I'm currently reading a paper regarding the random assignment problem and am stuck at the proof of a certain theorem.
In the random assignment problem you basically have $m$ workers and $n$ tasks to fulfill. Person $i$ carries out the task $j$ with cost $a_{i,j}$. The numbers $a_{i,j}$ form an $(m \times n)$-matrix. We also condition on a set $Z$ of matrix entries being $0$ while the remaining entries are $Exp(1)$-distributed. An $k$-assignment is a set of $k$ matrix entries, no two in the same row or column. The minimum $k$-assignment asks for the $k$-assignment with the least cost out of all possible $k$-assignments.
We condition on the values of all matrix elements except $a_{i,j}$. Let $X$ be the value of the minimal $k$-assignment which does not use $a_{i,j}$. Let $Y$ be the value of the minimal $k-1$-assignment which does not use row $i$ or column $j$. Then the minimum $k$-assignment either contains $(i,j)$ and has value $Y+a_{i,j}$ or does not contain $(i,j)$ and has value $X$. The probability that $a_{i,j}$ belongs to the optimal $k$-assignment is therefore equal to the probability $P[a_{i,j}<X-Y]$. Let $X>Y$ and define $\delta := X-Y$. Also suppose that $(i,j) \notin Z$.
Why does this equality hold: $$ E[\min\{\delta,a_{i,j}\}]=\delta e^{-\delta} + \int_0^\delta te^{-t}dt \text{ ?}$$
To provide more context; Im referencing theorem 2.10 in the following paper: https://arxiv.org/pdf/math/0006146.pdf. I do understand literally every step in the proof of this theorem, yet I don't understand why this equality holds. I'm not sure how to even approach this problem. Using the definition of the expected value doesn't quite seem to work here, so I guess there's something I'm overlooking (since I don't even know what im missing there might be a lot of unnecessary information). Any help would be much appreciated!