I need some help with evaluating a particular integral: $$\int_\theta^1 \int_0^1 \cdots \int_0^1 \left(x_1 - \max\{x_1,x_2-s, x_3-2\cdot s,\ldots,x_N-(N-1) \cdot s\}\right) \hspace{0.5ex}dx_N\cdots dx_1 $$
where $X_i$ $\overset{\text{iid}}{\sim} \operatorname{UNI}[0,1]$, N is an integer $\ge 2,$ $0< s < 0.5$, and assume that $\theta > 0$ is some real number such that $\theta < 1-N\cdot s$.
Let me show you my approach so far. First, let's rewrite the integral: $$\int_\theta^1 \int_0^1 \cdots \int_0^1 (x_1 - \max\{x_1,\underbrace{\max\{x_2-s, x_3-2\cdot s,\ldots,x_N-(N-1) \cdot s\}}_Z\}) \hspace{0.5ex} dx_N\, dx_1 $$ Next lets derive the CDF of Z, which is given by \begin{align} P(Z < z) &\stackrel{\text{def.}}{=} P(\max\{X_2-s, X_3-2\cdot s,\ldots,X_N-(N-1) \cdot s\} < z) \\ &\stackrel{\text{iid}}{=} P(X_2< z + s) \cdot P(X_3< z + 2\cdot s) \cdots P(X_N < z+(N-1)\cdot s) \\ &\stackrel{\mathrm{UNI}}{=} (z+s)\cdot (z + 2\cdot s) \cdots (z+(N-1)\cdot s) \\ &\stackrel{\text{def.}}{=} s^{N-1}\frac{\Gamma(N+\frac{z}{s})}{\Gamma(1+\frac{z}{s})} := G(z) \end{align} where $\Gamma(\cdot)$ is the Gamma function. Next, lets derive the density of Z given by $$ g(z) = \frac{dG(z)}{dz} = s^{N-2} \frac{\Gamma(N+\frac{z}{s})}{\Gamma(1+\frac{z}{s})}\cdot \left(\Psi\left(N+\frac{z}{s}\right)-\Psi\left(1+\frac{z}{s}\right)\right)$$ where $\Psi(\cdot)$ is the digamma function.
So far I think everything is correct. What I am unsure about, is the support of Z. I wonder whether I can just say that it is given by [-s , 1-s]? If yes, then let's rewrite the integral such that: $$ \int_\theta^1 \int_{-s}^{1-s} (x_1-\max\{x_1,z\})g(z) \hspace{0.5ex} \, dz\,dx_1$$
Whenever we have that $Z < X_1$ or $X_1 > 1-s$ the expression is $0$, such that we only consider the support $[\theta, 1-s]$ for $X_1$ and $[X_1, 1-s]$ for $Z$. Now the integral can be written as $$ \int_\theta^{1-s} \int_{x_1}^{1-s} (x_1- z)g(z)\hspace{0.5ex} dz\,dx_1$$ Using integration by parts w.r.t the integral of $Z$ yields $$ \int_\theta^{1-s} \left[\mathop{\big|} \limits_{x_1}^{1-s}(x_1-z)G(z) - \int_{x_1}^{1-s} (-1) \cdot G(z) \hspace{0.5ex}dz \right] \, dx_1$$ Since the upper boundary of the support of $Z$ is given by $1-s$ it follows that $G(1-s) = 1$ therefore we can further simplify the expression to get $$ \int_{\theta}^{1-s} (x_1+s-1) \hspace{0.5ex} dx_1 + \int_\theta^{1-s} \int_{x_1}^{1-s}G(z) \hspace{0.5ex}dz \, dx_1$$ Calculating the first expression and changing the order of integration for the second expression yields: $$ -\frac{(1-\theta-s)^2}{2} + \int_\theta^{1-s}\int_\theta^z G(z) \hspace{0.5ex}dx_1 \, dz = \frac{(1-\theta-s)^2}{2} + \int_\theta^{1-s} (z-\theta)G(z) \hspace{0.5ex} dz$$ I am unsure, whether my derivation so far was correct, especially because I am not sure whether it was valid to simply assume that the support of $Z$ was given by $[-s, 1-s]$. Lets assume it was correct, than inserting the expression of $G(Z)$ yields: $$-\frac{(1-\theta-s)^2}{2} + s^{N-1}\int_{\theta}^{1-s} (z-\theta) \frac{\Gamma(N+\frac{z}{s})}{\Gamma(1+\frac{z}{s})} \hspace{0.5ex} dz$$ At this point I have two questions:
First, is there any chance of further simplifying this expression?
Second, is there anything I can say about the asymptotic behavior of the second part when $N$ goes to infinity? Obviously $s^{N-1}$ will tend to $0$ since s < 1. But what about the integral? My guess would be that it converges to some constant, but I have no clue of how I could show this. In addition, I think I have a problem when analysing the asymptotic behavior w.r.t. $N$ due to the assumption $\theta < 1-N\cdot s$.
Thank you guys for your help in advance! I would really appreciate it.
Update 30.01: Can anybody please confirm whether it was fine to define the support of Z by [-s, 1-s]?