$. But this is too inaccurate for my application. For example, the above formula would give $I_{100}\approx 0.001868$ for an exact value $\approx 0.002011$.
Would someone have an idea either for an exact solution (probably a reccurence relation ?) or a much better approximation ? 
Thanks in advance.
Edit
Thanks to heropup's exact solution and Winther's work around the asymtotics
$$I_n = \frac{1}{n\log(n)}\left[1 - \frac{\gamma}{\log(n)} + \frac{\gamma^2 + \frac{\pi^2}{6}}{\log^2(n)} + \frac{\psi ^{(2)}(1)-\gamma ^3-\frac{\gamma  \pi ^2}{2}}{\log^3(n)} +  \cdots \right]  $$ incredible progress has been done.
I give below a few numbers for illustration
 $$\left(
\begin{array}{ccc}
n & \text{exact} & \text{approx} \\
 100 & 0.00201125049 & 0.00198077851 \\
 200 & 0.00088234568 & 0.00087283768 \\
 300 & 0.00054857248 & 0.00054365188 \\
 400 & 0.00039259469 & 0.00038949113 \\
 500 & 0.00030329103 & 0.00030111383 \\
 600 & 0.00024583983 & 0.00024420737 \\
 700 & 0.00020596177 & 0.00020468064 \\
 800 & 0.00017675824 & 0.00017571890 \\
 900 & 0.00015450281 & 0.00015363806 \\
 1000 & 0.00013701199 & 0.00013627810
\end{array}
\right)$$
Using the exact values for  $100 \leq n \leq 1500$ (step $\Delta n=50$) and performing a linear regression based on the suggested model $$I_n\approx \frac{1}{n\log(n)} \sum_0^3\frac {a_i}{\log^i(n)}$$ leads to
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\
 a_0 & +0.99283 & 0.0005 & \{+0.992,+0.994\} \\
 a_1 & -0.48997 & 0.0076 & \{-0.506,-0.474\} \\
 a_2 & +1.82615 & 0.0418 & \{+1.740,+1.912\} \\
 a_3 & -4.52490 & 0.0755 & \{-4.681,-4.369\} \\
\end{array}$$ giving a maximum absolute error $<10^{-8}$. This is already an incredibly good fit.

REPLY [5 votes]: 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.