Double integral with fractional part

We recall the (by now) standard result $\int_0^1 t^m \big\{t^{-1}\big\}\,dt \; = \; \frac{1}{m} - \frac{\zeta(m+1)}{m+1} \hspace{2cm} m \in \mathbb{N}\;.$ Then, using the substitution $y = tx$ , $\begin{aligned} \int_0^1 \int_0^1 x^N y^N \big\{\tfrac{x}{y}\big\} \big\{ \tfrac{y}{x} \big\}\,dx\,dy & = 2\int_0^1\,dx \int_0^x\,dy\, x^N y^N \big\{\tfrac{x}{y}\big\} \tfrac{y}{x} \\ & = 2\int_0^1 \,dx \int_0^x\,dy \,x^{N-1}y^{N+1} \big\{\tfrac{x}{y}\big\} \; = \; 2\int_0^1\,dx \int_0^1\,x\,dt \,x^{N-1} t^{N+1}x^{N+1} \big\{ t^{-1}\big\} \\ & = 2\int_0^1 x^{2N+1}\,dx \int_0^1 t^{N+1} \big\{t^{-1}\big\}\,dt \; = \; \frac{1}{N+1}\int_0^1 t^{N+1}\big\{ t^{-1}\big\}\, dt \\ & = \frac{1}{(N+1)^2} - \frac{\zeta(N+2)}{(N+1)(N+2)} \end{aligned}$ With $N = 2017$ the answer is $(N+1) + (N+2) + (N+1)(N+2) \; = \; \boxed{4078379}$

See here My Solution... A = 2018 , B = 2019 , C = (2018)(2019) . Hence A+B+C = 4078379 .