Simple equivalent of $\int_0^\infty\frac{dx}{(x+1)(x+2)...(x+n)}$ when $n\to\infty$ 
Give a simple equivalent of $\displaystyle\int_0^\infty\frac{1}{(x+1)(x+2)...(x+n)}dx$ (when $n\rightarrow\infty$)

My attempt:
I proved by decomposition into simple elements that:
$$\int_0^\infty\frac{1}{(x+1)(x+2)...(x+n)}dx=\frac{1}{n!}\displaystyle\sum_{k=1}^n(-1)^kk{n\choose k}\ln k$$
 A: Note that
$$
\begin{align}
&\int_0^\infty\frac{(n-1)\,\mathrm{d}x}{(x+1)(x+2)\cdots(x+n)}\\
&=\int_0^\infty\frac{\mathrm{d}x}{(x+1)(x+2)\cdots(x+n-1)}
-\int_0^\infty\frac{\mathrm{d}x}{(x+2)(x+3)\cdots(x+n)}\\
&=\int_0^\infty\frac{\mathrm{d}x}{(x+1)(x+2)\cdots(x+n-1)}
-\int_1^\infty\frac{\mathrm{d}x}{(x+1)(x+2)\cdots(x+n-1)}\\
&=\int_0^1\frac{\mathrm{d}x}{(x+1)(x+2)\cdots(x+n-1)}\tag{1}
\end{align}
$$
Therefore,
$$
\begin{align}
\int_0^\infty\frac{\mathrm{d}x}{(x+1)(x+2)\cdots(x+n)}
&=\frac1{n-1}\int_0^1\frac{\mathrm{d}x}{(x+1)(x+2)\cdots(x+n-1)}\\
&=\frac1{n-1}\int_0^1\frac{\Gamma(x+1)}{\Gamma(x+n)}\mathrm{d}x\tag{2}
\end{align}
$$
For $x\in[0,1]$, Gautschi's Inequalty says that $\frac{n^{1-x}}{n!}\le\frac1{\Gamma(x+n)}\le\frac{(n+1)^{1-x}}{n!}$. Furthermore, $1-\gamma\,x\le\Gamma(x+1)\le1$ for $x\in[0,1]$. Thus, $(2)$ is greater than or equal to
$$
\frac1{(n-1)n!}\int_0^1(1-\gamma\,x)n^{1-x}\,\mathrm{d}x=\frac1{n!\log(n)}\left(1-\gamma\left(\frac1{\log(n)}-\frac1{n-1}\right)\right)\tag{3}
$$
and less than or equal to
$$
\frac1{(n-1)n!}\int_0^1(n+1)^{1-x}\,\mathrm{d}x=\frac{n}{n-1}\frac1{n!\log(n+1)}\tag{4}
$$
Therefore,
$$
\int_0^\infty\frac{\mathrm{d}x}{(x+1)(x+2)\cdots(x+n)}\sim\frac1{n!\log(n)}\tag{5}
$$
A: $Edit: Thank to @Did and @Robjohn I have succeeded:
I am going to use Lebesgue's Dominated Convergence Theorem.
Let In denote the nth integral and $H_n=\sum_{k=1}^n\frac{1}{k}$,
$$I_n=\frac{1}{n!}\displaystyle\int_0^\infty\frac{1}{(1+x)(1+\frac{x}{2})...(1+\frac{x}{n})}dx=\frac{1}{n!H_n}\displaystyle\int_0^\infty\frac{1}{(1+\frac{u}{H_n})...(1+\frac{u}{nH_n})}dx$$ with $u=xH_n$
Then we have  $(1+\frac{u}{H_n})...(1+\frac{u}{nH_n})\xrightarrow[n\to+\infty]. e^u\tag{1}$   
Proof: $$ln((1+\frac{u}{H_n})...(1+\frac{u}{nH_n}))=\sum_{k=1}^n ln(1+\frac{u}{kH_n})$$
Yet we know that : $\forall u\in[0,1],  u-\frac{u^2}{2}\leq ln(1+u)\leq u$
$$\sum_{k=1}^n(\frac{u}{kH_n}-\frac{u^2}{2k^2H_n^2})\leq\sum_{k=1}^n ln(1+\frac{u}{kH_n})\leq \frac{u}{H_n}\sum_{k=1}^n \frac{1}{k}$$
$$u-\frac{u^2}{2H_n^2}\sum_{k=1}^n\frac{1}{k^2}\leq\sum_{k=1}^n ln(1+\frac{u}{kH_n})\leq u$$
Furthermore, the series $\sum\frac{1}{k^2}< \infty$, so $\sum_{k=1}^\infty\frac{1}{k^2}=S$ 
Therefore, $$e^u\leq \lim_{n\rightarrow\infty} \prod_{k=1}^n (1+\frac{u}{kH_n})\leq e^u$$ QED
The last step: 
$$(1+\frac{u}{H_n})...(1+\frac{u}{nH_n})\geq 1+u+\frac{u^2}{H_n^2}(\sum_{1\leq i<j\leq n} \frac1{ij})$$
$$=1+u+\frac{u^2}{H_n^2}(H_n^2-\sum_{1}^n\frac1{i^2})$$
$$\geq 1+u+u^2(1-\frac{\pi^2}{6H_n^2})$$
$$\geq 1+u+\frac{u^2}{2}\tag{2}$$ 
(1)+(2)+Lebesgue's Dominated Convergence Theorem $\Longrightarrow \int_0^\infty\frac{\mathrm{d}x}{(x+1)(x+2)\cdots(x+n)}\sim\frac1{n!\ln(n)}$
A: Let $I_n$ denote the $n$th integral, then the lower bound $1/(1+u)\geqslant\mathrm e^{-u}$, valid for every $u\geqslant0$, yields
$$
\frac1{(x+1)\cdots(x+n)}=\frac1{n!}\prod_{k=1}^n\frac1{1+\frac{x}k}\geqslant\frac1{n!}\mathrm e^{-xH_n},
$$
where $H_n=\sum\limits_{k=1}^n\frac1k$ denotes the $n$th harmonic number, hence $H_n\sim\log n$. Integrating this, one gets, for every $n\geqslant2$,
$$
I_n\geqslant \frac1{n!H_n},
$$
and I would be inclined to think, but  did not prove at the moment, that $1/(n!\log n)$ is also the correct equivalent of $I_n$ when $n\to\infty$. 
The best I can prove at the moment uses the lower bound $(x+1)\cdots(x+n)\geqslant\frac12n!(x+1)(x+2)$, valid for every $x\geqslant0$. Thus, for every $n\geqslant2$,
$$
I_n\leqslant\frac2{n!}I_2=\frac{2\log2}{n!}.
$$
