Integral Identity for Permanent of a Matrix Say we have an $N\times N$ matrix $A$ with elements $A_{ij} = \alpha_i^{(1- \delta_{ij})}$. For example, for $N=3$ we have
$$ A = \left[ \begin{array}{ccc} 
1 & \alpha_1 & \alpha_1 \\ 
\alpha_2 & 1 &  \alpha_2 \\ 
\alpha_3 & \alpha_3 &  1  \\
 \end{array}\right]$$
How can we show that $A$ satisfies the identity
\begin{equation}
 \text{perm}(A) =  \int^{\infty}_{0} ds\, e^{-s} \prod_{\ell=1}^{n} \Big[1+(s-1) \alpha_i\Big]
\end{equation}
More generally are there any other known integral identities for the permanent of a (perhaps more general) matrix?
 A: For the particular question, do something like this
$$Perm(A)=\sum _{\sigma \in S_n}\prod _{i=1}^nA_{i,\sigma (i)}=\sum _{\sigma \in S_n}\prod _{i=1}^n\alpha _i^{1-\delta _{i,\sigma (i)}},$$
so we have to know exactly where there are fixed points, for a permutation $\sigma, $call $Fix(\sigma)=\{i:\sigma (i)=i\}.$. Notice then than fixing a set $X\subseteq [n]=\{1,\cdots ,n\}$ we can iterate over this.
$$perm(A)=\sum _{X\subseteq [n]}\sum _{\substack{\sigma \in S_n\\ Fix(\sigma)=X}}\prod _{i=1}^{n}\alpha _i^{1-\delta _{i,\sigma (i)}}=\sum _{X\subseteq [n]}\sum _{\substack{\sigma \in S_n\\ Fix(\sigma)=X}}\prod _{x\in [n]\setminus X}\alpha _x=\sum _{X\subseteq [n]}D_{n-|X|}\prod _{x\in [n]\setminus X}\alpha _x,$$
where the last step is noticing that the argument of the sum does not depend really on the sum index. The number of permutations with exactly $|X|$ fixed points are counted by Derangements, denoted by $D_n$.
Now, in the alternative universe of the right hand side,
notice that the product can be seen as
$$\prod _{i=1}^n(1+(s-1)\alpha _i)=\sum _{X\subseteq [n]}(s-1)^{|X|}\prod _{x\in X}\alpha _x,$$
so the integral becomes
$$\sum _{X\subseteq [n]}\prod _{x\in X}\alpha _x\int _{0}^{\infty}(s-1)^{|X|}e^{-s}ds,$$
so the only thing left to check is that $$D_n=\int _{0}^{\infty}(s-1)^{n}e^{-s}ds.$$
I will suggest either expanding $(s-1)^n$ using the binomial expansion and deducing that these are indeed the Derangements or trying integration by parts to deduce a recursion.
For your follow up question, this will help.
