Alphabet Soup

For $xy>(x+y)$ we must have $xy-x-y+1>1$ or $(x-1)(y-1)>1$ . Letting $m=x-1$ and $n=y-1$ $P(a)$ is the probability that $mn>1$ or $m>\frac{1}{n}$ for $m, n$ chosen uniformly at random in the interval $[-a-1,a - 1]$ . For $P(\frac{1}{2})$ this is $m>\frac{1}{n}$ in the interval $[-\frac{3}{2},-\frac{1}{2}]$ which, as both will be negative is the same as considering the interval $[\frac{1}{2},\frac{3}{2}]$ . Graphically this is simply the integral $\int^{\frac{3}{2}}_{\frac{2}{3}} (\frac{3}{2}-\frac{1}{n}) dn = \frac{5}{4}-ln(3)+2ln(2)$ divided by the total area contained by this interval which is $(\frac{3}{2}-\frac{1}{2})^{2}=1$ so $P(\frac{1}{2})=\frac{5}{4}-ln(3)+2ln(2)$ . For $P(3)$ we can apply the same concept, splitting it into cases. Here we have $m, n$ in the interval $[-4,2]$ it is clear that for their product to be over $1$ they must either both be positive or both be negative. When they are both positive with $m, n$ in the interval $[0,2]$ the integral is graphically $\int^{2}_{\frac{1}{2}} (2-\frac{1}{n}) dn = 3-2ln(2)$ . When they are both negative in the interval $[-4,0]$ this is equivalent to the interval $[0,4]$ , thus the integral is $\int^{4}_{\frac{1}{4}} (4-\frac{1}{n}) dn = 15-2ln(4)$ . These two cases summed is $18-6ln(2)$ but this must be considered out of the total area contained by the interval $[-4,2]$ which is $(2-(-4))^{2}=36$ so $P(3)=\frac{18-6ln(2)}{36}=\frac{1}{2}-\frac{1}{6}ln(2)$ . Subtracting this from $P(\frac{1}{2})$ gives $P(\frac{1}{2})-P(3)=\frac{3}{4}-2ln(3)+\frac{13}{6}ln(2)$ so $A+B+C+D+E=3+4+2+13+6=\boxed{28}$