Let $\theta$ in $[0,1]$ and defined for $u,v\in [0,1]$,
$$C_\theta (u,v)=\begin{cases}\min(u.v),&&|v-u|\ge \theta\\ \max(u+v-1,0), && |u+v-1|\ge 1-\theta\\ (u+v-\theta)/2,&& other,\end{cases}$$
Let $P_\alpha$ the measure given by $x^\alpha$ for $\alpha >0$.Define
$$C'(u.v)=\int C_{\theta}(u,v)P_\alpha (d\theta),$$
show that
$$C'(u,v)=\max (u+v-1,0)+ \frac{1}{2(\alpha+1)}\left\{[1-|u+v-1|]^{\alpha+1}-|u-v|^{\alpha+1}\right\}.$$
Sorry I can't see how even begin this. There are a lot cases to check and I'm not sure how to integrate with respect to $P_\alpha$. Are there other alternative?
This is my incomplete try (my definition is probably wrong and seems not getting the answer)
$$ \begin{align} &~ C'(u,v) \\ = &~\int_0^1 C_{\theta}(u,v)\theta^{\alpha}d\theta \\ = &~\int_0^{|v-u|} \min(u,v)\theta^{\alpha}d\theta + \int_{|v-u|}^{1-|u+v-1|} \max(u+v-1,0)\theta^{\alpha}d\theta + \int_{1-|u+v-1|}^1 \frac {u+v-\theta} {2}\theta^{\alpha}d\theta \\ = &~ \min(u,v)\frac {|v-u|^{\alpha + 1}} {\alpha + 1} + \max(u+v-1,0)\frac {(1-|u+v-1|)^{\alpha+1}-|v-u|^{\alpha+1} } {\alpha + 1}\\ &~ + \frac {u+v} {2} \frac {1 - (1-|u+v-1|)^{\alpha+1}} {\alpha + 1} - \frac {1 - (1-|u+v-1|)^{\alpha+2}} {2(\alpha + 2)} \\ = &~ \frac {|v-u|^{\alpha + 1}} {\alpha + 1} \left[\min(u,v)-\max(u+v-1,0)\right] \\ &~ + \frac {(1-|u+v-1|)^{\alpha+1}} {\alpha + 1} \left[\max(u+v-1,0)-\frac {u+v} {2}\right] \\ &~ + \frac {u+v} {2(\alpha+1)} - \frac {1 - (1-|u+v-1|)^{\alpha+2}} {2(\alpha + 2)} \\ \end{align}$$
Now we make use of the fact that $$ \min(u,v) = \frac {u+v - |v-u|} {2}, \max(u+v-1,0) = \frac {u+v-1+|u+v-1|} {2}$$
So we have
$$ \begin{align} &~ C'(u,v) \\ = &~ \frac {|v-u|^{\alpha + 1}} {\alpha + 1} \left[\frac {1 - |u+v-1| - |v-u|} {2}\right] - \frac {(1-|u+v-1|)^{\alpha+2}} {2(\alpha + 1)} \\ &~ + \frac {u+v} {2(\alpha+1)} - \frac {1 - (1-|u+v-1|)^{\alpha+2}} {2(\alpha + 2)} \\ = &~ \frac {|v-u|^{\alpha + 1}(1 - |u+v-1|) - |v-u|^{\alpha + 2}} {2(\alpha + 1)} - \frac {(1-|u+v-1|)^{\alpha+2}} {2(\alpha + 1)(\alpha+2)}\\ &~ + \frac {u+v} {2(\alpha+1)} - \frac {1} {2(\alpha + 2)} \\ \end{align}$$