Following a previous related question of mine, regarding this paper on bribing in a second price auctions.
I am interested in calculating the chance for inefficiency to occur (i.e. the chance that the bribing player will win the auction although he doesn't have the highest valuation)
Inefficiency will occur if the bribing player will offer the bribe ($\theta_j > B^b)$ and the bribed player will accept the bribe $(\theta_i > A^b)$ and $ \theta_i > \theta_j$.
In the case of uniform distribution this happens with probability of $\frac{(A^b - B^b)^2}{2}$
Assuming $b$ is also uniformly distributed, I want to calculate
$$\int_0^\frac{1}{2} \frac{(A^b - B^b)^2}{2} db$$
I split this integral into 2 cases, where $A=1$, and where $A<1$ The case where $A=1$ is rather simple and I was able to calculate it
But when $A<1$ we have $$(B-b)^2(1-B)b=\frac18B^2(B-2b)^2$$ $$A=\frac{B^2}{2(B-b)}$$ I am not able to isolate the formula for $B$ and therefore cant manage to calculate this integral.